Heat in joules must be added to 1.15 kg of beryllium to change it from a solid at 700°C to a liquid at 1285°C the values: Q1 = 1.15 kg * 1820 J/kg°C * (1285°C - 700°C)
Q2 = 1.15 kg * 1.35 × 10^6 J/kg
To calculate the heat required to change the temperature of beryllium from a solid at 700°C to a liquid at 1285°C, we need to consider the heat required for two processes: heating the solid beryllium from 700°C to its melting point and then melting it at its melting point.
First, let's calculate the heat required to heat the solid beryllium:
Q1 = m * c * ΔT1
Where:
m = mass of beryllium = 1.15 kg
c = specific heat capacity of beryllium = 1820 J/kg°C
ΔT1 = change in temperature = (melting point - initial temperature) = (1285°C - 700°C)
Q1 = 1.15 kg * 1820 J/kg°C * (1285°C - 700°C)
Next, let's calculate the heat required to melt the beryllium at its melting point:
Q2 = m * Lf
Where:
Lf = latent heat of fusion of beryllium = 1.35 × 10^6 J/kg
Q2 = 1.15 kg * 1.35 × 10^6 J/kg
Finally, the total heat required is the sum of Q1 and Q2:
Total heat = Q1 + Q2
Note: Since the temperature is given in degrees Celsius, we don't need to convert it to Kelvin as the temperature difference remains the same.
Calculate the values:
Q1 = 1.15 kg * 1820 J/kg°C * (1285°C - 700°C)
Q2 = 1.15 kg * 1.35 × 10^6 J/kg
Total heat = Q1 + Q2
Evaluate the expression to find the total heat required in joules.
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An automobile traveling 76.0 km/h has tires of 70.0 cm diameter. (a) What is the angular speed of the tires about their axles? (b) If the car is brought to a stop uniformly in 39.0 complete turns of the tires, what is the magnitude of the angular acceleration of the wheels? (c) How far does the car move during the braking? (
(a) Angular speed: 60.3 rad/s
(b) Angular acceleration: 0.244 rad/s²
(c) Distance moved: 5182.4 meters
(a) To find the angular speed of the tires about their axles, we can use the formula:
Angular speed (ω) = Linear speed (v) / Radius (r)
First, let's convert the speed from km/h to m/s:
76.0 km/h = (76.0 km/h) * (1000 m/km) * (1/3600 h/s) ≈ 21.1 m/s
The radius of the tire is half of its diameter:
Radius (r) = 70.0 cm / 2 = 0.35 m
Now we can calculate the angular speed:
Angular speed (ω) = 21.1 m/s / 0.35 m ≈ 60.3 rad/s
Therefore, the angular speed of the tires about their axles is approximately 60.3 rad/s.
(b) To find the magnitude of the angular acceleration of the wheels, we can use the formula:
Angular acceleration (α) = Change in angular velocity (Δω) / Time (t)
The change in angular velocity can be found by subtracting the initial angular velocity (ω_i = 60.3 rad/s) from the final angular velocity (ω_f = 0 rad/s), as the car is brought to a stop:
Δω = ω_f - ω_i = 0 rad/s - 60.3 rad/s = -60.3 rad/s
The time (t) is given as 39.0 complete turns of the tires. One complete turn corresponds to a full circle, or 2π radians. Therefore:
Time (t) = 39.0 turns * 2π radians/turn = 39.0 * 2π rad
Now we can calculate the magnitude of the angular acceleration:
Angular acceleration (α) = (-60.3 rad/s) / (39.0 * 2π rad) ≈ -0.244 rad/s²
The magnitude of the angular acceleration of the wheels is approximately 0.244 rad/s².
(c) To find the distance the car moves during the braking, we can use the formula:
Distance (d) = Linear speed (v) * Time (t)
The linear speed is given as 21.1 m/s, and the time is the same as calculated before:
Time (t) = 39.0 turns * 2π radians/turn = 39.0 * 2π rad
Now we can calculate the distance:
Distance (d) = 21.1 m/s * (39.0 * 2π rad) ≈ 5182.4 m
Therefore, the car moves approximately 5182.4 meters during the braking.
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An LRC circuit has L=15.4mH and R=3.50Ω. Part A What value must C have to produce resonance at 4600 Hz ?
The answer is the value of capacitance required to produce resonance at 4600 Hz is approximately 9.13 × 10^(-9) F. As we know, for an LRC (inductance, resistance, capacitance) circuit, the resonant frequency is given by: f = 1 / (2π√(LC))
Here, we are given L = 15.4 mH and R = 3.50 Ω, and we need to find the value of C for resonance at 4600 Hz.
Substituting the values in the formula: 4600 = 1 / (2π√(15.4×10^(-3)C))
Squaring both sides and rearranging, we get:
C = (1 / (4π²×15.4×10^(-3)×4600²))
C ≈ 9.13 × 10^(-9) F
Therefore, the value of capacitance required to produce resonance at 4600 Hz is approximately 9.13 × 10^(-9) F.
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What is the effect of increasing the tension in the vibrating string to the frequency if
linear mass density & vibrating length are held constant?
Increasing the tension in a vibrating string while keeping the linear mass density and vibrating length constant will result in an increase in the frequency of vibration.
This is because the frequency of vibration in a string is directly proportional to the square root of the tension in the string. By increasing the tension, the restoring force in the string increases, leading to faster vibrations and a higher frequency.
Therefore, increasing the tension in the vibrating string will result in a higher frequency of vibration.
The frequency of vibration in a string is determined by various factors, including tension, linear mass density, and vibrating length. When the linear mass density and vibrating length are held constant, changing the tension has a direct impact on the frequency.
Increasing the tension increases the restoring force in the string, causing the string to vibrate more rapidly and resulting in a higher frequency of vibration.
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Green light at 520 nm is diffracted by a grating with 3200 lines per cm The light is normally incident on the diffraction grating. Through what angle is the light diffracted in the first order? Express your answer in degrees. Through what angle is the light diffracted in the fifth order? Express your answer in degrees.
a) The angle of diffraction at which the light is diffracted in the first order is 9.52 °. b) The angle at which the light is diffracted in the fifth order is 55.77 °.
