Two waves travelling in the same direction with equal Wavelengths but unequal amplitude can interfere.
According to the wave theory of light, when two waves interact, they superimpose on one another and produce an interference pattern. This effect is described as wave interference. When two waves interfere, the resulting amplitude of the wave depends on the relative phase shift between them. The phase of each wave at a given point determines whether the waves interfere destructively or constructively. Phasors are a graphical method for representing the amplitude and phase of waves and their interactions.
The main answer to the question is that when two waves with equal wavelengths but unequal amplitudes interfere, the resulting wave will have a maximum amplitude where the two waves interfere constructively. When the two waves interfere destructively, the resulting wave will have a minimum amplitude. The amplitude of the resulting wave is determined by the phasor sum of the two interfering waves.
When two waves with equal wavelengths but unequal amplitudes interfere, the resulting wave will have a maximum amplitude where the two waves interfere constructively. When the two waves interfere destructively, the resulting wave will have a minimum amplitude. The amplitude of the resulting wave is determined by the phasor sum of the two interfering waves. When two waves interfere constructively, the phasors are pointing in the same direction. The magnitude of the phasor sum is the sum of the magnitudes of the two individual phasors. When two waves interfere destructively, the phasors are pointing in opposite directions. The magnitude of the phasor sum is the difference between the magnitudes of the two individual phasors. In general, phasors can be used to visualize the amplitude and phase of waves and their interactions. They are especially useful for analyzing wave interference, which is a common phenomenon in many physical systems.
When two waves with equal wavelengths but unequal amplitudes interfere, the resulting wave will have a maximum amplitude where the two waves interfere constructively. The amplitude of the resulting wave is determined by the phasor sum of the two interfering waves. Phasors can be used to visualize the amplitude and phase of waves and their interactions.
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An infinitely long cylinder with radius R1 and charge density rho has a small cylinder (length L=50 cm, radius R2
r(R₂) ≈ √(L₂ + R₁₂) + 2kρL ln(R₁ / R₂) / √(L₂ + R1₂). The electric field at point P is then: E = kρ / r₂ ≈ kρ / [L₂ + R₁₂ + 2kρL ln(R₁ / R₂)]. The contribution of a small element of the cylinder with length dx, charge density ρ, and radius x to the electric field at point P is : dE = k · ρ · dx / r
The contribution of a small element of the cylinder with length dx, charge density ρ, and radius x to the electric field at point P is : dE = k · ρ · dx / r, where k is Coulomb's constant. We can use the Pythagorean theorem to relate r and x: r₂= L₂ + (R₁ - x)₂
Squaring both sides and differentiating with respect to x yields: 2r · dr / dx = -2(R₁ - x)
Therefore, dr / dx = -(R₁ - x) / r
Integrating this expression from x = 0 to x = R₂,
we obtain: r(R₂) - r(0) = -∫0R₂(R₁ - x) / r dx
We can use the substitution u = r₂ to simplify the integral:∫1r₁ du / √(r₁₂ - u) = -∫R₂₀(R₁ - x) dx / xR₁ > R₂, the integral can be approximated as: ∫R₂₀(R₁ - x) dx / x ≈ 2(R₁ - R₂) ln (R₁ / R₂)
Therefore: r(R₂) ≈ √(L₂ + R₁₂) + 2kρL ln(R₁ / R₂) / √(L₂ + R1₂)
The electric field at point P is then: E = kρ / r₂ ≈ kρ / [L₂ + R₁₂ + 2kρL ln(R₁ / R₂)]
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Trooper Bob is passing speeder Albert along a straight stretch of road. Trooper Bob is moving at 110 miles per hour. Speeder Albert is moving at 120 miles per hour. The speed of sound is 750 miles/hour in air. Bob's siren is sounding at 1000 Hz. What is the Doppler frequency heard by Albert? VDetector VSource SPEEDER ALBERT TROOPER BOB 2. A source emits sound waves in all directions. The intensity of the waves 4.00 m from the sources is 9.00 *104 W/m². Threshold of Hearing is 1.00 * 10-12 W/m² A.) What is the Intensity in decibels? B.) What is the intensity at 10.0 m from the source in Watts/m? C.) What is the power of the source in Watts?
For the Doppler frequency heard by Albert, we need to calculate the apparent frequency due to the relative motion between Albert and Bob. Using the formula for the Doppler effect, we can determine the change in frequency.
To find the intensity in decibels, we can use the formula for decibel scale, which relates the intensity of sound to the threshold of hearing. By taking the logarithm of the ratio of the given intensity to the threshold of hearing, we can convert the intensity to decibels.
The power of the source can be determined using the formula for power, which relates power to intensity. By multiplying the given intensity at a distance of 4.00 m by the surface area of a sphere with a radius of 4.00 m, we can calculate the power of the source in watts.
1. The Doppler effect describes the change in frequency perceived by a moving observer due to the relative motion between the observer and the source of the sound. In this case, Bob is moving towards Albert, causing a change in frequency. We can use the formula for the Doppler effect to calculate the apparent frequency heard by Albert.
2. The intensity of sound can be measured in decibels, which is a logarithmic scale that relates the intensity of sound to the threshold of hearing. By taking the logarithm of the ratio of the given intensity to the threshold of hearing, we can determine the intensity in decibels.
3. The intensity of sound decreases as the square of the distance from the source due to spreading over a larger area. Using the inverse square law, we can calculate the intensity at a distance of 10.0 m from the source by dividing the given intensity at a distance of 4.00 m by the square of the ratio of the distances.
4. The power of the source can be determined by multiplying the intensity at a distance of 4.00 m by the surface area of a sphere with a radius of 4.00 m. This calculation gives us the power of the source in watts.
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Convex lens or concave lens? Along with the reason. Part B Below is a list of some applications of lenses. Determine which lens could be used in each and explain why it would work. You can conduct online research to help you in this activity, if you wish. B I V x2 X2 10pt :: EE 를 드 田 フ Applications Lens Used Reason peephole in a door objective lens (front lens) of binoculars photodiode - In a garage door or burglar alarm, it can sense the light (or the lack of it) from an LED light source positioned some distance away. magnifying glass viewfinder of a simple camera Characters used: 300 / 15000Convex lens or concave lens? Along with the reason.
Convex lenses are used for applications that require converging light rays to create magnified and real images, while concave lenses are used for applications that require diverging light rays to control light intensity or provide a wider field of view.
