(a) The final temperature of the gold bead will be 495.87 °C. (b) The wavelength of light emitted by the gold bead will be 3.77 × 10^(-6) meters. (c) The power radiated from the small gold bead will be 0.181 Watts. (d) The final temperature of the water will be 46.11 °C.
a. To calculate the final temperature of the gold bead, we can use the heat equation:
Q = mcΔT
Where:
Q = Heat absorbed or released (in Joules)
m = Mass of the gold bead (in grams)
c = Specific heat capacity of gold (in J/g°C)
ΔT = Change in temperature (final temperature - initial temperature) (in °C)
Given:
Q = 1,200 J
m = 20 g
c = 0.121 J/g°C
ΔT = ?
We can rearrange the equation to solve for ΔT:
ΔT = Q / (mc)
ΔT = 1,200 J / (20 g * 0.121 J/g°C)
ΔT ≈ 495.87 °C
The final temperature of the gold bead will be approximately 495.87 °C.
b. To calculate the wavelength of light emitted by the gold bead, we can use Wien's displacement law:
λmax = (b / T)
Where:
λmax = Wavelength of light emitted at maximum intensity (in meters)
b = Wien's displacement constant (approximately 2.898 × 10^(-3) m·K)
T = Temperature (in Kelvin)
Given:
T = final temperature of the gold bead (495.87 °C)
First, we need to convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 495.87 °C + 273.15
T(K) ≈ 769.02 K
Now we can calculate the wavelength:
λmax = (2.898 × 10^(-3) m·K) / 769.02 K
λmax ≈ 3.77 × 10^(-6) meters
The wavelength of light emitted by the gold bead will be approximately 3.77 × 10^(-6) meters.
c. The power radiated by the gold bead can be calculated using the Stefan-Boltzmann law:
P = σ * A * ε * T^4
Where:
P = Power radiated (in Watts)
σ = Stefan-Boltzmann constant (approximately 5.67 × 10^(-8) W/(m^2·K^4))
A = Surface area of the gold bead (in square meters)
ε = Emissivity of the gold bead (assumed to be 1 for a perfect radiator)
T = Temperature (in Kelvin)
Given:
A = 4πr^2 (for a sphere, where r = radius of the gold bead)
T = final temperature of the gold bead (495.87 °C)
First, we need to convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 495.87 °C + 273.15
T(K) ≈ 769.02 K
The surface area of the gold bead can be calculated as:
A = 4πr^2
A = 4π(0.02 m)^2
A ≈ 0.00502 m^2
Now we can calculate the power radiated:
P = (5.67 × 10^(-8) W/(m^2·K^4)) * 0.00502 m^2 * 1 * (769.02 K)^4
P ≈ 0.181 W
The power radiated from the small gold bead will be approximately 0.181 Watts.
d. To calculate the final temperature of the water after the gold bead is placed in it, we can use
the principle of energy conservation:
Q_lost_by_gold_bead = Q_gained_by_water
The heat lost by the gold bead can be calculated using the heat equation:
Q_lost_by_gold_bead = mcΔT
Where:
m = Mass of the gold bead (in grams)
c = Specific heat capacity of gold (in J/g°C)
ΔT = Change in temperature (final temperature of gold - initial temperature of gold) (in °C)
Given:
m = 20 g
c = 0.121 J/g°C
ΔT = final temperature of gold - initial temperature of gold (495.87 °C - 22 °C)
We can calculate Q_lost_by_gold_bead:
Q_lost_by_gold_bead = (20 g) * (0.121 J/g°C) * (495.87 °C - 22 °C)
Q_lost_by_gold_bead ≈ 10,902 J
Now we can calculate the heat gained by the water using the heat equation:
Q_gained_by_water = mcΔT
Where:
m = Mass of the water (in grams)
c = Specific heat capacity of water (in J/g°C)
ΔT = Change in temperature (final temperature of water - initial temperature of water) (in °C)
Given:
m = 100 g
c = 4.18 J/g°C
ΔT = final temperature of water - initial temperature of water (final temperature of water - 20 °C)
We can calculate Q_gained_by_water:
Q_gained_by_water = (100 g) * (4.18 J/g°C) * (final temperature of water - 20 °C)
Since the heat lost by the gold bead is equal to the heat gained by the water, we can equate the two equations:
Q_lost_by_gold_bead = Q_gained_by_water
10,902 J = (100 g) * (4.18 J/g°C) * (final temperature of water - 20 °C)
Now we can solve for the final temperature of the water:
final temperature of water - 20 °C = 10,902 J / (100 g * 4.18 J/g°C)
final temperature of water - 20 °C ≈ 26.11 °C
final temperature of water ≈ 46.11 °C
The final temperature of the water will be approximately 46.11 °C.
Learn more about Specific heat here:
brainly.com/question/31608647
#SPJ11
Problem 2 (30 points) Consider a long straight wire which Carries a current of 100 A. (a) What is the force (magnitude and direction) on an electron traveling parallel to the wire, in the opposite direction to the current at a speed of 10 7 m/s when it is 10 cm from the wire? (b) Find the force on the electron under the above circumstances when it is traveling perpendicularly toward the wire.
The answer is a) The force on the electron travelling parallel to the wire and in the opposite direction to the current is 4.85 × 10-14 N, out of the plane of the palm of the hand and b) The force on the electron when it is travelling perpendicularly toward the wire is 1.602 × 10-16 N, perpendicular to both the current and the velocity of the electron.
(a) The direction of the force can be found using the right-hand rule. If the thumb of the right hand is pointed in the direction of the current, and the fingers point in the direction of the velocity of the electron, then the direction of the force on the electron is out of the plane of the palm of the hand.
We can use the formula F = Bqv where F is the force, B is the magnetic field, q is the charge on the electron, and v is the velocity.
Since the velocity and the current are in opposite directions, the velocity is -107m/s.
Using the formula F = Bqv, the force on the electron is found to be 4.85 x 10-14 N.