To determine the angle of diffraction for a given order of diffraction, we can use the formula:
sinθ = mλ/d
Where:
θ is the angle of diffraction,
m is the order of diffraction,
λ is the wavelength of light, and
d is the spacing between the grating lines.
a) For the first order of diffraction:
m = 1
λ = 520 nm = 520 × 10^(-9) m
d = 1 cm / 3200 lines = 1 × 10^(-2) m / 3200 = 3.125 × 10^(-6) m
Plugging in the values:
sinθ = (1) × (520 × 10^(-9) m) / (3.125 × 10^(-6) m)
sinθ ≈ 0.1664
To find the angle θ, we take the inverse sine of the value:
θ ≈ arcsin(0.1664)
θ ≈ 9.52 degrees
Therefore, the light is diffracted at an angle of approximately 9.52 degrees in the first order.
b) For the fifth order of diffraction:
m = 5
λ = 520 nm = 520 × 10^(-9) m
d = 1 cm / 3200 lines = 1 × 10^(-2) m / 3200 = 3.125 × 10^(-6) m
Plugging in the values:
sinθ = (5) × (520 × 10^(-9) m) / (3.125 × 10^(-6) m)
sinθ ≈ 0.832
To find the angle θ, we take the inverse sine of the value:
θ ≈ arcsin(0.832)
θ ≈ 55.77 degrees
Therefore, the light is diffracted at an angle of approximately 55.77 degrees in the fifth order.
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An electron microscope produces electrons with a wavelength of 2.8 pm
d= 2.8 pm
If these are passed through a 0.75 um single slit, at what angle (in degrees) will the first diffraction minimum be found?
For an electron microscope produces electrons with a wavelength of 2.8 pm d= 2.8 pm, if these are passed through a 0.75 the diffraction can be calculated. The angle at which the first diffraction minimum will be found is approximately 0.028 degrees.
To calculate the angle at which the first diffraction minimum occurs, we can use the formula for the angular position of the minima in single-slit diffraction:
θ = λ / (2d)
Where:
θ is the angle of the diffraction minimum,
λ is the wavelength of the electrons, and
d is the width of the single slit.
Given that the wavelength of the electrons is 2.8 pm (2.8 × [tex]10^{-12}[/tex] m) and the width of the single slit is 0.75 μm (0.75 × [tex]10^{-6}[/tex] m), we can substitute these values into the formula to find the angle:
θ = (2.8 × [tex]10^{-12}[/tex] m) / (2 × 0.75 × [tex]10^{-6}[/tex] m)
Simplifying the expression, we have:
θ = 0.028
Therefore, the angle at which the first diffraction minimum will be found is approximately 0.028 degrees.
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6) Write the expressions for the electric and magnetic fields, with their corresponding directions, of an electromagnetic wave that has an electric field parallel to the axis and whose amplitude is 300 V/m. Also, this wave has a frequency of 3.0 GHz and travels in the +y direction.
The electric field (E) is along the y-axis and given by E(y, t) = 300 sin(2π(3.0 GHz)t) V/m. The magnetic field (B) is along the x-axis and given by B(y, t) = (300 V/m) / (3.0 x 10^8 m/s) sin(2π(3.0 GHz)t).
The general expression for an electromagnetic wave in free space can be written as:
E(x, t) = E0 sin(kx - ωt + φ)
where:
E(x, t) is the electric field as a function of position (x) and time (t),
E0 is the amplitude of the electric field,
k is the wave number (related to the wavelength λ by k = 2π/λ),
ω is the angular frequency (related to the frequency f by ω = 2πf),
φ is the phase constant.
For the given wave with an electric field parallel to the axis (along the y-axis) and traveling in the +y direction, the expression can be simplified as:
E(y, t) = E0 sin(ωt)
where:
E(y, t) is the electric field as a function of position (y) and time (t),
E0 is the amplitude of the electric field,
ω is the angular frequency (related to the frequency f by ω = 2πf).
In this case, the electric field remains constant in magnitude and direction as it propagates in the +y direction. The amplitude of the electric field is given as 300 V/m, so the expression becomes:
E(y, t) = 300 sin(2π(3.0 GHz)t)
Now let's consider the magnetic field associated with the electromagnetic wave. The magnetic field is perpendicular to the electric field and the direction of wave propagation (perpendicular to the y-axis). Using the right-hand rule, the magnetic field can be determined to be in the +x direction.
The expression for the magnetic field can be written as:
B(y, t) = B0 sin(kx - ωt + φ)
Since the magnetic field is perpendicular to the electric field, its amplitude (B0) is related to the amplitude of the electric field (E0) by the equation B0 = E0/c, where c is the speed of light. In this case, the wave is propagating in free space, so c = 3.0 x 10^8 m/s.
Therefore, the expression for the magnetic field becomes:
B(y, t) = (E0/c) sin(ωt)
Substituting the value of E0 = 300 V/m and c = 3.0 x 10^8 m/s, the expression becomes:
B(y, t) = (300 V/m) / (3.0 x 10^8 m/s) sin(2π(3.0 GHz)t)
To summarize:
- The electric field (E) is along the y-axis and given by E(y, t) = 300 sin(2π(3.0 GHz)t) V/m.
- The magnetic field (B) is along the x-axis and given by B(y, t) = (300 V/m) / (3.0 x 10^8 m/s) sin(2π(3.0 GHz)t).
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Destructive interference of two superimposed waves requires the waves to travel in opposite directions. Select one: True False
The given statement, "Destructive interference of two superimposed waves requires the waves to travel in opposite directions" is false because destructive interference of two superimposed waves requires the waves to be traveling in the same direction and having a phase difference of π or an odd multiple of π.
In destructive interference, the two waves will have a phase difference of either an odd multiple of π or an odd multiple of 180 degrees. When the phase difference is an odd multiple of π, it results in a complete cancellation of the two waves in the region where they are superimposed and the resultant wave has zero amplitude. In constructive interference, the two waves will have a phase difference of either an even multiple of π or an even multiple of 180 degrees. When the phase difference is an even multiple of π, it results in a reinforcement of the two waves in the region where they are superimposed and the resultant wave has maximum amplitude.