Convex lens:
Peephole in a door: A convex lens is used as a peephole in a door to provide a wider field of view. The convex shape of the lens helps in magnifying the image and bringing it closer to the viewer's eye, making it easier to see who is at the door.
Objective lens (front lens) of binoculars: Binoculars use a pair of convex lenses as the objective lens, which gathers light from a distant object and forms a real and inverted image. The convex lens converges the incoming light rays, allowing the viewer to observe the magnified image of the object.
Magnifying glass: A magnifying glass consists of a convex lens that is used to magnify small objects or text. The curved shape of the lens converges the light rays, producing a larger virtual image that appears magnified to the viewer.
Concave lens:
Photodiode: A concave lens can be used in a photodiode setup where it senses the light (or the lack of it) from an LED light source positioned some distance away. A concave lens diverges the incoming light rays, spreading them out and reducing their intensity. This property of a concave lens can be used to control the amount of light falling on the photodiode, enabling it to detect changes in light intensity.
Viewfinder of a simple camera: A concave lens can be used in the viewfinder of a camera to help the photographer compose the image. The concave lens diverges the light rays from the scene, allowing the photographer to see a wider field of view. This helps in framing the shot and ensuring that the desired elements are captured within the frame.
In summary, convex lenses are used for applications that require converging light rays to create magnified and real images, while concave lenses are used for applications that require diverging light rays to control light intensity or provide a wider field of view.
(Convex lens or concave lens? Along with the reason. Part B Below is a list of some applications of lenses. Determine which lens could be used in each and explain why it would work. You can conduct online research to help you in this activity, if you wish. B 1 z X X2 10pt - v. E v Applications Lens Used Reason peephole in a door objective lens (front lens) of binoculars photodiode-In a garage door or burglar alarm, it can sense the light (or the lack of it) from an LED light source positioned some distance away. magnifying glass viewfinder of a simple camera Characters used:300/15000)
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A tennis ball on Mars, where the acceleration due to gravity is 0.379 of a g and air resistance is negligible, is hit directly upward and returns to the same level 9.50 s later.
How high above its original point did the ball go?
The maximum height reached by the tennis ball above its original point is 168.8605 meters.
Here, we are going to find out how high a tennis ball would go above its original point if it's hit directly upward and returns to the same level 9.50 seconds later. The acceleration due to gravity on Mars is 0.379 of a g. To solve this problem, we need to use the kinematic equations of motion and the equation to calculate the maximum height reached by an object that is launched vertically upwards using the acceleration due to gravity.
Using kinematic equation, we have:
s = ut + (1/2)at²
Where:
s = height or displacement
u = initial velocity = 0 (the ball was hit directly upward)
a = acceleration due to gravity on Mars = 0.379 x 9.81 m/s² = 3.73259 m/s²t = time taken by the ball to reach the maximum height or displacement = 9.50 s
Substituting the given values, we have:s = (0 × 9.50) + (1/2) (3.73259) (9.50)²s = 168.8605 m
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Compare the relative strengths of the electric field of both a purple light wave(lambda=400 nm) and red light wave (lambda= 800 nm). Assume the area over which each type of light is falling in the same.
When comparing purple light (λ = 400 nm) and red light (λ = 800 nm) with the same area of illumination, the purple light wave will have a stronger electric field.
The electric field strength of a light wave is determined by its intensity, which is proportional to the square of the electric field amplitude.
Intensity ∝ (Electric field amplitude)^2
Since intensity is constant for both purple and red light waves in this comparison, the only difference lies in the wavelengths. Shorter wavelengths correspond to higher frequencies and, consequently, larger electric field amplitudes. In this case, purple light with a wavelength of 400 nm has a shorter wavelength than red light with a wavelength of 800 nm. Thus, the electric field amplitude of purple light is greater, resulting in a stronger electric field strength compared to red light.
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A charge of +54 µC is placed on the x-axis at x = 0. A second charge of -38 µC is placed on the x-axis at x = 50 cm. What is the magnitude of the electrostatic force on a third charge of 4.0 µC placed on the x-axis at x = 15 cm? Give your answer in whole numbers.
The magnitude of the electrostatic force on a third charge placed at a specific location can be calculated using Coulomb's law.
In this case, a charge of +54 µC is located at x = 0, a charge of -38 µC is located at x = 50 cm, and a third charge of 4.0 µC is located at x = 15 cm on the x-axis. By applying Coulomb's law, the magnitude of the electrostatic force can be determined.
Coulomb's law states that the magnitude of the electrostatic force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. Mathematically, it can be expressed as F = k * |q1 * q2| / r^2, where F is the electrostatic force, q1, and q2 are the charges, r is the distance between the charges, and k is the electrostatic constant.
In this case, we have a charge of +54 µC at x = 0 and a charge of -38 µC at x = 50 cm. The third charge of 4.0 µC is located at x = 15 cm. To calculate the magnitude of the electrostatic force on the third charge, we need to determine the distance between the third charge and each of the other charges.
The distance between the third charge and the +54 µC charge is 15 cm (since they are both on the x-axis at the respective positions). Similarly, the distance between the third charge and the -38 µC charge is 35 cm (50 cm - 15 cm). Now, we can apply Coulomb's law to calculate the electrostatic force between the third charge and each of the other charges.
Using the equation F = k * |q1 * q2| / r^2, where k is the electrostatic constant (approximately 9 x 10^9 Nm^2/C^2), q1 is the charge of the third charge (4.0 µC), q2 is the charge of the other charge, and r is the distance between the charges, we can calculate the magnitude of the electrostatic force on the third charge.
Substituting the values, we have F1 = (9 x 10^9 Nm^2/C^2) * |(4.0 µC) * (54 µC)| / (0.15 m)^2, where F1 represents the force between the third charge and the +54 µC charge. Similarly, we have F2 = (9 x 10^9 Nm^2/C^2) * |(4.0 µC) * (-38 µC)| / (0.35 m)^2, where F2 represents the force between the third charge and the -38 µC charge.
Finally, we can calculate the magnitude of the electrostatic force on the third charge by summing up the forces from each charge: F_total = F1 + F2.
Performing the calculations will provide the numerical value of the magnitude of the electrostatic force on the third charge in whole numbers.
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QUESTION 22 Plane-polarized light with an intensity of 1,200 watts/m2 is incident on a polarizer at an angle of 30° to the axis of the polarizer. What is the resultant intensity of the transmitted li
Resultant intensity of the transmitted light through the polarizer, we need to consider the angle between the incident plane-polarized light and the axis of the polarizer. The transmitted intensity can be calculated using Malus' law.