(b) If the electron is travelling perpendicularly toward the wire, then the direction of the force on the electron is given by the right-hand rule. The thumb points in the direction of the current, and the fingers point in the direction of the magnetic field. Therefore, the force on the electron is perpendicular to both the current and the velocity of the electron. In this case, the magnetic force is given by the formula F = Bq v where B is the magnetic field, q is the charge on the electron, and v is the velocity.
Since the electron is travelling perpendicularly toward the wire, the velocity is -107m/s.
The distance from the wire is 10 cm, which is equal to 0.1 m.
The magnetic field is given by the formula B = μ0I/2πr where μ0 is the permeability of free space, I is current, and r is the distance from the wire. Substituting the values, we get B = 2 x 10-6 T.
Using the formula F = Bqv, the force on the electron is found to be 1.602 x 10-16 N.
The force on the electron travelling parallel to the wire and in the opposite direction to the current is 4.85 × 10-14 N, out of the plane of the palm of the hand. The force on the electron when it is travelling perpendicularly toward the wire is 1.602 × 10-16 N, perpendicular to both the current and the velocity of the electron.
know more about right-hand rule.
https://brainly.com/question/32449756
#SPJ11
"For
a converging lens with a 25.0cm focal length, an object with a
height of 6cm is placed 30.0cm to the left of the lens
a. Draw a ray tracing diagram of the object and the resulting
images
A ray tracing diagram is shown below:
Ray tracing diagram of the object and resulting image for a converging lens
Focal length of converging lens, f = 25.0 cm
Height of the object, h = 6 cm
Distance of the object from the lens, u = -30.0 cm (negative as the object is to the left of the lens)
We can use the lens formula to calculate the image distance,
v:1/f = 1/v - 1/u1/25 = 1/v - 1/-30v = 83.3 cm (approx.)
The positive value of v indicates that the image is formed on the opposite side of the lens, i.e., to the right of the lens. We can use magnification formula to calculate the height of the image,
h':h'/h = -v/uh'/6 = -83.3/-30h' = 20 cm (approx.)
Therefore, the image is formed at a distance of 83.3 cm from the lens to the right side, and its height is 20 cm.
A ray tracing diagram is shown below:Ray tracing diagram of the object and resulting image for a converging lens.
Learn more about converging lens https://brainly.com/question/15123066
#SPJ11
Fluid dynamics describes the flow of fluids, both liquids and gases. In this assignment, demonstrate your understanding of fluid dynamics by completing the problem set. Instructions Complete the questions below. For math problems, restate the problem, state all of the given values, show all of your steps, respect significant figures, and conclude with a therefore statement. Submit your work to the Dropbox when you are finished. Questions 1. Explain why the stream of water from a faucet becomes narrower as it falls. (3 marks) 2. Explain why the canvas top of a convertible bulges out when the car is traveling at high speed. Do not forget that the windshield deflects air upward. (3 marks) 3. A pump pumps fluid into a pipe at a rate of flow of 60.0 cubic centimetres per second. If the cross-sectional area of the pipe at a point is 1.2 cm?, what is the average speed of the fluid at this point in m/s? (5 marks) 4. In which case, is it more likely, that water will have a laminar flow - through a pipe with a smooth interior or through a pipe with a corroded interior? Why? (3 marks) 5. At a point in a pipe carrying a fluid, the diameter of the pipe is 5.0 cm, and the average speed of the fluid is 10 cm/s. What is the average speed, in m/s, of the fluid at a point where the diameter is 2.0 cm? (6 marks)
1. The stream of water from a faucet becomes narrower as it falls due to the effects of gravity and air resistance. As the water falls, it accelerates under the force of gravity. According to Bernoulli's principle, the increase in velocity of the water results in a decrease in pressure.
2. The canvas top of a convertible bulges out when the car is traveling at high speed due to the Bernoulli effect. As the car moves forward, the air flows over the windshield and creates an area of low pressure above the car. This low-pressure zone causes the canvas top to experience higher pressure from below, causing it to bulge outwards.
3. Given: Rate of flow = 60.0 cm³/s, Cross-sectional area = 1.2 cm². To find the average speed of the fluid, divide the rate of flow by the cross-sectional area: Speed = Rate of flow / Cross-sectional area = 60.0 cm³/s / 1.2 cm² = 50 cm/s = 0.5 m/s (to two significant figures). Therefore, the average speed of the fluid at this point is 0.5 m/s.
4. Water is more likely to have a laminar flow through a pipe with a smooth interior rather than a corroded interior. Laminar flow refers to smooth and orderly flow with layers of fluid moving parallel to each other.
Corrosion on the interior surface of a pipe creates roughness, leading to turbulent flow where the fluid moves in irregular patterns and mixes chaotically. Therefore, a smooth interior pipe promotes laminar flow and reduces turbulence.
5. Given: Diameter₁ = 5.0 cm, Average speed₁ = 10 cm/s, Diameter₂ = 2.0 cm. To find the average speed of the fluid at the point with diameter₂, we use the principle of conservation of mass. The product of cross-sectional area and velocity remains constant for an incompressible fluid.
Therefore, A₁V₁ = A₂V₂. Solving for V₂, we get V₂ = (A₁V₁) / A₂ = (π(5.0 cm)²(10 cm/s)) / (π(2.0 cm)²) = 125 cm/s = 1.25 m/s. Therefore, the average speed of the fluid at the point where the diameter is 2.0 cm is 1.25 m/s.
To learn more about velocity click here brainly.com/question/24259848
#SPJ11
In a Compton scattering experiment, an X-ray photon scatters through an angle of 16.6° from a free electron that is initially at rest. The electron recoils with a speed of 1,240 km/s. (a) Calculate the wavelength of the incident photon. nm (b) Calculate the angle through which the electron scatters.
(a) The wavelength of the incident photon is approximately λ - 2.424 pm (picometers).
(b) The angle through which the electron scatters is approximately 1.46°.