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1. Pressure is described as ___ per unit area. a. Flow b. Pounds c. Force d. Inches 2. Pressure is increased when: a. The number of molecules per unit area is decreased Heavier molecules per unit area are introduced b. c. Molecules begin to move faster d. The number of molecules are spread out over a larger ur 3. Atmospheric pressure at sea level is__ _psia? a. 0 b. 2 C. 14.7 d. 27.73
1. Pressure is described as ___ per unit area.
a. Flow
b. Pounds
c. Force
d. Inches
The correct answer is c. Force. Pressure is the force exerted per unit area.
2. Pressure is increased when:
a. The number of molecules per unit area is decreased
b. Heavier molecules per unit area are introduced
c. Molecules begin to move faster
d. The number of molecules are spread out over a larger area
The correct answer is c. Molecules begin to move faster. When molecules move faster, they collide with surfaces more frequently and with greater force, resulting in an increase in pressure.
Atmospheric pressure at sea level is __ psia?
a. 0
b. 2
c. 14.7
d. 27.73
The correct answer is c. 14.7. Atmospheric pressure at sea level is approximately 14.7 pounds per square inch absolute (psia).
2 Magnetic Domain Theory. Answer each of the following questions a) When a bar magnet is broken into two pieces, the two pieces actually become two independent magnets instead of a north-pole magnet and a south-pole magner. Explain this phenomenon b) When a magnet is heated up, it loses it magnetization power. However, when the temperature cools back down, the magnetism power returns (assuming the temperature is lower than the Curie point).
a) When a bar magnet is broken into two pieces, the two pieces become two independent magnets, and not a north-pole magnet and a south-pole magnet. This is because each piece contains its own magnetic domain, which is a region where the atoms are aligned in the same direction. The alignment of atoms in a magnetic domain creates a magnetic field. In a magnet, all the magnetic domains are aligned in the same direction, creating a strong magnetic field.
When a magnet is broken into two pieces, each piece still has its own set of magnetic domains and thus becomes a magnet itself. The new north and south poles of the pieces will depend on the arrangement of the magnetic domains in each piece.
b) When a magnet is heated up, the heat energy causes the atoms in the magnet to vibrate more, which can disrupt the alignment of the magnetic domains. This causes the magnetization power to decrease. However, when the temperature cools back down, the atoms in the magnet stop vibrating as much, and the magnetic domains can re-align, causing the magnetism power to return. This effect is assuming that the temperature is lower than the Curie point, which is the temperature at which a material loses its magnetization permanently.
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Suppose you are on another planet and you want to measure its acceleration of gravity so you drop an object from rest. It hits the ground, traveling a distance of 0.8 min 0.5 second and then bounces back up and stops exactly where it started from. a) Please calculate the acceleration of gravity on this planet. b) Taking downward to be positive, how does the ball's average speed compare to the magnitude of its average velocity on the way down? c) Taking the beginning of the motion as the time the ball was dropped, how does its average speed compare to the magnitude of its average velocity on the way up? d) with what speed did the ball hit the ground? e) When distance is divided by time the result is 1.6 m/sec
Given that an object is dropped from rest on another planet and hits the ground, travelling a distance of 0.8 m in 0.5 s and bounces back up and stops exactly where it started from.
Let's find out the acceleration of gravity on this planet. Step-by-step explanation: a) To calculate the acceleration of gravity on this planet, we use the formula d = 1/2 gt².Using this formula, we get0.8 m = 1/2 g (0.5 s)²0.8 m = 0.125 g0.125 g = 0.8 mg = 0.8/0.125g = 6.4 m/s²The acceleration of gravity on this planet is 6.4 m/s².b) Taking downward to be positive, the ball's average speed is equal to its magnitude of average velocity on the way down.
Therefore, the average speed of the ball is equal to the magnitude of its average velocity on the way down.c) The ball's initial speed (when dropped) is zero, so the magnitude of its average velocity on the way up is equal to its final velocity divided by the time taken to stop. Using the formula v = u + gt where v = 0 m/s and u = -6.4 m/s² (negative because the ball is moving up), we get0 = -6.4 m/s² + g*t t = 6.4/gt = √(0.8 m/6.4 m/s²)t = 0.2 seconds.
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You are asked to change a racecar's properties to make it accelerate faster. You have two options: decrease the car's drag coefficient and use better tires so that its net horizontal force is 25% larger, or remove unnecessary items and use lighter weight materials so that the car's mass is 25% smaller. Which of those changes will produce the largest acceleration? Hint: careful! Try some numbers out. Increasing the net force by 25% Decreasing the mass by 25% It doesn't matter: both of these choices will produce the same effect on the car's acceleration Not enough information
Option 2 will produce the largest acceleration.
To calculate the changes that will produce the largest acceleration, let us first consider the following formula:
F = ma
where,
F = force applied
m = mass
a = acceleration
We can assume that the force applied will be constant; hence, by reducing the drag coefficient or the mass of the car, we can observe an increase in the car's acceleration.
Option 2 will produce the largest acceleration if we consider the formula.
When we change the racecar's mass by 25% by removing unnecessary items and using lighter weight materials, we decrease the mass.
If the mass of the car is reduced, acceleration will increase accordingly.
The second option, which is to remove unnecessary items and use lighter weight materials so that the car's mass is 25% smaller, will produce the largest acceleration.
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A dipole is formed by point charges +3.5 μC and -3.5 μC placed on the x axis at (0.30 m , 0) and (-0.30 m , 0), respectively.
At what positions on the x axis does the potential have the value 7.3×105 V ?
x1, x2 = _____ m
A dipole is formed by point charges +3.5 μC and -3.5 μC placed on the x axis at (0.30 m , 0) and (-0.30 m , 0), respectively.The expression for the electric potential due to the point charges along the x-axis is given by;V=kq1/x1+kq2/x2where,k=9.0×10^9 Nm²/C²q1=+3.5 μCq2=-3.5 μCV=7.3×105 VX-axis coordinates of the charges are x1=0.30 m and x2=-0.30 m.