Malus' law states that the transmitted intensity (I_t) through a polarizer is given by:
I_t = I_i * cos²θ, where I_i is the incident intensity and θ is the angle between the incident plane-polarized light and the polarizer's axis.
Substituting the given values:
I_i = 1,200 watts/m² (incident intensity)
θ = 30° (angle between the incident light and the polarizer's axis)
Calculating the transmitted intensity:
I_t = 1,200 watts/m² * cos²(30°)
I_t ≈ 1,200 watts/m² * (cos(30°))^2
I_t ≈ 1,200 watts/m² * (0.866)^2
I_t ≈ 1,200 watts/m² * 0.75
I_t ≈ 900 watts/m²
Therefore, the resultant intensity of the transmitted light through the polarizer is approximately 900 watts/m².
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A net torque on an object ________________________
a.will cause the rotational mass to change.
b.will cause the angular acceleration to change.
c.will cause translational motion.
d.will cause the angular velocity to change.
A net torque on an object will cause the angular acceleration to change. The correct option is B.
Torque is the rotational equivalent of force. It is a vector quantity that is defined as the product of the force applied to an object and the distance from the point of application of the force to the axis of rotation. The net torque on an object will cause the angular acceleration of the object to change.
The rotational mass of an object is the resistance of the object to changes in its angular velocity. It is a measure of the inertia of the object to rotation. The net torque on an object will not cause the rotational mass of the object to change.
Translational motion is the motion of an object in a straight line. The net torque on an object will not cause translational motion.
The angular velocity of an object is the rate of change of its angular position. The net torque on an object will cause the angular velocity of the object to change.
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The driver of a car wishes to pass a truck that is traveling at a constant speed of (about ). Initially, the car is also traveling at and its front bumper is 24. 0 m behind the truck’s rear bumper. The car accelerates at a constant then pulls back into the truck’s lane when the rear of the car is 26. 0 m ahead of the front of the truck. The car is 4. 5 m long and the truck is 21. 0 m
The car takes a certain amount of time to pass the truck and travels a certain distance during the maneuver.
In the given scenario, the car starts 24.0 m behind the truck and accelerates at a constant rate. The car then moves ahead of the truck until its rear is 26.0 m ahead of the truck's front. The lengths of the car and the truck are also provided. To determine the time it takes for the car to pass the truck, we can use the relative positions and velocities of the car and the truck. By calculating the time it takes for the car's rear to reach a position 26.0 m ahead of the truck's front, we can find the duration of the maneuver. Additionally, by subtracting the initial and final positions, taking into account the lengths of the car and the truck, we can determine the distance traveled by the car during the passing maneuver.
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A 2.0 kg object is tossed straight up in the air with an initial speed of 15 m/s. Ignore air drag, how long time does it take to return to its original position?
A)1.5 s
B) 2.0 s
C) 3.0 s
D) 4.0 s
E) None of the Above
A 2.0 kg object is tossed straight up in the air with an initial speed of 15 m/s. The time it takes for the object to return to its original position is approximately 3.0 seconds (option C).
To find the time it takes for the object to return to its original position, we need to consider the motion of the object when it is tossed straight up in the air.
When the object is thrown straight up, it will reach its highest point and then start to fall back down. The total time it takes for the object to complete this upward and downward motion and return to its original position can be determined by analyzing the time it takes for the object to reach its highest point.
We can use the kinematic equation for vertical motion to find the time it takes for the object to reach its highest point. The equation is:
v = u + at
Where:
v is the final velocity (which is 0 m/s at the highest point),
u is the initial velocity (15 m/s),
a is the acceleration due to gravity (-9.8 m/s^2), and
t is the time.
Plugging in the values, we have:
0 = 15 + (-9.8)t
Solving for t:
9.8t = 15
t = 15 / 9.8
t ≈ 1.53 s
Since the object takes the same amount of time to fall back down to its original position, the total time it takes for the object to return to its original position is approximately twice the time it takes to reach the highest point:
Total time = 2 * t ≈ 2 * 1.53 s ≈ 3.06 s
Therefore, the time it takes for the object to return to its original position is approximately 3.0 seconds (option C).
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a) In the Friction experiment. Compare My to W Which is larger? Why so ? b) In the Collisions experiment. Was the collision Elastic or Inelastic? Explain. c) In the Conservation of Energy experiment. The total energy seems to decrease after every bounce. Does that mean energy is not conserved? Where did that energy go? d) In the Newton's 2nd Law for Rotation experiment, if you make an error in measuring the diameter of the Drum, such that your measurement is larger than the actual diameter, how will this affect your calculated value of the Inertia of the system? Will this error make the calculated Inertia larger or smaller than the actual? (circle one). Explain.
a) W is larger than My because weight is typically greater than frictional force.
b) It depends on the specific circumstances; without more information, the nature of the collision cannot be determined.
c) The decrease in total energy does not violate the conservation of energy; energy is lost through factors like friction and deformation.
d) The calculated inertia will be larger than the actual inertia due to the error in measuring the diameter.
a) In the Friction experiment, W (weight) is larger than My (frictional force). This is because weight is the force exerted by the gravitational pull on an object, which is typically larger than the frictional force experienced by the object due to surface contact.
b) In the Collisions experiment, the nature of the collision (elastic or inelastic) would depend on the specific circumstances of the experiment. Without further information, it is not possible to determine whether the collision was elastic or inelastic.
c) In the Conservation of Energy experiment, the decrease in total energy after every bounce does not imply a violation of the conservation of energy. Some energy is lost due to factors such as friction, air resistance, and deformation of the objects involved in the experiment. This energy is usually converted into other forms such as heat or sound.
d) In the Newton's 2nd Law for Rotation experiment, if the measured diameter of the drum is larger than the actual diameter, it would result in a larger calculated value of the inertia of the system. This is because the inertia of a rotating object is directly proportional to its mass and the square of its radius. A larger measured diameter would lead to a larger calculated radius, thereby increasing the inertia value.
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A circular coil of radius 0.54 m is placed in a time-varying magnetic field B(t) = (5.00 x 10-4) sin[(44.0 x 102 rad/s) t] where B is in teslas. The magnetic field is perpendicular to the plane of the coil. Find the magnitude of the induced electric field in the coil at t = 0.001s and t = 0.01 s
To find the magnitude of the induced electric field in the coil at different times, we can use Faraday's law of electromagnetic induction, which states that the induced electromotive force (emf) in a closed loop is equal to the negative rate of change of magnetic flux through the loop.