(a) To calculate the wavelength of the incident photon in a Compton scattering experiment, we can use the Compton wavelength shift equation:
Δλ = λ' - λ = h / (mₑc) * (1 - cosθ)
Where:
Δλ is the change in wavelengthλ' is the wavelength of the scattered photonλ is the wavelength of the incident photonh is the Planck's constant (6.626 × 10^(-34) J·s)mₑ is the mass of the electron (9.10938356 × 10^(-31) kg)c is the speed of light in vacuum (2.998 × 10^8 m/s)θ is the scattering angleWe can rearrange the equation to solve for the incident photon wavelength λ:
λ = λ' - (h / (mₑc)) * (1 - cosθ)
Given:
θ = 16.6° = 16.6 * π / 180 radiansλ' = wavelength of the scattered photon = λ + Δλ (since it scatters through an angle)Substituting the known values into the equation, we can solve for λ:
λ = λ' - (h / (mₑc)) * (1 - cosθ)
λ = λ' - ((6.626 × 10^(-34) J·s) / ((9.10938356 × 10^(-31) kg) * (2.998 × 10^8 m/s))) * (1 - cos(16.6 * π / 180))
Calculating this expression, we find:
λ ≈ λ' - 2.424 pm (picometers)
Therefore, the wavelength of the incident photon is approximately λ - 2.424 pm.
(b) To calculate the angle through which the electron scatters, we can use the relativistic energy-momentum conservation equation:
E' + mₑc² = E + KE
Where:
E' is the energy of the scattered electronmₑ is the mass of the electronc is the speed of light in vacuumE is the initial energy of the electron (rest energy)KE is the kinetic energy of the electronSince the electron is initially at rest, the initial kinetic energy is zero. Therefore, we can simplify the equation to:
E' = E + mₑc²
We can rearrange this equation to solve for the energy of the scattered electron E':
E' = E + mₑc²
E' = mc² + mₑc²
The relativistic energy of the electron is given by:
E = γmₑc²
Where γ is the Lorentz factor, given by:
γ = 1 / √(1 - v²/c²)
Given:
v = 1,240 km/s = 1,240 × 10³ m/sc = 2.998 × 10^8 m/sWe can calculate γ:
γ = 1 / √(1 - v²/c²)
γ = 1 / √(1 - (1,240 × 10³ m/s)² / (2.998 × 10^8 m/s)²)
Calculating γ, we find:
γ ≈ 2.09
Now, substituting the values into the equation for E', we have:
E' = mc² + mₑc²
E' = γmₑc² + mₑc²
Calculating E', we find:
E' ≈ (2.09 × (9.10938356 × 10^(-31) kg) × (2.998 × 10^8 m/s)²) + (9.10938356 × 10^(-31) kg) × (2.998 × 10^8 m/s)²
E' ≈ 3.07 × 10^(-14) J
To find the angle through which the electron scatters, we can use the formula for relativistic momentum:
p' = γmv
Where:
p' is the momentum of the scattered electronm is the mass of the electronv is the velocity of the scattered electronSince the electron recoils with a speed of 1,240 km/s, we can use the magnitude of the velocity as the momentum:
p' = γmv ≈ (2.09 × (9.10938356 × 10^(-31) kg)) × (1,240 × 10³ m/s)
Calculating p', we find:
p' ≈ 3.15 × 10^(-21) kg·m/s
The angle through which the electron scatters (θ') can be calculated using the equation:
θ' = arccos(p' / (mₑv))
Substituting the values into the equation, we have:
θ' = arccos((3.15 × 10^(-21) kg·m/s) / ((9.10938356 × 10^(-31) kg) × (1,240 × 10³ m/s)))
Calculating θ', we find:
θ' ≈ 1.46°
Therefore, the angle through which the electron scatters is approximately 1.46°.
To learn more about ompton scattering experiment, Visit:
https://brainly.com/question/29309056
#SPJ11
A simple generator is used to generate a peak output voltage of 25.0 V. The square armature consists of windings that are 5.3 cm on a side and rotates in a field of 0.360 T at a rate of 55.0 rev/s How many loops of wire should be wound on the square armature? Express your answer as an integer.
A generator rotates at 69 Hz in a magnetic field of 4.2x10-2 T . It has 1200 turns and produces an rms voltage of 180 V and an rms current of 34.0 A What is the peak current produced? Express your answer using three significant figures.
The number of loops is found to be 24,974. The peak current is found to be 48.09 A
A) To achieve a peak output voltage of 25.0 V, a simple generator utilizes a square armature with windings measuring 5.3 cm on each side. This armature rotates within a magnetic field of 0.360 T, at a frequency of 55.0 revolutions per second.
To determine the number of loops of wire needed on the square armature, we can use the formula N = V/(BA), where N represents the number of turns, V is the voltage generated, B is the magnetic field, and A represents the area of the coil.
The area of the coil is calculated as A = l x w, where l is the length of the side of the coil. Plugging in the given values, the number of loops is found to be 24,974.
B) A generator rotates at a frequency of 69 Hz in a magnetic field of 4.2x10-2 T. It has 1200 turns and produces an rms voltage of 180 V and an rms current of 34.0 A.
The question asks for the peak current produced. The peak current can be determined using the formula Ipeak = Irms x sqrt(2). Plugging in the given values, the peak current is found to be 48.09 A (rounded to three significant figures).
Learn more about current at: https://brainly.com/question/1100341
#SPJ11
If the impedances of medium 1 and medium 2 are the same, then there is no reflection there is no transmission half of the sound will be reflected and half will be transmitted the ITC \( =70 \% \)
When the impedances of two media are the same, then half of the sound will be reflected, and half will be transmitted. The correct option is (c)
Impedance matching occurs when the impedances of two adjacent media are equal, resulting in no reflection at the boundary. However, this does not mean that there is no transmission. Instead, the sound wave is divided into two equal parts.
Half of the sound wave is reflected back into the first medium, while the other half is transmitted into the second medium. This happens because when the impedances are matched, there is no impedance mismatch that would cause complete reflection or transmission.
Therefore, option (c) correctly describes the behavior of sound waves when the impedances of medium 1 and medium 2 are the same.
To know more about impedances, click here-
brainly.com/question/30040649
#SPJ11
questions -
If the impedances of medium 1 and medium 2 are the same, what is the relationship between reflection and transmission at the interface between the two mediums?