Substitute the given values in the above expression, V=kq1/x1+kq2/x2=9.0×10^9×3.5×10⁻⁶/|x1|+9.0×10^9×3.5×10⁻⁶/|x2|=9.0×10^9×3.5×10⁻⁶(|x1|+|x2|)/|x1x2|=7.3×10⁵On simplifying, we get,(|x1|+|x2|)/|x1x2|=8.11x1x2=x1(x1+x2)=9.0×10^9×3.5×10⁻⁶/7.3×10⁵=4.32×10⁻⁴Solve for x2,x2=-x1-x2=-0.3-0.3= -0.6mx1+x2=0.432x1-0.6=0x1=1.39m. Substitute the value of x1 in x1+x2=0.432,We get,x2= -1.39m.Thus, x1=1.39m and x2=-1.39m.
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Two forces are acting on an object. I 250 N at an angle of 49 degrees and FB is 125 N at an angle of 128 degrees. What are the force and angle of the equilibrium force?
The force of equilibrium force is approximately 303.05 N at an angle of 70.5 degrees.
To find the force and angle of the equilibrium force, we need to calculate the resultant force by adding the two given forces.
Let's break down the given forces into their horizontal and vertical components:
Force FA = 250 N at an angle of 49 degrees
Force FB = 125 N at an angle of 128 degrees
For FA:
Horizontal component FAx = FA * cos(49 degrees)
Vertical component FAy = FA * sin(49 degrees)
For FB:
Horizontal component FBx = FB * cos(128 degrees)
Vertical component FBy = FB * sin(128 degrees)
Now, let's calculate the horizontal and vertical components:
FAx = 250 N * cos(49 degrees) ≈ 160.39 N
FAy = 250 N * sin(49 degrees) ≈ 189.88 N
FBx = 125 N * cos(128 degrees) ≈ -53.05 N (Note: The negative sign indicates the direction of the force)
FBy = 125 N * sin(128 degrees) ≈ 93.82 N
To find the resultant force (FR) in both horizontal and vertical directions, we can sum the respective components:
FRx = FAx + FBx
FRy = FAy + FBy
FRx = 160.39 N + (-53.05 N) ≈ 107.34 N
FRy = 189.88 N + 93.82 N ≈ 283.7 N
The magnitude of the resultant force (FR) can be calculated using the Pythagorean theorem:
|FR| = √(FRx^2 + FRy^2)
|FR| = √((107.34 N)^2 + (283.7 N)^2)
≈ √(11515.3156 N^2 + 80349.69 N^2)
≈ √(91864.0056 N^2)
≈ 303.05 N
The angle of the resultant force (θ) can be calculated using the inverse tangent function:
θ = atan(FRy / FRx)
θ = atan(283.7 N / 107.34 N)
≈ atan(2.645)
θ ≈ 70.5 degrees
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#9 Magnetic field strength in the center of a ring Suppose a conductor in the shape of a perfectly circular ring bears a current of \( 0.451 \) Amperes, If the conductor has a radius of \( 0.0100 \) m
The distance between the plates decreases, the force exerted on the positive plate of the capacitor increases and vice versa. Given, Speed of parallel plate capacitor = v = 34 m/s
Magnetic field = B = 4.3 TArea of each plate = A = 9.3 × 10⁻⁴ m²
Electric field within the capacitor = E = 220 N/C
Let the distance between the plates of the capacitor be d.
Now, the magnitude of the magnetic force exerted on the positive plate of the capacitor is given by
F = qVB sinθ
where q = charge on a plate = C/d
V = potential difference between the plates = Edsinθ = 1 (since velocity is perpendicular to the magnetic field)
Thus,
F = qVB
Putting the values, we get
F = qVB
= (C/d) × (E/d) × B
= (EA)/d²= (220 × 9.3 × 10⁻⁴)/d²
= 0.2046/d²
Since d is not given, we cannot calculate the exact value of the magnetic force. However, we can say that the force is inversely proportional to the square of the distance between the plates.
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How many moles of acetic acid would you need to add to 2.00 l of water to make a solution with a ph of 2.25?
Approximately 0.005623 moles of acetic acid would be needed to achieve a solution with a pH of 2.25 in 2.00 liters of water.
To determine the number of moles of acetic acid needed to achieve a pH of 2.25 in a solution, we first need to understand the relationship between pH, concentration, and dissociation of the acid.
Acetic acid (CH3COOH) is a weak acid that partially dissociates in water. The dissociation can be represented by the equation: CH3COOH ⇌ CH3COO- + H+
The pH of a solution is a measure of its acidity and is defined as the negative logarithm (base 10) of the concentration of hydrogen ions (H+). The pH scale ranges from 0 to 14, where a pH of 7 is considered neutral, below 7 is acidic, and above 7 is basic.
In the case of acetic acid, we need to calculate the concentration of H+ ions that corresponds to a pH of 2.25. The concentration can be determined using the formula:
[H+] = 10^(-pH)
[H+] = 10^(-2.25)
Once we have the concentration of H+ ions, we can assume that the concentration of acetic acid (CH3COOH) will be equal to the concentration of the H+ ions, as the acid partially dissociates.
Now, to calculate the number of moles of acetic acid needed, we multiply the concentration (in moles per liter) by the volume of the solution. In this case, the volume is given as 2.00 liters.
Number of moles of acetic acid = Concentration (in moles/L) * Volume (in liters)
Substitute the concentration of H+ ions into the equation and calculate the number of moles of acetic acid.
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"Two converging lenses with the same focal length of 10 cm are 40
cm apart. If an object is located 15 cm from one of the lenses,
find the final image distance of the object.
a. 0 cm
b. 5 cm
c. 10 cm
d 15 cm
The final image distance of the object, if the object is located 15 cm from one of the lenses is 6 cm. So none of the options are correct.