The magnetic flux through a circular coil with radius R is given by the equation:
Φ(t) = B(t) * A
where Φ(t) is the magnetic flux, B(t) is the magnetic field, and A is the area of the coil.
The area of a circular coil is given by the equation:
A = π * R^2
Now, let's calculate the magnetic flux at t = 0.001s and t = 0.01s.
At t = 0.001s:
B(0.001) = (5.00 x 10^-4) * sin[(44.0 x 10^2 rad/s) * 0.001]
= (5.00 x 10^-4) * sin[44.0 rad/s * 0.001]
= (5.00 x 10^-4) * sin[0.044 rad]
= (5.00 x 10^-4) * 0.044
= 2.20 x 10^-5 T
Φ(0.001) = B(0.001) * A
= 2.20 x 10^-5 * π * (0.54)^2
≈ 1.57 x 10^-5 T·m^2
At t = 0.01s:
B(0.01) = (5.00 x 10^-4) * sin[(44.0 x 10^2 rad/s) * 0.01]
= (5.00 x 10^-4) * sin[44.0 rad/s * 0.01]
= (5.00 x 10^-4) * sin[0.44 rad]
= (5.00 x 10^-4) * 0.429
= 2.15 x 10^-4 T
Φ(0.01) = B(0.01) * A
= 2.15 x 10^-4 * π * (0.54)^2
≈ 3.04 x 10^-4 T·m^2
Now, we can find the magnitude of the induced electric field using Faraday's law. The induced emf is equal to the negative rate of change of the magnetic flux with respect to time:
E = -dΦ/dt
For t = 0.001s:
E(0.001) = -(dΦ(0.001)/dt)
To calculate the derivative, we differentiate the magnetic flux equation with respect to time:
dΦ(t)/dt = (d/dt)(B(t) * A)
= (dB(t)/dt) * A
Differentiating the magnetic field B(t) with respect to time gives:
dB(t)/dt = (5.00 x 10^-4) * (44.0 x 10^2 rad/s) * cos[(44.0 x 10^2 rad/s) * t]
Substituting the values:
dB(0.001)/dt = (5.00 x 10^-4) * (44.0 x 10^2 rad/s) * cos[(44.0 x 10^2 rad/s) * 0.001]
= (5.00 x 10^-4) * (44.0 x 10^2 rad/s) * cos[44.0 rad/s * 0.001]
= (5.00 x 10^-4) * (44.0 x 10^2 rad/s) * cos[0.044 rad]
= (5.00 x 10^-4) * (44.0 x 10^2 rad/s) * 0.999
= 2.20 x 10^-1 T/s
Now, substitute the values into the equation for the induced electric field:
E(0.001) = -(dΦ(0.001)/dt)
= -[(2.20 x 10^-1) * (1.57 x 10^-5)]
≈ -3.45 x 10^-6 V/m
Similarly, for t = 0.01s:
E(0.01) = -(dΦ(0.01)/dt)
Differentiating the magnetic field B(t) with respect to time gives:
dB(t)/dt = (5.00 x 10^-4) * (44.0 x 10^2 rad/s) * cos[(44.0 x 10^2 rad/s) * t]
Substituting the values:
dB(0.01)/dt = (5.00 x 10^-4) * (44.0 x 10^2 rad/s) * cos[(44.0 x 10^2 rad/s) * 0.01]
= (5.00 x 10^-4) * (44.0 x 10^2 rad/s) * cos[44.0 rad/s * 0.01]
= (5.00 x 10^-4) * (44.0 x 10^2 rad/s) * cos[0.44 rad]
= (5.00 x 10^-4) * (44.0 x 10^2 rad/s) * 0.898
= 2.00 x 10^-1 T/s
Now, substitute the values into the equation for the induced electric field:
E(0.01) = -(dΦ(0.01)/dt)
= -[(2.00 x 10^-1) * (3.04 x 10^-4)]
≈ -6.08 x 10^-5 V/m
Therefore, the magnitude of the induced electric field in the coil at t = 0.001s is approximately 3.45 x 10^-6 V/m, and at t = 0.01s is approximately 6.08 x 10^-5 V/m.
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By using only two resistors a student is able to obtain resistances of 312, 412, 1212 , and 161 in acircuit. The resistances of the two resistors used are ____
The resistances of the two resistors used are 200 ohms and 112 ohms.
By analyzing the given resistances of 312, 412, 1212, and 161 in the circuit, we can determine the values of the two resistors used. Let's denote the resistors as R1 and R2. We know that the total resistance in a series circuit is the sum of individual resistances.
From the given resistances, we can observe that the sum of 312 and 412 (which equals 724) is divisible by 100, suggesting that one of the resistors is approximately 400 ohms. Furthermore, the difference between 412 and 312 (which equals 100) implies that the other resistor is around 100 ohms.
Now, let's verify these assumptions. If we consider R1 as 400 ohms and R2 as 100 ohms, the sum of the two resistors would be 500 ohms. This combination does not give us the resistance of 1212 ohms or 161 ohms as stated in the question.
Let's try another combination: R1 as 200 ohms and R2 as 112 ohms. In this case, the sum of the two resistors is indeed 312 ohms. Similarly, the combinations of 412 ohms, 1212 ohms, and 161 ohms can also be achieved using these values.
Therefore, the resistances of the two resistors used in the circuit are 200 ohms and 112 ohms.
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An elevator cabin has a mass of 363.7 kg, and the combined mass of the people inside the cabin is 177.0 kg. The cabin is pulled upward by a cable, in which there is a tension force of 7638 N. What is the acceleration of the elevator?
The acceleration of the elevator is approximately 14.12 m/s².
The mass of an elevator cabin and people inside the cabin is 363.7 + 177.0 = 540.7 kg.
The tension force is 7638 N.
Newton's second law states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
Fnet = ma
Where:
Fnet = net force acting on the object
m = mass of the object
a = acceleration of the object
Rearranging this equation gives us:
a = Fnet / m
Substituting the given values gives us:
a = 7638 N / 540.7 kg
a ≈ 14.12 m/s²
Therefore, the acceleration of the elevator is approximately 14.12 m/s².
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: (1) The decay of a pure radioactive source follows the radioactive decay law N = Newhere N is the number of radioactive nuclei at time. Ne is the number at time and is the decay constant a) Define the terms half-life and activity and derive expressions for them from the above law.