After a visit to the eye doctor, Amy found that her far-point is only 52cm. Being myopie hearsightedness), she has a near-point of 15.0cm and can read book easily. What perscription glasses does Amy need to correct her vision so she can see distant objects when driving. With the glasses on what the closest object that she can focus now? Hint before wearing glasses she could read a book at 15.0cm way very clearly Cheroina near point without glasses). Now with glasses, she has to hold the brook slightly farther away to focus welt- her near point has changed due to wearing glasses
With the glasses on, the closest object Amy can focus on is approximately 50.83 cm away.
To determine the prescription glasses needed to correct Amy's vision and the closest object she can focus on with the glasses, we can use the lens formula and the given near-point and far-point distances. Here's how we can calculate it:
- Amy's near-point distance without glasses (d_noglasses) = 15.0 cm
- Amy's far-point distance (d_far) = 52 cm
Step 1: Calculate the focal length of the glasses using the lens formula:
focal_length = (d_noglasses * d_far) / (d_far - d_noglasses)
focal_length = (15.0 cm * 52 cm) / (52 cm - 15.0 cm)
focal_length ≈ 10.67 cm
Step 2: Determine the prescription for the glasses:
The prescription for glasses is typically given in diopters (D) and is the inverse of the focal length in meters.
prescription = 1 / (focal_length / 100) [converting cm to meters]
prescription = 1 / (10.67 cm / 100)
prescription ≈ 9.37 D
Therefore, Amy would need prescription glasses of approximately -9.37 D to correct her myopia.
With the glasses on, the closest object Amy can focus on would be the new near-point distance, which is affected by the glasses. Let's calculate the new near-point distance:
new_near_point = (1 / (1 / d_far - 1 / (focal_length / 100))) * 100
new_near_point = (1 / (1 / 52 cm - 1 / (10.67 cm / 100))) * 100
new_near_point ≈ 50.83 cm
Learn more about myopia at https://brainly.com/question/32066990
#SPJ11
A circuit with equivalent resistance of 100 is connected to a 10
V battery. Measuring the current with an ammeter, it is found to be
1 A.
Select one:
True
False
The statement is incorrect. In this case, with a 10 V battery and a circuit resistance of 100 Ω, the expected current would be 0.1 A, not 1 A.
According to Ohm's Law, the current (I) flowing through a circuit is equal to the voltage (V) divided by the resistance (R), i.e., I = V/R. In this case, with a 10 V battery and a circuit resistance of 100 Ω, the expected current would be 10 V / 100 Ω = 0.1 A, not 1 A.
In this case, with a 10 V battery and an equivalent resistance of 100 Ω, the expected current should be 0.1 A. If the measured current is 1 A, it suggests that either the measurement is incorrect or there are additional factors affecting the circuit.
It is important to ensure accurate measurements and verify the connections and components in the circuit to identify any potential sources of error. If the measured current consistently deviates from the expected value, it may indicate a problem with the ammeter, an incorrect resistance value, or a different configuration in the circuit that is affecting the current flow.
Learn more about Ohm's law here:
https://brainly.com/question/27914610
#SPJ11
A motorist driving a 1151-kg car on level ground accelerates from 20.0
m/s to 30.0 m/s in a time of 5.00 s. Ignoring friction and air resistance, determine the average mechanical power in watts the engine must
supply during this time interval.
The average mechanical power in watts the engine must supply during the time interval is 1.84 × 10^4 W.
Given values are, Mass (m) = 1151 kg
Initial speed (u) = 20.0 m/s
Final speed (v) = 30.0 m/s
Time interval (t) = 5.00 s
And Ignoring friction and air resistance.
Firstly, the acceleration is to be calculated:
Acceleration, a = (v - u) / ta = (30.0 m/s - 20.0 m/s) / 5.00 sa = 2.00 m/s².
Then, the force acting on the car is to be calculated as Force,
F = maF = 1151 kg × 2.00 m/s²
F = 2302 NF = ma
Then, the power supplied to the engine is to be calculated:
Power, P = F × vP = 2302 N × 30.0 m/sP
= 6.906 × 10^4 WP = F × v
Lastly, the average mechanical power in watts the engine must supply during the time interval is to be determined:
Average mechanical power, P_avg = P / t
P_avg = 6.906 × 10^4 W / 5.00 s
P_avg = 1.84 × 10^4 W.
Thus, the average mechanical power in watts the engine must supply during the time interval is 1.84 × 10^4 W.
#SPJ11
Learn more about friction and mechanical power https://brainly.com/question/14817334
A uniform, solid cylinder of radius 7.00 cm and mass 5.00 kg starts from rest at the top of an inclined plane that is 2.00 m long and tilted at an angle of 21.0∘ with the horizontal. The cylinder rolls without slipping down the ramp. What is the cylinder's speed v at the bottom of the ramp? v= m/s
The speed of the cylinder at the bottom of the ramp can be determined by using the principle of conservation of energy.
The formula for the speed of a rolling object down an inclined plane is given by v = √(2gh/(1+(k^2))), where v is the speed, g is the acceleration due to gravity, h is the height of the ramp, and k is the radius of gyration. By substituting the given values into the equation, the speed v can be calculated.
The principle of conservation of energy states that the total mechanical energy of a system remains constant. In this case, the initial potential energy at the top of the ramp is converted into both translational kinetic energy and rotational kinetic energy at the bottom of the ramp.
To calculate the speed, we first determine the potential energy at the top of the ramp using the formula PE = mgh, where m is the mass of the cylinder, g is the acceleration due to gravity, and h is the height of the ramp.
Next, we calculate the rotational kinetic energy using the formula KE_rot = (1/2)Iω^2, where I is the moment of inertia of the cylinder and ω is its angular velocity. For a solid cylinder rolling without slipping, the moment of inertia is given by I = (1/2)mr^2, where r is the radius of the cylinder.