To determine the final image distance of the object in the given setup of two converging lenses, we can use the lens formula:
1/f = 1/di - 1/do
Where: f is the focal length of the lens, di is the image distance, do is the object distance.
Given that both lenses have the same focal length of 10 cm, we can consider them as a single lens with an effective focal length of 10 cm. The lenses are 40 cm apart, and the object distance (do) is 15 cm.
Using the lens formula, we can rearrange it to solve for di:
1/di = 1/f + 1/do
1/di = 1/10 cm + 1/15 cm
= (15 + 10) / (10 * 15) cm⁻¹
= 25 / 150 cm⁻¹
= 1 / 6 cm⁻¹
di = 1 / (1 / 6 cm⁻¹) = 6 cm
Therefore, the final image distance of the object is 6 cm. So, none of the options are correct.
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A hose is connected to a faucel and used to fill a 4.0-L. container in a time of 45 s
Determine the volume flow rate in m.
The volume flow rate in m is 8.89 × 10⁻⁵ m³/s.
The volume flow rate is the measure of how much fluid is flowing through a section of a pipeline per unit time. In this case, a hose is connected to a faucet and is used to fill a 4.0-L container in 45 s. To determine the volume flow rate, we need to find out how much water is flowing through the hose per unit time.Volume flow rate = volume of water/time taken
The volume of water that flows through the hose is equal to the volume of water that fills the container.
Therefore, Volume of water = 4.0 L = 4.0 × 10⁻³ m³
Time taken = 45 s
Using the above formula,
Volume flow rate = volume of water/time taken
= 4.0 × 10⁻³ m³/45 s
= 0.0889 × 10⁻³ m³/s
= 8.89 × 10⁻⁵ m³/s
Therefore, the volume flow rate in m is 8.89 × 10⁻⁵ m³/s.
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The point chargest 7 cm apart have an electric pohler501 The total change is 29 nC What are the two charges?
The problem involves two point charges that are 7 cm apart and have a total charge of 29 nC.
To determine the values of the individual charges, we can set up a system of equations based on Coulomb's law and solve for the unknown charges.
Coulomb's law states that the electric force between two charged objects 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|) /[tex]r^2[/tex], where F is the force, k is the electrostatic constant, q1 and q2 are the charges, and r is the distance between the charges.
In this problem, we are given that the charges are 7 cm apart (r = 7 cm) and the total charge is 29 nC. Let's denote the two unknown charges as q1 and q2.
Since the total charge is positive, we know that the charges on the two objects must have opposite signs. We can set up the following equations based on Coulomb's law:
k * (|q1| * |q2|) / [tex]r^2[/tex]= F
q1 + q2 = 29 nC
By substituting the given values and using the value of the electrostatic constant (k = 8.99x10^9 N [tex]m^2[/tex]/[tex]c^2[/tex]), we can solve the system of equations to find the values of q1 and q2, which represent the two charges.
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QUESTION 16 A parallel-plate capacitor consists of two identical, parallel, conducting plates each with an area of 2.00 cm2 and a charge of +4.00 nC. What is the potential energy stored in this capaci
The potential energy stored in the capacitor is [tex]7.03 * 10^{(-10)} J[/tex].
The potential energy stored in a capacitor can be calculated using the formula:
[tex]U = (1/2) * C * V^2,[/tex]
where U is the potential energy, C is the capacitance, and V is the potential difference (voltage) across the capacitor.
The capacitance of a parallel-plate capacitor is given by:
C = (ε0 * A) / d,
where ε0 is the permittivity of free space, A is the area of each plate, and d is the separation between the plates.
Given:
Area of each plate (A) = [tex]2.00 cm^2[/tex] = [tex]2.00 * 10^{(-4)} m^2[/tex],
Charge on each plate = +4.00 nC = [tex]+4.00 * 10^{(-9)} C[/tex],
Plate separation (d) = 0.300 mm =[tex]0.300 * 10^{(-3)} m[/tex].
First, we need to calculate the capacitance:
C = (ε0 * A) / d.
The permittivity of free space (ε0) is approximately [tex]8.85 * 10^{(-12) }F/m[/tex].
Substituting the values:
[tex]C = (8.85 * 10^{(-12)} F/m) * (2.00 * 10^{(-4)} m^2) / (0.300 * 10^{(-3)} m).[/tex]
[tex]C = 1.18 * 10^{(-8)} F.[/tex]
Next, we can calculate the potential energy:
[tex]U = (1/2) * C * V^2.[/tex]
The potential difference (V) is given by:
V = Q / C,
where Q is the charge on the capacitor.
Substituting the values:
[tex]V = (+4.00 * 10^{(-9)} C) / (1.18 * 10^{(-8)} F).[/tex]
V = 0.34 V.
Now, we can calculate the potential energy:
[tex]U = (1/2) * (1.18 * 10^{(-8)} F) * (0.34 V)^2.[/tex]
[tex]U = 7.03 * 10^{(-10)} J.[/tex]
Therefore, the potential energy stored in the capacitor is [tex]7.03 * 10^{(-10)}J[/tex]The closest option is a. [tex]1.77 * 10^{(-9)} J[/tex].
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The complete question is:
A parallel-plate capacitor consists of two identical , parallel, conducting plates each with an area of 2.00 cm2 and a charge of + 4.00 nC. What is the potential energy stored in this capacitor if the plate separation is 0.300 mm?
a. 1.77
b.1.36
c. 2.43
d. 3.764
e. 1.04
A sprinter starts from rest and accelerates to her maximum speed of 9.5 m/s In a distance of 9.0 m. (a) What was her acceleration, if you assume it to be constant? 9.5 m/s X Dimensionally incorrect. Please check the type or dimension of your unit. (b) If this maximum speed is maintained for another 81.9 m, how long does it take her to run 90.9 m?
(a) The acceleration of the sprinter is approximately 5.014 m/s². (b) It takes approximately 17.284 seconds for the sprinter to run 90.9 m.