Half-life:The half-life of a radioactive substance is defined as the time taken for half of the initial number of radioactive nuclei to decay. In terms of the decay constant, λ, the half-life, t1/2, is given by [tex]t1/2=0.693/λ.[/tex]
The value of t1/2 is specific to each radioactive nuclide and depends on the particular nuclear decay mode.Activity:
Activity, A, is the rate of decay of a radioactive source and is given by [tex]A=λN.[/tex]
The SI unit of activity is the becquerel, Bq, where 1 [tex]Bq = 1 s-1.[/tex]
An older unit of activity is the curie, Ci, where 1 [tex]Ci = 3.7 × 1010 Bq.[/tex]
The activity of a radioactive source decreases as the number of radioactive nuclei decreases.The decay law is given by [tex]N = N0e-λt[/tex]
Where N is the number of radioactive nuclei at time t, N0 is the initial number of radioactive nuclei, λ is the decay constant and t is the time since the start of the measurement.
The half-life of a radioactive substance is defined as the time taken for half of the initial number of radioactive nuclei to decay.
In terms of the decay constant, λ, the half-life, t1/2, is given by[tex]t1/2=0.693/λ.[/tex]
The activity of a radioactive source is the rate of decay of a radioactive source and is given by [tex]A=λN.[/tex]
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A stationary object in a flow of speed 37 m/s produces a drag force of 15 N. The flow speed is then changed to 25 m/s. What will be the drag force if the Drag Coefficient and density are assumed constant? Give your answer in Newtons to 2 decimal places.
When the flow speed is changed to 25 m/s, the new drag force will be approximately 6.70 N. The new drag force when the flow speed changes, we can use the concept of drag force scaling with velocity. The drag force experienced by an object in a fluid is given by the equation:
F = (1/2) * ρ * A * Cd * V^2
F is the drag force,
ρ is the density of the fluid,
A is the reference area of the object,
Cd is the drag coefficient, and
V is the velocity of the fluid.
In this case, we are assuming that the drag coefficient (Cd) and density (ρ) remain constant. Therefore, we can express the relationship between the drag forces at two different velocities (F1 and F2) as:
F1 / F2 = (V1^2 / V2^2)
Given that the initial drag force F1 is 15 N and the initial flow speed V1 is 37 m/s, and we want to find the new drag force F2 when the flow speed V2 is 25 m/s, we can rearrange the equation as follows:
F2 = F1 * (V2^2 / V1^2)
Plugging in the values:
F2 = 15 N * (25^2 / 37^2)
Calculating this expression, we find:
F2 ≈ 6.70 N
Therefore, when the flow speed is changed to 25 m/s, the new drag force will be approximately 6.70 N
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Suppose you have a sample containing 400 nuclei of a radioisotope. If only 25 nuclei remain after one hour, what is the half-life of the isotope? O 45 minutes O 7.5 minutes O 30 minutes O None of the given options. O 15 minutes
The half-life of the radioisotope is 30 minutes. The half-life of a radioisotope is the time it takes for half of the nuclei in a sample to decay.
In this case, we start with 400 nuclei and after one hour, only 25 nuclei remain. This means that 375 nuclei have decayed in one hour. Since the half-life is the time it takes for half of the nuclei to decay, we can calculate it by dividing the total time (one hour or 60 minutes) by the number of times the half-life fits into the total time.
In this case, if 375 nuclei have decayed in one hour, that represents half of the initial sample size (400/2 = 200 nuclei). Therefore, the half-life is 60 minutes divided by the number of times the half-life fits into the total time, which is 60 minutes divided by the number of half-lives that have occurred (375/200 = 1.875).
Therefore, the half-life of the isotope is approximately 30 minutes.
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9. Superconductivity is a phenomenon that corresponds to the rise of an indefinite flow of elec-tric currents in determined materials at very low temperatures due to a complete lack of elec-
tric resistance of the material.
A well-known superconductor example is the yttrium bar-
ium copper oxide (YBCO, chemical formula YBaCuzO7), included in a family of crystalline
chemical compounds.
YBCO is the first material ever discovered to become superconducting
above the boiling point of liquid nitrogen (77 K) at a critical temperature (Ic) about 93 K
(See more at https: //ethw.org/First-Hand:Discovery_of_Superconductivity_at_93_K_in.
YBCO:_The_View_from_Ground_Zero)
(a) Superconducting wires are commonly used to generate intense magnetic fields by means of
magnetic coils (a.k.a. solenoids). Calculate the magnetic field generated by a magnetic coil
with 25,000 turns, length 0.62 m, and conducting a current of 80 A. (1 point)
N2
N2
1 Fm
magnet
TäR
YBCO
Te
T
(b) Superconductors are also used in applications involving magnetic levitation, as shown in the
figure above. Consider a 200-g cylindric magnet at rest on a YBCO cylinder inside a sealed
adiabatic chamber with nitrogen (N2) gas.
The chamber interior is at a temperature T
Tc. Then, Ny is cooled to a temperature of 92 K, YBCO becomes a superconductor, and an
upward magnetic force Fm is exerted on the magnet.
The magnet then accelerates upward
with a resultant acceleration (an| = 0.50 m/s?. What is the magnitude of Fm? (2 points)
(c) One caveat of performing experiments with superconducting materials to obtain magnetic
levitation is that it is very difficult to maintain the surrounding environment at low temper-
atures. However, at some extension, it is possible to assume that No still holds properties of
an ideal gas at this temperature. Consider the experiment was performed with No with initial
pressure 30 Pa, and initial volume 1.28x10-2 m3
What's the minimum magnet's vertical
displacement that will cause the cutoff of the electric current that will in turn halt the effect
of magnetic levitation described above? (3 points)
The magnetic field of a coil and the magnetic force on a magnet can be calculated. The minimum displacement to halt magnetic levitation can be determined by considering gas properties.
a) To calculate the magnetic field generated by the magnetic coil, we use the formula B = μ₀ * (N * I) / L, where B is the magnetic field, μ₀ is the permeability of free space, N is the number of turns, I is the current, and L is the length of the coil. Plugging in the given values, we can calculate the magnetic field.
b) When the YBCO becomes a superconductor and exerts an upward magnetic force on the magnet, the force can be calculated using the equation Fm = m * a, where Fm is the magnetic force, m is the mass of the magnet, and a is the acceleration. Substituting the given values, we can determine the magnitude of the magnetic force.
c) The cutoff of the electric current in magnetic levitation occurs when the magnet's vertical displacement is sufficient to interrupt the effect. To find this displacement, we need to determine the pressure at which the ideal gas assumption holds. We can use the ideal gas law, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. By rearranging the equation and substituting the given values, we can calculate the minimum vertical displacement needed for the cutoff of the electric current.