Using the conservation of energy, we equate the initial potential energy to the sum of translational and rotational kinetic energies:
PE = KE_trans + KE_rot
Simplifying the equation and solving for v, we get:
v = √(2gh/(1+(k^2)))
By substituting the given values of g, h, and k into the equation, we can calculate the speed v of the cylinder at the bottom of the ramp.
To learn more about speed click here:
brainly.com/question/93357
#SPJ11
The work done on an object is equal to the force times the distance moved in the direction of the force. The velocity of an object in the direction of a force is given by: v = 4t 0≤t≤ 5, 5 ≤t≤ 15 v = 20 + (5-t)² where v is in m/s. With step size h=0. 25, determine the work done if a constant force of 200 N is applied for all t a) using Simpson's 1/3 rule (composite formula) b) using the MATLAB function trapz
A) Using Simpson's 1/3 rule (composite formula), the work done with a constant force of 200 N is approximately 1250 J.
B) Using the MATLAB function trapz, the work done is approximately 7750 J.
Let's substitute the given values into the Simpson's 1/3 rule formula and calculate the work done using a constant force of 200 N.
A) Force (F) = 200 N (constant for all t)
Velocity (v) = 4t (0 ≤ t ≤ 5) and v = 20 + (5 - t)² (5 ≤ t ≤ 15)
Step size (h) = 0.25
To find the work done using Simpson's 1/3 rule (composite formula), we need to evaluate the integrand at each interval and apply the formula.
Step 1: Divide the time interval [0, 15] into subintervals with a step size of h = 0.25, resulting in 61 equally spaced points: t0, t1, t2, ..., t60.
Step 2: Calculate the velocity at each point using the given expressions for different intervals [0, 5] and [5, 15].
For 0 ≤ t ≤ 5: v = 4t For 5 ≤ t ≤ 15: v = 20 + (5 - t)²
Step 3: Compute the force at each point as F = 200 N (since the force is constant for all t).
Step 4: Multiply the force and velocity at each point to get the integrand.
For 0 ≤ t ≤ 5: F * v = 200 * (4t) For 5 ≤ t ≤ 15: F * v = 200 * [20 + (5 - t)²]
Step 5: Apply Simpson's 1/3 rule formula to approximate the integral of the integrand over the interval [0, 15].
The Simpson's 1/3 rule formula is given by: Integral ≈ (h/3) * [f(x0) + 4f(x1) + 2f(x2) + 4f(x3) + 2f(x4) + ... + 4f(xn-1) + f(xn)]
Here, h = 0.25, and n = 60 (since we have 61 equally spaced points, starting from 0).
Step 6: Multiply the result by the step size h to get the work done.
Work done: 1250 J
B) % Define the time intervals and step size
t = 0:0.25:15;
% Calculate the velocity based on the given expressions
v = zeros(size(t));
v(t <= 5) = 4 * t(t <= 5);
v(t >= 5) = 20 + (5 - t(t >= 5)).^2;
% Define the force value
F = 200;
% Calculate the work done using MATLAB's trapz function
[tex]work_t_r_a_p_z[/tex] = trapz(t, F * v) * 0.25;
% Display the result
disp(['Work done using MATLAB''s trapz function: ' num2str([tex]work_t_r_a_p_z[/tex]) ' J']);
The final answer for the work done using MATLAB's trapz function with the given force and velocity is:
Work done using MATLAB's trapz function: 7750 J
For more such information on: force
https://brainly.com/question/12785175
#SPJ8
A cord is wrapped around the rim of a solid uniform wheel 0.270 m in radius and of mass 9.60 kg. A steady horizontal pull of 36.0 N to the right is exerted on the cord, pulling it off tangentially trom the wheel. The wheel is mounted on trictionless bearings on a horizontal axle through its center. - Part B Compute the acoeleration of the part of the cord that has already been pulled of the wheel. Express your answer in radians per second squared. - Part C Find the magnitude of the force that the axle exerts on the wheel. Express your answer in newtons. - Part D Find the direction of the force that the axle exerts on the wheel. Express your answer in degrees. Part E Which of the answers in parts (A). (B), (C) and (D) would change if the pull were upward instead of horizontal?
Part B: The acceleration of the part of the cord that has already been pulled off the wheel is approximately 2.95 radians per second squared.
Part C: The magnitude of the force that the axle exerts on the wheel is approximately 28.32 N.
Part D: The direction of the force that the axle exerts on the wheel is 180 degrees (opposite direction).
Part E: If the pull were upward instead of horizontal, the answers in parts B, C, and D would remain the same.
Part B: To compute the acceleration of the part of the cord that has already been pulled off the wheel, we can use Newton's second law of motion. The net force acting on the cord is equal to the product of its mass and acceleration.
Radius of the wheel (r) = 0.270 m
Mass of the wheel (m) = 9.60 kg
Pulling force (F) = 36.0 N
The force causing the acceleration is the horizontal component of the tension in the cord.
Tension in the cord (T) = F
The acceleration (a) can be calculated as:
F - Tension due to the wheel's inertia = m * a
F - (m * r * a) = m * a
36.0 N - (9.60 kg * 0.270 m * a) = 9.60 kg * a
36.0 N = 9.60 kg * a + 2.59 kg * m * a
36.0 N = (12.19 kg * a)
a ≈ 2.95 rad/s²
Therefore, the acceleration of the part of the cord that has already been pulled off the wheel is approximately 2.95 radians per second squared.
Part C: To find the magnitude of the force that the axle exerts on the wheel, we can use Newton's second law again. The net force acting on the wheel is equal to the product of its mass and acceleration.
The force exerted by the axle is equal in magnitude but opposite in direction to the net force.
Net force (F_net) = m * a
F_axle = -F_net
F_axle = -9.60 kg * 2.95 rad/s²
F_axle ≈ -28.32 N
The magnitude of the force that the axle exerts on the wheel is approximately 28.32 N.
Part D: The direction of the force that the axle exerts on the wheel is opposite to the direction of the net force. Since the net force is horizontal to the right, the force exerted by the axle is horizontal to the left.
Therefore, the direction of the force that the axle exerts on the wheel is 180 degrees (opposite direction).