To find the acceleration of the sprinter, we can use the kinematic equation;
v² = u² + 2as
where;
v = final velocity = 9.5 m/s
u = initial velocity = 0 m/s (starting from the rest)
s = distance covered = 9.0 m
Rearranging the equation to solve for acceleration (a), we have;
Plugging in the values;
a = (9.5² - 0²) / (2 × 9.0)
a = 90.25 / 18
a ≈ 5.014 m/s²
Therefore, the acceleration of the sprinter is approximately 5.014 m/s².
a = (v² - u²) / (2s)
If the sprinter maintains the maximum speed of 9.5 m/s for another 81.9 m, we can use the equation:
s = ut + (1/2)at²
where;
s = total distance covered = 90.9 m
u = initial velocity = 9.5 m/s
a = acceleration = 0 m/s² (since the speed is maintained)
t = time taken
Rearranging the equation to solve for time (t), we have;
t = (2s) / u
Plugging in the values;
t = (2 × 81.9) / 9.5
t ≈ 17.284 seconds
Therefore, it takes approximately 17.284 seconds for the sprinter to run 90.9 m.
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When the Venera 14 probe landed on Venus's surface, its barometer measured an air pressure of 9.5 MPa. The surface acceleration due to gravity was measured to be 8.87 m/s2. If Earth's atmosphere with a pressure of 101 kPa raises mercury 0.760 m where gravitational acceleration is 9.81 m/s2. To what height in m to two significant digits would Venus's atmosphere raise liquid mercury?
The height to which Venus's atmosphere would raise liquid mercury is determined based on the given air pressure and surface acceleration due to gravity. The calculation involves comparing the pressure in Venus's atmosphere to Earth's atmosphere and using the difference to determine the height of the mercury column.
To calculate the height to which Venus's atmosphere would raise liquid mercury, we can use the principle of hydrostatic pressure. The pressure difference between two points in a fluid column is directly proportional to the difference in height.Given that Earth's atmosphere raises mercury to a height of 0.760 m when the pressure is 101 kPa and the acceleration due to gravity is 9.81 m/s^2, we can set up a proportion to find the height in Venus's atmosphere.
The ratio of pressure to height is constant, so we can write:
(9.5 MPa / 101 kPa) = (8.87 m/s^2 / 9.81 m/s^2) * (h / 0.760 m)
Solving for h, we can find the height to which Venus's atmosphere would raise liquid mercury.
By rearranging the equation and substituting the given values, we can calculate the height to two significant digits.
Therefore, the height to which Venus's atmosphere would raise liquid mercury can be determined using the given air pressure and surface acceleration due to gravity.
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QUESTION 5 Which of the following is NOT true? The sum of two vectors of the same magnitude cannot be zero The location of a vector on a grid has no impact on its meaning The magnitude of a vector quantity is considered a scalar quantity Any vector can be expressed as the sum of two or more vectors QUESTION 6 What would be the distance from your starting position if you were to follow the directions: "Go North 10 miles, then East 4 miles and then South 7 miles" 7 miles 5 miles 21 miles 14 miles
QUESTION 5 is: The magnitude of a vector quantity is considered a scalar quantity. This statement is NOT true.
QUESTION 6 is: 7 miles.
The answer to QUESTION 5 is: The magnitude of a vector quantity is considered a scalar quantity. This statement is NOT true. The magnitude of a vector represents its size or length and is always considered a scalar quantity.
The answer to QUESTION 6 is: 7 miles.
If you start at a certain position and go North 10 miles, you would move 10 miles in the North direction. Then, if you go East 4 miles, you would move 4 miles in the East direction. Finally, if you go South 7 miles, you would move 7 miles in the South direction.
Since the 7-mile Southward movement cancels out the initial 7-mile Northward movement, the net displacement in the North-South direction is zero. The remaining 4-mile Eastward movement determines the final distance from the starting position, which is 4 miles.
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QUESTION 5. The statement "The sum of two vectors of the same magnitude cannot be zero" is NOT true.
QUESTION 6. The distance from the starting position after following the directions "Go North 10 miles, then East 4 miles, and then South 7 miles" would be 7 miles.
QUESTION 5
The statement "The sum of two vectors of the same magnitude cannot be zero" is incorrect. In fact, the sum of two vectors of the same magnitude can be zero. This occurs when the two vectors have equal magnitudes but are in opposite directions. In such cases, their combined effect cancels out, resulting in a net sum of zero.
QUESTION 6
To calculate the distance from the starting position after following the directions "Go North 10 miles, then East 4 miles, and then South 7 miles," we need to determine the net displacement. Starting from the initial point and moving North by 10 miles, we establish a displacement of 10 miles in the North direction. Then, moving East by 4 miles adds a displacement of 4 miles in the East direction. However, when we move South by 7 miles, we have a displacement in the opposite direction of the initial North direction.
Taking these displacements into account, we find that the net displacement is given by 10 miles (North) + 4 miles (East) - 7 miles (South). Simplifying this expression, we get a net displacement of 7 miles.
Therefore, the correct option for the distance from the starting position is 7 miles.
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A
body whose density is 2500 kg/m' weighs 98 N in air and 66.64 N
submerged in a liquid. N. Find the density of the liquid
Answer: the density of the liquid is approximately 2499.2 kg/m³
Explanation:
To find the density of the liquid, we can use Archimedes' principle, which states that the buoyant force experienced by an object submerged in a fluid is equal to the weight of the fluid displaced by the object.
The weight of the body in air is given as 98 N, and the weight of the body submerged in the liquid is given as 66.64 N. The difference in weight between the two states represents the weight of the liquid displaced by the body.