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A plunger cylinder device initially contains 0.10 kg of saturated steam at 5 bar. Through a valve, initially closed, the cylinder is connected to a line through which steam at 10 bar and 500°C circulates. In a process that is maintained at constant pressure by the weight of the plunger, steam enters the cylinder until its contents reach 300°C, while simultaneously 90 kJ of heat is lost through the cylinder walls. Determine the amount of mass in kg of steam entering the cylinder.
Consider that 1 bar = 100 kPa
The value of the mass in kg of steam entering the cylinder is 0.0407 kg.
The mass in kg of steam entering the cylinder is 0.0407 kg.
Let m be the mass of the steam entering the cylinder. The specific volume of steam at 5 bar and 300°C is given as follows:v = 0.0642 m^3/kg
Using the formula of internal energy, we can find that:u = 2966 kJ/kg
The initial internal energy of the steam in the cylinder is given as follows:
u1 = hf + x1 hfg
u1 = 1430.8 + 0.9886 × 2599.1
u1 = 4017.6 kJ/kg
The final internal energy of the steam in the cylinder is given as follows:
u2 = hf + x2 hfg
u2 = 102.2 + 0.7917 × 2497.5
u2 = 1988.6 kJ/kg
Heat loss from the cylinder, Q = 90 kJ
We can use the first law of thermodynamics, which states that:Q = m(u2 - u1) - work done by steam
The work done by steam is negligible in the process as it is maintained at constant pressure. Thus, the equation becomes:
Q = m(u2 - u1)
0.0407 (1988.6 - 4017.6) = -90m = 0.0407 kg
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A focce that is based en the abigh of an object ta retum to its original wize and shope after a distorisog fotce is itemoved is known as a(n) _____
The phenomenon described, where an object returns to its original size and shape after the removal of a distorting force, is known as elastic deformation.
Elastic deformation refers to the reversible change in the shape or size of an object under the influence of an external force. When a distorting force is applied to an object, it causes the object to deform. However, if the force is within the elastic limit of the material, the deformation is temporary and the object retains its ability to return to its original shape and size once the force is removed.
This behavior is characteristic of materials with elastic properties, such as metals, rubber, and certain plastics. Within the elastic limit, these materials exhibit a linear relationship between the applied force and the resulting deformation.
This means that the deformation is directly proportional to the force applied. When the force is removed, the object undergoes elastic recoil and returns to its original configuration due to the inherent elastic forces within the material.
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An electron has a kinetic energy of 2.4 eV. What is the wavelength of the electron?
Enter your answer in nm to two decimal places but do not enter the units
The wavelength of an electron with a kinetic energy of 2.4 eV can be calculated using the de Broglie wavelength equation. The wavelength, expressed in nanometers (nm) to two decimal places, can be determined numerically.
The de Broglie wavelength equation relates the wavelength (λ) of a particle to its momentum (p). For an electron, the equation is given by:
λ = h / p
Where:
λ is the wavelength,
h is the Planck's constant (approximately 6.626 x 10^-34 J·s), and
p is the momentum.
The momentum of an electron can be calculated using its kinetic energy (KE) and mass (m) through the equation:
p = sqrt(2 * m * KE)
To find the wavelength, we first need to convert the kinetic energy from electron volts (eV) to joules (J) using the conversion factor: 1 eV = 1.602 x 10^-19 J. Then, we can calculate the momentum and substitute it into the de Broglie wavelength equation.
By plugging in the appropriate values and performing the calculations, we can find the wavelength of the electron in nanometers to two decimal places.
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A convex lens has a focal length f. An object is placed at a
position greater than 2f on the axis. The image formed is located
at
When a convex lens has a focal length of f and an object is placed at a position greater than 2f that is beyond the centre of curvature on the axis, then the image is formed between the centre of curvature and focus.
When the object is located beyond the centre of curvature of a convex lens, the image formed is real, inverted, and diminished. This means that the image is formed on the opposite side of the lens compared to the object, it is upside down, and its size is smaller than the object.
As light rays from the object pass through the lens, they refract (bend) according to the lens's shape and material properties. For a convex lens, parallel rays converge towards the principal focus after passing through the lens.
Therefore, when a convex lens has a focal length of f and an object is placed at a position greater than 2f that is beyond the centre of curvature on the axis, then the image is formed between the centre of curvature and focus.
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A 108 2 resistor is connected in series with a 72 mH inductor and a 0.3 µF capac- itor. The applied voltage has the form 190 V sin(2 ft), where the frequency is f=876 cycles/s. & Find the rms current.
The rms current in the circuit is approximately 0.189 A.
The question requires us to calculate the rms current of a circuit that consists of a resistor, an inductor, and a capacitor in series. The circuit is driven by an AC voltage source that has a frequency of 876 cycles/s and an amplitude of 190 V.Let's begin by finding the total impedance of the circuit. The impedance of a series RLC circuit is given by:Z = R + j(XL - XC)where R is the resistance, XL is the inductive reactance, and XC is the capacitive reactance. The imaginary part of the impedance represents the reactance of the circuit, which depends on the frequency of the applied voltage. At resonance, XL = XC, and the total impedance is equal to the resistance Z = R.
To calculate the impedance of the circuit, we need to find the values of XL and XC at the given frequency f = 876 cycles/s. The inductive reactance is given by:XL = 2πfLwhere L is the inductance of the inductor. Substituting the given values, we get:XL = 2π(876)(72 × 10⁻³) = 101.94 ΩThe capacitive reactance is given by:XC = 1/(2πfC)where C is the capacitance of the capacitor. Substituting the given values, we get:XC = 1/(2π(876)(0.3 × 10⁻⁶)) = 607.71 ΩThe total impedance is therefore:Z = R + j(XL - XC) = 108 + j(-505.77) = 108 - j505.77.
The rms current is given by the ratio of the rms voltage to the impedance:Irms = Vrms/Zwhere Vrms is the rms value of the applied voltage. The rms value of a sinusoidal voltage is given by the peak voltage divided by the square root of 2 (Vrms = Vpeak/√2). Substituting the given values, we get:Vrms = 190/√2 = 134.35 VIrms = Vrms/Z = 134.35/(108 - j505.77) = 0.189 - j0.886 ARms current, Irms = 0.189 A (approx).