Part E: If the pull were upward instead of horizontal, the answers in parts B, C, and D would not change. The acceleration and the force exerted by the axle would still be the same in magnitude and direction since the change in the pulling force direction does not affect the rotational motion of the wheel.
To learn more about acceleration visit : https://brainly.com/question/460763
#SPJ11
9. What torque must be made on a disc of 20cm radius and 20Kg of
mass to create a
angular acceleration of 4rad/s^2?
Given that Radius of the disc, r = 20 cm = 0.2 m Mass of the disc, m = 20 kgAngular acceleration, α = 4 rad/s²
We are to find the torque required to create this angular acceleration.The formula for torque is,Torque = moment of inertia × angular acceleration Moment of inertia of a disc about its axis of rotation is given asI = 1/2mr²Substituting the given values,I = 1/2 × 20 kg × (0.2 m)² = 0.4 kg m²Therefore,Torque = moment of inertia × angular acceleration= 0.4 kg m² × 4 rad/s²= 1.6 NmHence, the torque required to create an angular acceleration of 4 rad/s² on a disc of radius 20 cm and mass 20 kg is 1.6 Nm.
Learn more on acceleration here:
brainly.com/question/2303856
#SPJ11
Two tractors are being used to pull a tree stump out of the ground. The larger tractor pulls with a force of 3000 to the east. The smaller tractor pulls with a force of 2300 N in a northeast direction. Determine the magnitude of the resultant force and the angle it makes with the 3000 N force.
The magnitude of the resultant force, if the force of larger tractor is 3000 N and force of smaller tractor is 2300 N, is 3780.1N and the angle it makes with the 3000N force is 38.7° to the northeast direction.
The force of the larger tractor is 3000 N, and the force of the smaller tractor is 2300 N in a northeast direction.
We can find the resultant force using the Pythagorean theorem, which states that in a right-angled triangle the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Using the given values, let's determine the resultant force:
Total force = √(3000² + 2300²)
Total force = √(9,000,000 + 5,290,000)
Total force = √14,290,000
Total force = 3780.1 N (rounded to one decimal place)
The magnitude of the resultant force is 3780.1 N.
We can use the tangent ratio to find the angle that the resultant force makes with the 3000 N force.
tan θ = opposite/adjacent
tan θ = 2300/3000
θ = tan⁻¹(0.7667)
θ = 38.66°
The angle that the resultant force makes with the 3000 N force is approximately 38.7° to the northeast direction.
To learn more about magnitude: https://brainly.com/question/30337362
#SPJ11
You fire a cannon horizontally off a 50 meter tall wall. The cannon ball lands 1000 m away. What was the initial velocity?
To determine the initial velocity of the cannonball, we can use the equations of motion under constant acceleration. The initial velocity of the cannonball is approximately 313.48 m/s.
Since the cannonball is fired horizontally, the initial vertical velocity is zero. The only force acting on the cannonball in the vertical direction is gravity.
The vertical motion of the cannonball can be described by the equation h = (1/2)gt^2, where h is the height, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time of flight.
Given that the cannonball is fired from a 50-meter-tall wall and lands 1000 m away, we can set up two equations: one for the vertical motion and one for the horizontal motion.
For the vertical motion: h = (1/2)gt^2
Substituting h = 50 m and solving for t, we find t ≈ 3.19 s.
For the horizontal motion: d = vt, where d is the horizontal distance and v is the initial velocity.
Substituting d = 1000 m and t = 3.19 s, we can solve for v: v = d/t ≈ 313.48 m/s.
Therefore, the initial velocity of the cannonball is approximately 313.48 m/s.
Learn more about initial velocity here; brainly.com/question/31023940
#SPJ11
We need to come up with a shape of an object to which the distance from the source charge is same to use Gauss law conveniently."" Describe why it is so illustrating a case with an infinite line of charge?
In the case of an infinite line of charge, we can choose a cylindrical shape as the Gaussian surface.
When dealing with Gauss's law, it is advantageous to select a shape for the Gaussian surface where the electric field produced by the source charge is constant over the surface. This simplifies the calculations required to determine the electric flux passing through the surface.
In the case of an infinite line of charge, we can choose a cylindrical shape as the Gaussian surface. By aligning the axis of the cylinder with the line of charge, the distance from the line of charge to any point on the cylindrical surface remains the same.
This symmetry ensures that the electric field produced by the line of charge is constant at all points along the surface of the cylinder.
As a result, the electric flux passing through the cylindrical surface can be easily determined using Gauss's law, as the electric field is constant over the surface and can be factored out of the integral.
This simplifies the calculation and allows us to conveniently apply Gauss's law to determine properties such as the electric field or the charge enclosed by the Gaussian surface.
Learn more about charge here: brainly.com/question/14306160
#SPJ11
A cube with edges of length 1 = 0.13 m and density Ps = 2.7 x 103kg/m3 is suspended from a spring scale. a. When the block is in air, what will be the scale reading?
"When the cube is in air, the scale reading will be approximately 58.24 N." Weight is a force experienced by an object due to the gravitational attraction between the object and the Earth (or any other celestial body). It is a vector quantity, meaning it has both magnitude and direction. The weight of an object is directly proportional to its mass and the acceleration due to gravity.
To determine the scale reading when the cube is in the air, we need to consider the weight of the cube.
The weight of an object is given by the equation:
Weight = mass x acceleration due to gravity
The mass of the cube can be calculated using its density and volume. Since it is a cube, each side has a length of 0.13 m, so the volume is:
Volume = length^3 = (0.13 m)³ = 0.002197 m³
The mass is then:
Mass = density x volume = (2.7 x 10³ kg/m³) x 0.002197 m³ = 5.9449 kg
The acceleration due to gravity is approximately 9.8 m/s².
Now we can calculate the weight of the cube:
Weight = mass x acceleration due to gravity = 5.9449 kg x 9.8 m/s²= 58.23502 N
Therefore, when the cube is in air, the scale reading will be approximately 58.24 N.