Weight of the liquid displaced = Weight in air - Weight submerged = 98 N - 66.64 N = 31.36 N
Now, we can use the formula for density:
Density = (Weight of the liquid displaced) / (Volume of the liquid displaced)
Since the weight of the liquid displaced is 31.36 N and the density of the body is given as 2500 kg/m³, we can rearrange the formula to solve for the volume of the liquid displaced:
Volume of the liquid displaced = (Weight of the liquid displaced) / (Density of the body)
Volume of the liquid displaced = 31.36 N / 2500 kg/m³ = 0.012544 m³
Now, we can find the density of the liquid:
Density of the liquid = (Weight of the liquid displaced) / (Volume of the liquid displaced)
Density of the liquid = 31.36 N / 0.012544 m³ ≈ 2499.2 kg/m³
3) A wire runs above the ground, carrying a large current. In the picture shown below, the current comes out of the page. K The Long Wire, Viewed head on The Ground A) If you stand on the ground directly underneath the wire, which way will a compass point? (Ignore the field of the Earth.) B) The wire is sagging downward. You realize that by using additional magnets, you can counteract the force of gravity on the wire, so that it doesn't sag. What direction magnetic field will be required to do this? (Hint: a current is just moving charge!) C) Show how to position bar magnet(s) near the wire to accomplish your answer from part B. (If you don't have an answer for part B, just guess a direction so you can get credit here.)
Using the concept of the magnetic field generated by current-carrying wire:
(A) The compass needle will point anticlockwise. if you are standing right below it.
(B)The magnets should be directed vertically upward.
(C) The north pole of the bar magnet should point downward.
A straight current-carrying wire generates a circular magnetic field around it as the axis.
A) The compass needle will point anticlockwise if you are standing right underneath the wire. The right-hand rule can be used to figure this out. When viewed from above, the magnetic field produced by the current will move anticlockwise around the wire if the current is exiting the page. The compass needle will point anticlockwise because its north pole lines up with the magnetic field lines.
B) The magnetic field created by the extra magnets should be directed vertically upward to oppose the pull of gravity on the wire and prevent sagging. The upward magnetic force can counterbalance the downward gravitational attraction by positioning the magnetic field in opposition to the gravitational pull.
C) You can place bar magnets in a precise way to provide the necessary upward magnetic field close to the wire. The north poles of the bar magnets should be pointed downward as you position them vertically above the wire. The magnets' south poles should be facing up. By positioning the bar magnets in this way, their magnetic fields will interact to produce an upward magnetic field close to the wire that will work to fight gravity and stop sagging.
Therefore, Using the concept of the magnetic field generated by a current-carrying wire:
(A) The compass needle will point anticlockwise. if you are standing right below it.
(B)The magnets should be directed vertically upward.
(C) The north pole of the bar magnet should point downward.
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A positively-charged object with a mass of 0.191 kg oscillates at the end of a spring, generating ELF (extremely low frequency) radio waves that have a wavelength of 4.40×107 m. The frequency of these radio waves is the same as the frequency at which the object oscillates. What is the spring constant of the spring? Number Units
The spring constant of the spring is approximately 1.90 × 10⁻¹⁷ N/m. This value is obtained by substituting the mass of the object (0.191 kg) and the time period of oscillation (4.35536 × 10¹⁴ s²) into the formula for the spring constant (k = (4π²m) / T²).
According to the information provided, a positively-charged object with a mass of 0.191 kg oscillates at the end of a spring, generating ELF (extremely low frequency) radio waves that have a wavelength of 4.40×10^7 m.
The frequency of these radio waves is the same as the frequency at which the object oscillates. We have to determine the spring constant of the spring. The formula for calculating the spring constant is given as below;k = (4π²m) / T²
Wherek = spring constant
m = mass of the object
T = time period of oscillation
Therefore, first we need to find the time period of oscillation. The formula for time period is given as below;T = 1 / f
Where T = time period
f = frequency
Thus, substituting the given values, we get;
T = 1 / f = 1 / (f (same for radio waves))
Now, to find the spring constant, we substitute the known values of mass and time period into the formula of the spring constant: k = (4π²m) / T²k = (4 x π² x 0.191 kg) / (4.35536 x 10¹⁴ s²) k = 1.90 × 10⁻¹⁷ N/m
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Part A Superman throws a boulder of weight 2700 N at an adversary. What horizontal force must Superman apply to the boulder to give it a horizontal acceleration of 11.4 m/s²? Express your answer in newtons. 15. ΑΣΦ SAEED ? F = Submit Request Answer N
Superman must apply a horizontal force of approximately 3142.09 N to the boulder.
To find the horizontal force that Superman must apply to the boulder we can use Newton's second law of motion.
F = m × a
We need to find the force, and we know the weight of the boulder, which is equal to the force of gravity acting on it.
The weight (W) is given as 2700 N.
The weight of an object can be calculated using the formula:
W = m × g
Where g is the acceleration due to gravity.
g= 9.8 m/s².
Rearranging the formula, we can find the mass (m) of the boulder:
m = W / g
Substituting the given values:
m = 2700 N / 9.8 m/s²
= 275.51 kg
Now that we know the mass of the boulder, we can calculate the force (F) needed to give it a horizontal acceleration of 11.4 m/s²:
F = m × a
F = 275.51 kg× 11.4 m/s²
= 3142.09 N
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The Earth is 1.49x108km from the Sun, and its period of revolution is 1.0a. Venus is 1.08x108km from the Sun, on average. Use Kepler's third law to calculate the length of a Venus year in Earth years.
The length of a Venus year is approximately 0.615 Earth years.Kepler's third law states that the square of the orbital period (T) of a planet is proportional to the cube of its average distance from the Sun (r).
Mathematically, it can be written as:
T² = k * r³
where T is the orbital period, r is the average distance from the Sun, and k is a constant.
Let's denote the Earth's orbital period as TE, the Earth-Sun distance as RE, the Venus's orbital period as TV, and the Venus-Sun distance as RV.
According to the problem:
RE = 1.49 × 10⁸ km
TE = 1.0 Earth year
RV = 1.08 × 10⁸ km
We can set up the following equation using Kepler's third law:
(TV)² = k * (RV)³
To find the length of a Venus year in Earth years, we need to find the ratio TV/TE.