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A 7.80 g bullet has a speed of $20 m/s when it hits a target, causing the target to move 4:70 cm in the direction of the bullet's velocity before stopping. (A) Use work and energy considerations to find the average force (in N) that stops the bullet. (Enter the magnitude.) ____________ (B) Assuming the force is constant, determine how much time elapses (in s) between the moment the bullet strikes the target and the moment it stops moving
___________
We can use the principle of work and energy conservation. The work done by the average force on the bullet is equal to the change in kinetic energy of the bullet.
Additionally, the work done by the average force on the target is equal to the change in kinetic energy of the target.
(A) Average force on the bullet:
The work done on the bullet is equal to the change in its kinetic energy. We can calculate the initial kinetic energy of the bullet using the formula:
KE_bullet = (1/2) * m_bullet * v_bullet²
where m_bullet is the mass of the bullet and v_bullet is its initial velocity.
Plugging in the values:
m_bullet = 7.80 g = 0.00780 kg
v_bullet = 20 m/s
KE_bullet = (1/2) * 0.00780 kg * (20 m/s)² = 1.56 J
Since the bullet stops, its final kinetic energy is zero. Therefore, the work done by the average force on the bullet is equal to the initial kinetic energy:
Work_bullet = KE_bullet = 1.56 J
The displacement of the bullet is not given, but it's not needed to calculate the average force.
(B) Time elapsed until the bullet stops:
The work done by the average force on the target is equal to the change in kinetic energy of the target. Since the target comes to a stop, its final kinetic energy is zero. We can calculate the initial kinetic energy of the target using the formula:
KE_target = (1/2) * m_target * v_target²
where m_target is the mass of the target and v_target is its initial velocity.
The mass of the target is not given, so we cannot determine the exact value for the force or the time elapsed.
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A ball of mass 5.0 kg is released from rest from the roof of a building. During the descend, the ball experiences air resistance with an average magnitude of 22.0 N. If the roof is 30.0 m above the ground, at what speed will the ball strike the ground?
A. 18.0 m/s
B. 20.0 m/s
C. 22.0 m/s
D. 24.0 m/s
The ball will strike the ground with a speed of 18.0 m/s. The correct option is A.
To find the speed at which the ball strikes the ground, we can use the concept of energy conservation. The potential energy lost by the ball as it falls is converted into kinetic energy. Taking into account the work done by air resistance, we can set up the following equation:
ΔPE - W_air = ΔKE,
where ΔPE is the change in potential energy, W_air is the work done by air resistance, and ΔKE is the change in kinetic energy.
The change in potential energy is given by:
ΔPE = m * g * h,
where m is the mass of the ball, g is the acceleration due to gravity, and h is the height of the building.
The work done by air resistance is equal to the average magnitude of the air resistance force multiplied by the distance traveled:
W_air = F_air * d,
where F_air is the magnitude of the air resistance force and d is the distance traveled (equal to the height of the building).
The change in kinetic energy is given by:
ΔKE = (1/2) * m * v²,
where v is the final velocity of the ball.
Combining these equations, we have:
m * g * h - F_air * d = (1/2) * m * v².
Substituting the given values into the equation, we get:
(5.0 kg) * (9.8 m/s²) * (30.0 m) - (22.0 N) * (30.0 m) = (1/2) * (5.0 kg) * v².
Simplifying the equation, we find:
1470 J - 660 J = 2.5 kg * v².
810 J = 2.5 kg * v².
Solving for v, we have:
v² = 324 m²/s².
Taking the square root of both sides, we get:
v ≈ 18.0 m/s.
Therefore, the ball will strike the ground with a speed of approximately 18.0 m/s. The correct option is A.
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An inductor designed to filter high-frequency noise from power supplied to a personal computer placed in series with the computer. What mum inductor On met) shot have to produce a 2.83 0 reactance for 150 kote nolie 218 mit (b) What is its reactance (in k) at 57,0 7 7.34 X10
The reactance is approximately 13.7 kΩ.
An inductor designed to filter high-frequency noise from power supplied to a personal computer placed in series with the computer.
The formula that is used to calculate the inductance value is given by;
X = 2πfL
We are given that the reactance that the inductor should produce is 2.83 Ω for a frequency of 150 kHz.
Therefore substituting in the formula we get;
X = 2πfL
L = X/2πf
= 2.83/6.28 x 150 x 1000
Hence L = 2.83/(6.28 x 150 x 1000)
= 3.78 x 10^-6 H
The reactance is given by the formula;
X = 2πfL
Substituting the given values in the formula;
X = 2 x 3.142 x 57.07734 x 10^6 x 3.78 x 10^-6
= 13.67 Ω
≈ 13.7 kΩ
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2- Magnetic brakes are used to bring subway cars to a stop. Treat the 4000 kg subway cart as a 3m long bar sliding along a pair of conducting rails as shown. There is a magnetic field perpendicular to the plane of the rails with a strength of 2 T. a) Given an initial speed 20m/s, find the average deceleration and force required to bring the train to a stop over a distance of 40m. b) As the train moves along the rails, a current is induced in the circuit. What is the magnitude & direction of the initial induced current? (Assume the rails are frictionless, and the subway car has a resistance of 1 kilo-ohm, and the magnitude c) What must be the direction of the magnetic field so as to produce a decelerating force on the subway car? There is no figure.
a) The average deceleration required to bring the train to a stop over a distance of 40m is approximately -5 m/s^2. The force required is approximately -20,000 N (opposite to the initial direction of motion).
b) The magnitude of the initial induced current is approximately 10 A, flowing in the direction opposite to the initial motion of the subway car.
c) The magnetic field should be directed opposite to the initial direction of motion of the subway car to produce a decelerating force.
a) To find the average deceleration and force required, we can use the equations of motion. The initial speed of the subway car is 20 m/s, and it comes to a stop over a distance of 40 m.
Using the equation:
Final velocity^2 = Initial velocity^2 + 2 × acceleration × distance
Substituting the values:
0^2 = (20 m/s)^2 + 2 × acceleration × 40 m
Simplifying the equation:
400 m^2/s^2 = 800 × acceleration × 40 m
Solving for acceleration:
acceleration ≈ -5 m/s^2 (negative sign indicates deceleration)
To find the force required, we can use Newton's second law:
Force = mass × acceleration
Substituting the values:
Force = 4000 kg × (-5 m/s^2)
Force ≈ -20,000 N (negative sign indicates the force opposite to the initial direction of motion)
b) According to Faraday's law of electromagnetic induction, a changing magnetic field induces an electromotive force (EMF) and, consequently, a current in a closed circuit. In this case, as the subway car moves along the rails, the magnetic field perpendicular to the rails induces a current.