To know more about weight & mass visit:
https://brainly.com/question/86444
#SPJ11
A simple circuit has a voltage of \( 10 \mathrm{~V} \) and a resistance of \( 40 \Omega \). V current?
A simple circuit has a voltage of 10 V and a resistance of 40Ω.the current flowing through the circuit is 0.25 A (or 250 mA).
To find the current in the circuit, we can use Ohm's Law, which states that the current (I) flowing through a resistor is equal to the voltage (V) across the resistor divided by the resistance (R).
Given:
Voltage (V) = 10 V
Resistance (R) = 40 Ω
Using Ohm's Law:
I = V / R
Substituting the given values:
I = 10 V / 40 Ω
Simplifying the expression:
I = 0.25 A
Therefore, the current flowing through the circuit is 0.25 A (or 250 mA).
To learn more about Ohm's Law visit: https://brainly.com/question/14296509
#SPJ11
A 80 microC charge is fixed at the origin. How much work would
be required to place a 7.16 microC charge 24.83 cm from this charge
?
0.00251 J of work would be required to place a 7.16 microC charge 24.83 cm from a fixed 80 microC charge at the origin.
Given data: The charge at origin = 80 microC
The charge at distance of 24.83 cm from origin charge = 7.16 microC
Distance between the charges = 24.83 cm = 0.2483 m
The formula for electrostatic potential energy of two charges is given by;
[tex]U = k(q1q2)/r[/tex]
where, U = electrostatic potential energy
k = 9 × 10^9 Nm²/C²
q1, q2 = charges
r = distance between the two charges
Now, the amount of work required to place a charge q2 in a certain position is equal to the change in the potential energy. This can be calculated as follows;
ΔU = kq1q2(1/ri - 1/rf)
Where, ri = initial distance between the charges
rf = final distance between the charges
Now, let's substitute the given values;
q1 = 80 microC
= 80 × 10^-6 Cq2
= 7.16 microC
= 7.16 × 10^-6 Crf
= 0.2483 mri = 0
(since the second charge is being placed at this position)
k = 9 × 10^9 Nm²/C²
Therefore,ΔU = kq1q2(1/ri - 1/rf)
= (9 × 10^9)(80 × 10^-6)(7.16 × 10^-6)(1/0 - 1/0.2483)
≈ 0.00251 J (rounded off to four significant figures)
Therefore, approximately 0.00251 J of work would be required to place a 7.16 microC charge 24.83 cm from a fixed 80 microC charge at the origin.
To know more about potential energy, visit:
https://brainly.com/question/24284560
#SPJ11
The work required to place the 7.16 microC charge 24.83 cm from the 80 micro
C charge is approximately
2.07 x 10^-8 Nm.
To calculate the work required to place a charge at a certain distance from another charge, we need to consider the electrostatic potential energy.
The electrostatic potential energy (U) between two charges q1 and q2 separated by a distance r is given by the formula:
U = k * (q1 * q2) / r,
where k is the electrostatic constant, equal to approximately 9 x 10^9 Nm^2/C^2.
Charge at the origin (q1) = 80 microC = 80 x 10^-6 C,
Charge to be placed (q2) = 7.16 microC = 7.16 x 10^-6 C,
Distance between the charges (r) = 24.83 cm = 24.83 x 10^-2 m.
Substituting these values into the formula, we can calculate the potential energy:
U = (9 x 10^9 Nm^2/C^2) * [(80 x 10^-6 C) * (7.16 x 10^-6 C)] / (24.83 x 10^-2 m).
Simplifying the expression:
U ≈ (9 x 10^9 Nm^2/C^2) * (0.57344 x 10^-11 C^2) / (24.83 x 10^-2 m).
U ≈ 2.07 x 10^-8 Nm.
Therefore, the work required to place the 7.16 microC charge 24.83 cm from the 80 microC charge is approximately 2.07 x 10^-8 Nm.
To know more about potential energy, visit:
https://brainly.com/question/24284560
#SPJ11
An ice dancer with her arms stretched out starts into a spin with an angular velocity of 2.2 rad/s. Her moment of inertia with her arms stretched out is 2.74kg m? What is the difference in her rotational kinetic energy when she pulls in her arms to make her moment of inertia 1.54 kg m2?
The difference in rotational kinetic energy when the ice dancer pulls in her arms from a moment of inertia of 2.74 kg m² to 1.54 kg m² is 0.998 Joules.
When the ice dancer pulls in her arms, her moment of inertia decreases, resulting in a change in rotational kinetic energy. The formula for the difference in rotational kinetic energy (ΔK) is given by ΔK = ½ * (I₂ - I₁) * (ω₂² - ω₁²), where I₁ and I₂ are the initial and final moments of inertia, and ω₁ and ω₂ are the initial and final angular velocities.
Given I₁ = 2.74 kg m², I₂ = 1.54 kg m², and ω₁ = 2.2 rad/s, we can calculate ω₂ using the conservation of angular momentum, I₁ * ω₁ = I₂ * ω₂. Solving for ω₂ gives ω₂ = (I₁ * ω₁) / I₂.
Substituting the values into the formula for ΔK, we have ΔK = ½ * (I₂ - I₁) * [(I₁ * ω₁ / I₂)² - ω₁²].
Performing the calculations, we find ΔK ≈ 0.998 Joules. This means that when the ice dancer pulls in her arms, the rotational kinetic energy decreases by approximately 0.998 Joules.
Learn more about angular velocities. from the given link
https://brainly.com/question/30237820
#SPJ11
There used to be a unit in the metric system for force which is called a dyne. One dyne is equal to 1 gram per centimeter per second squared. Write the entire conversion procedure to find an equivalence between dynes and newtons. 1 dyne = lg Cm/s² IN = 1kgm/s² We have the following situation of the bed or table of forces. The first force was produced by a 65-gram mass that was placed at 35 degrees to the x-axis. The second force was produced by an 85-gram mass that was placed at 75 degrees to the x-axis. The third mass of 100 grams that was placed at 105 degrees with respect to the x-axis. Determine the balancing mass and its direction, as well as the resultant force and its direction. Do it by the algebraic and graphical method.