Dividing both sides of the equation by (TE)², we get:
(TV/TE)² = (k/TE²) * (RV)³
Let's denote k/TE² as a constant C:
(TV/TE)² = C * (RV)³
To find the value of C, we can use the information given for Earth:
(TE)² = k * (RE)³
Dividing both sides by (RE)³:
(TE/RE)² = k
Since (TE/RE) is known, we can substitute this value into the equation:
(TV/TE)² = (TE/RE)² * (RV)³
Now we can substitute the given values:
(TV/1.0)² = (1.0/1.49)² * (1.08)³
Simplifying:
(TV)² = (1/1.49)² * (1.08)³
Taking the square root of both sides:
TV = √[(1/1.49)² * (1.08)³]
TV ≈ 0.615 Earth years
Therefore, the length of a Venus year is approximately 0.615 Earth years
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1. (a) At what temperature do the Fahrenheit and Celsius scales have the same numerical value? (b) At what temperature do the Fahrenheit and Kelvin scales have the same numerical value? 1. How large an expansion gap should be left between steel railroad rails if they may reach a maximum temperature 30 deg C greater than when they were laid? Their 1 original length is 12.5 m. Use a=1.2x10-5 O m
The point at which the Fahrenheit and Celsius scales have the same numerical value is -40°C. The point at which the Fahrenheit and Kelvin scales have the same numerical value is 459.67°F the expansion gap that should be left between the steel railroad rails is 0.0045 m or 4.5 mm.
(a) The point at which the Fahrenheit and Celsius scales have the same numerical value is -40°C. This is because this temperature is equivalent to -40°F. At this temperature, both scales intersect and meet the same numerical value.
(b) The point at which the Fahrenheit and Kelvin scales have the same numerical value is 459.67°F. At this temperature, both scales intersect and meet the same numerical value.
For the second part of the question:
Given that the original length of the steel railroad rails is 12.5m, the maximum temperature rise is 30℃, and the coefficient of linear expansion (a) is 1.2×10⁻⁵/℃.
Therefore, the expansion ΔL can be calculated as:
ΔL = L×a×ΔT
Where L is the original length of the steel railroad rails, a is the coefficient of linear expansion, and ΔT is the temperature rise.
Substituting the given values, we have:
ΔL = 12.5×1.2×10⁻⁵×30
ΔL = 0.0045 m
Therefore, the expansion gap that should be left between the steel railroad rails is 0.0045 m or 4.5 mm. This gap allows the rails to expand without buckling or bending.
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Spaceman Spiff is on a distant planet. He observed a large bird drop a large nut onto a rock to break the shell. The nut has a mass of 6.0 kg. (I told you, it's a large bird and a large nut.) Using his handy-dandy quadricorder, Spiff is able to measure the velocity of the nut to be 19.4 m/s when it hits the ground. If the bird is at a height of 30 meters and air resistance isn't a factor, what is the acceleration due to gravity on this planet? Later, a small bird drops a small nut from the same height. The mass of this nut is 0.75 kg. Now air resistance does work on the nut as it falls. If the work done by the air resistance is 20% of the initial potential energy, what is the speed of the small nut when it hits the ground?
Part 1: The acceleration due to gravity on this planet is approximately 6.27 m/s^2.
Part 2: The speed of the small nut when it hits the ground, taking into account air resistance, is approximately 8.66 m/s.
** Part 1: To calculate the acceleration due to gravity on the distant planet, we can use the equation of motion for free fall:
v^2 = u^2 + 2as
where v is the final velocity (19.4 m/s), u is the initial velocity (0 m/s), a is the acceleration due to gravity, and s is the displacement (30 m).
Rearranging the equation, we have:
a = (v^2 - u^2) / (2s)
a = (19.4^2 - 0^2) / (2 * 30)
a = 376.36 / 60
a ≈ 6.27 m/s^2
Therefore, the acceleration due to gravity on this planet is approximately 6.27 m/s^2.
** Part 2: Considering air resistance, we need to account for the work done by air resistance, which is equal to the change in mechanical energy.
The initial potential energy of the small nut is given by:
PE = mgh
where m is the mass of the nut (0.75 kg), g is the acceleration due to gravity (6.27 m/s^2), and h is the height (30 m).
PE = 0.75 * 6.27 * 30
PE = 141.675 J
Since the work done by air resistance is 20% of the initial potential energy, we can calculate it as:
Work = 0.2 * PE
Work = 0.2 * 141.675
Work = 28.335 J
The work done by air resistance is equal to the change in kinetic energy of the nut:
Work = ΔKE = KE_final - KE_initial
KE_final = KE_initial + Work
Since the initial kinetic energy is 0, the final kinetic energy is equal to the work done by air resistance:
KE_final = 28.335 J
Using the kinetic energy formula:
KE = (1/2)mv^2
v^2 = (2 * KE_final) / m
v^2 = (2 * 28.335) / 0.75
v^2 ≈ 75.12
v ≈ √75.12
v ≈ 8.66 m/s
Therefore, the speed of the small nut when it hits the ground, taking into account air resistance, is approximately 8.66 m/s.
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I want to check the answers
A man pulls a sled along a rough horizontal surface by applying a constant force at an angle above the horizontal. In pulling the sled a horizontal distance d, the work done by the man is: Fd/cos 0 Fd
The work done by the man in pulling the sled a horizontal distance d is Fd/cos θ. Understanding this relationship allows us to calculate the work done in various scenarios involving forces applied at angles relative to the displacement.
When a force is applied at an angle above the horizontal to pull an object, the work done is calculated as the product of the force applied, the displacement of the object, and the cosine of the angle between the force and the displacement vectors.
In this case, the force applied by the man is F, and the displacement of the sled is d. The angle between the force and the displacement vectors is given as θ. Therefore, the work done can be calculated as:
Work = Force × Displacement × cos θ
Substituting the values, we have:
Work = F × d × cos θ
Thus, the work done by the man in pulling the sled a horizontal distance d is Fd/cos θ.
The work done by the man in pulling the sled a horizontal distance d is given by the formula Fd/cos θ, where F is the applied force, d is the displacement, and θ is the angle between the force and the displacement vectors. This formula takes into account the component of the force in the direction of displacement, which is determined by the cosine of the angle. Understanding this relationship allows us to calculate the work done in various scenarios involving forces applied at angles relative to the displacement.
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