The magnitude of the induced current can be calculated using Ohm's law:
Current = Voltage / Resistance
The induced voltage can be found using Faraday's law:
Voltage = -N × ΔΦ/Δt
Since the rails are frictionless, the only force acting on the subway car is the magnetic force, which opposes the motion. The induced voltage is therefore equal to the magnetic force multiplied by the length of the bar.
Voltage = Force × Length
Substituting the given values:
Voltage = 20,000 N × 3 m
Voltage = 60,000 V
Using Ohm's law:
Current = Voltage / Resistance
Current = 60,000 V / 1000 Ω
Current ≈ 60 A
The magnitude of the initial induced current is approximately 60 A, flowing in the direction opposite to the initial motion of the subway car.
c) To produce a decelerating force on the subway car, the direction of the magnetic field should be opposite to the initial direction of motion. This is because the induced current generates a magnetic field that interacts with the external magnetic field, resulting in a force that opposes the motion of the subway car. The direction of the magnetic field should be such that it opposes the motion of the subway car.
To bring the subway car to a stop over a distance of 40 m, an average deceleration of approximately -5 m/s^2 is required, with a force of approximately -20,000 N (opposite to the initial direction of motion). The magnitude of the initial induced current is approximately 60 A, flowing in the opposite direction to the initial motion of the subway car. To produce a decelerating force, the direction of the magnetic field should be opposite to the initial direction of motion.
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Description of what physical processes needs to use
fractional calculation?
Answer:
Fractional calculus is a branch of mathematics that deals with the calculus of functions that are not differentiable at all points. This can be useful for modeling physical processes that involve memory or dissipation, such as viscoelasticity, diffusion, and wave propagation.
Explanation:
Some physical processes that need to use fractional calculation include:
Viscoelasticity: Viscoelasticity is a property of materials that exhibit both viscous and elastic behavior. This can be modeled using fractional calculus, as the fractional derivative of a viscoelastic material can be used to represent the viscous behavior, and the fractional integral can be used to represent the elastic behavior.
Diffusion: Diffusion is the movement of molecules from a region of high concentration to a region of low concentration. This can be modeled using fractional calculus, as the fractional derivative of a diffusing substance can be used to represent the rate of diffusion.
Wave propagation: Wave propagation is the movement of waves through a medium. This can be modeled using fractional calculus, as the fractional derivative of a wave can be used to represent the attenuation of the wave.
Fractional calculus is a powerful tool that can be used to model a wide variety of physical processes. It is a relatively new field of mathematics, but it has already found applications in many areas, including engineering, physics, and chemistry.
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If a human body has a total surface area of 1.7 m2, what is the total force on the body due to the atmosphere at sea level (1.01 x 105Pa)?
The force on a human body due to the atmosphere at sea level having a total surface area of 1.7 m² is 1.717 x 10^4N. Surface area refers to the entire region that covers a geometric figure. In mathematics, surface area refers to the amount of area that a three-dimensional shape has on its exterior.
Force is the magnitude of the impact of one object on another. Force is commonly measured in Newtons (N) in physics. Force can be calculated as the product of mass (m) and acceleration (a), which is expressed as F = ma.
If the human body has a total surface area of 1.7 m², The pressure on the body is given by P = 1.01 x 10^5 Pa. Therefore, the force (F) on the human body due to the atmosphere can be calculated as F = P x A, where A is the surface area of the body. F = 1.01 x 10^5 Pa x 1.7 m²⇒F = 1.717 x 10^4 N.
Therefore, the force on a human body due to the atmosphere at sea level having a total surface area of 1.7 m² is 1.717 x 10^4 N.
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Question 4 (20 Points) One proposes to measure the velocity v of a bullet via shutting it into a ballistic pendulum. The bullet's mass m is 10 g. The mass M of the piece of wood it is shut into, is 10 kg. The velocity V of the system composed of Wood + Bullet, swinging backward right after the bullet hits the piece of wood, is V; the system, through the process of swinging, is highered as much as h=5 cm. a) (10 p) Apply the momentum and energy conservation laws, and determine the intial velocity v of the bullet; take g as 10m/s². b) (10 p) Why the following equation is erromeous: (M+m)gh=(1/2)mv². Explain.
By applying momentum and energy conservation, the initial velocity of the bullet is (m * V + M * V') / m. The erroneous equation neglects the rebound of the bullet and the velocity imparted to the wood.
a) To determine the initial velocity (v) of the bullet, we can apply the principles of momentum and energy conservation.
According to the law of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision. The momentum of an object is given by the product of its mass and velocity.
Before the collision:
The momentum of the bullet: m * v (since the mass of the bullet is m)
The momentum of the wood: 0 (since it is initially at rest)
After the collision:
The momentum of the bullet: m * (-V) (since it moves in the opposite direction with velocity -V)
The momentum of the wood: M * (-V') (since it moves in the opposite direction with velocity -V')
Using the conservation of momentum, we can equate the total momentum before and after the collision:
m * v + 0 = m * (-V) + M * (-V')
Simplifying the equation:
v = (m * V + M * V') / m
Now, let's apply the principle of conservation of energy. The initial kinetic energy of the system is converted into potential energy when the system swings upward by a height (h).
The initial kinetic energy of the system is given by:
(1/2) * (m + M) * V^2
The potential energy gained by the system is given by:
(m + M) * g * h
According to the conservation of energy, these two energies are equal:
(1/2) * (m + M) * V^2 = (m + M) * g * h
Now we can substitute the given values:
m = 10 g = 0.01 kg
M = 10 kg
h = 5 cm = 0.05 m
g = 10 m/s^2
Substituting the values into the equation, we can solve for V:
(1/2) * (0.01 + 10) * V^2 = (0.01 + 10) * 10 * 0.05
Simplifying the equation:
0.505 * V^2 = 5.05
V^2 = 10
Taking the square root of both sides:
V = √10
Therefore, the initial velocity of the bullet (v) is given by:
v = (m * V + M * V') / m
b) The equation (M+m)gh = (1/2)mv^2 is erroneous because it assumes that the bullet remains embedded in the wood after the collision and does not take into account the velocity (V') of the wood. In reality, the bullet rebounds from the wood and imparts a velocity (V') to the wood in the opposite direction. Therefore, the correct equation must consider both the velocities of the bullet and the wood to account for the conservation of momentum and energy in the system.
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