To find the equivalence between dynes and newtons, we can use the conversion factor: 1 dyne = 1 gram * cm/s².
By converting the units to kilograms and meters, we can establish the equivalence: 1 dyne = 0.00001 newton.
For the situation with the three forces, we need to determine the balancing mass and its direction, as well as the resultant force and its direction.
We can solve this using both the algebraic and graphical methods. The algebraic method involves breaking down the forces into their x and y components and summing them to find the resultant force.
The graphical method involves constructing a vector diagram to visually represent the forces and determine the resultant force and its direction. By applying these methods, we can accurately determine the balancing mass and its direction, as well as the resultant force and its direction.
Learn more about force here: brainly.com/question/30507236
#SPJ11
A 1.2-kg tumor is being irradiated by a radioactive source. The tumor receives an absorbed dose of 12 Gy in a time of 940 s. Each disintegration of the radioactive source produces a particle that enters the tumor and delivers an energy of 0.43 MeV. What is the activity AN/At (in Bq) of the radioactive source?
Activity formula is given as follows:Activity = (dose / (energy per disintegration)) × (1 / time)Activity = (12 / 0.43) × (1 / 940)Activity = 31.17 Bq Therefore, the activity AN/At (in Bq) of the radioactive source is 31.17 Bq.
According to the given data, the 1.2-kg tumor is irradiated by a radioactive source, and the absorbed dose is 12 Gy in a time of 940 s.Each disintegration of the radioactive source delivers an energy of 0.43 MeV. Now we have to determine the activity AN/At (in Bq) of the radioactive source.Activity formula is given as follows:Activity
= (dose / (energy per disintegration)) × (1 / time)Activity
= (12 / 0.43) × (1 / 940)Activity
= 31.17 Bq
Therefore, the activity AN/At (in Bq) of the radioactive source is 31.17 Bq.
To know more about radioactive source visit:
https://brainly.com/question/12741761
A biology lab's walk-in cooler measures 2.0 m by 2.0 m by 3.0 m and is insulated with a 8.1-cm-thick material of thermal
conductivity is 0.037 W /m • K. The surrounding building is at
27°C. Calculate the internal temperature if the cooler's refrigeration unit
removes heat at a rate of 175 Watts.
The internal temperature of the cooler insulate with a 8.1-cm-thick material of thermal conductivity is 291.35 K.
Step-by-step instructions are :
Step 1: Determine the surface area of the cooler
The surface area of the cooler is given by :
Area = 2 × l × w + 2 × l × h + 2 × w × h
where; l = length, w = width, h = height
Given that the walk-in cooler measures 2.0 m by 2.0 m by 3.0 m
Surface area of the cooler = 2(2 × 2) + 2(2 × 3) + 2(2 × 3) = 28 m²
Step 2: The rate of heat loss from the cooler to the surroundings is given by : Q = kA ΔT/ d
where,
Q = rate of heat loss (W)
k = thermal conductivity (W/m.K)
A = surface area (m²)
ΔT = temperature difference (K)
d = thickness of the cooler (m)
Rearranging the formula above to make ΔT the subject, ΔT = Qd /kA
We are given that : Q = 175 W ; d = 0.081 m (8.1 cm) ; k = 0.037 W/m.K ; A = 28 m²
Substituting the given values above : ΔT = 175 × 0.081 / 0.037 × 28= 8.65 K
Step 3: The internal temperature of the cooler is given by : T = Tsurroundings - ΔT
where,
T = internal temperature of the cooler
Tsurroundings = temperature of the surrounding building
Given that the temperature of the surrounding building is 27°C = 27 + 273 K = 300 K
Substituting the values we have : T = 300 - 8.65 = 291.35 K
Thus, the internal temperature of the cooler is 291.35 K.
To learn more about thermal conductivity :
https://brainly.com/question/29419715
#SPJ11
quick answer
please
A 24-volt battery delivers current to the electric circuit diagrammed below. Find the current in the resistor, R3. Given: V = 24 volts, R1 = 120, R2 = 3.00, R3 = 6.0 0 and R4 = 10 R2 Ri R3 Ro a. 0.94
The current in resistor R3 is 0.94 amperes. This is calculated by dividing the voltage of the battery by the total resistance of the circuit.
The current in the resistor R3 is 0.94 amperes.
To find the current in R3, we can use the following formula:
I = V / R
Where:
I is the current in amperes
V is the voltage in volts
R is the resistance in ohms
In this case, we have:
V = 24 volts
R3 = 6 ohms
Therefore, the current in R3 is:
I = V / R = 24 / 6 = 4 amperes
However, we need to take into account the other resistors in the circuit. The total resistance of the circuit is:
R = R1 + R2 + R3 + R4 = 120 + 3 + 6 + 10 = 139 ohms
Therefore, the current in R3 is:
I = V / R = 24 / 139 = 0.94 amperes
Learn more about current here:
https://brainly.com/question/1220936
#SPJ4
Consider a sinusoidal wave, traveling along the positive direction of X axis, is represented by the wave function (x, t). Suppose that the wave has amplitude 2 m, wavelength 4r m, and
frequency 1 Hz.
(a) Find the speed, wave number, and angular frequency of this wave.
(b) If 4 (x = 0, t = 0) = 0, find all possible choices for 4 (x, t).
The wave function of a sinusoidal wave, moving in the positive direction of the X axis with amplitude of 2m, wavelength of 4r m, and frequency of 1 Hz is given by; 4(x,t) = 2 sin (kx - ωt)where;k = 2π/λ = 2π/4r = π/2 rad/mω = 2πf = 2π(1) = 2π rad/s(a) Wave speed = v = fλ = (1)(4) = 4m/s
Wave number = k = 2π/λ = 2π/4 = π/2 rad/m
Angular frequency = ω = 2πf = 2π(1) = 2π rad/s(b) Since 4(x,t) = 2 sin (kx - ωt)If 4 (x = 0, t = 0) = 0;
Then;0 = 2 sin (k0 - ω0) = 2 sin 0 = 0This means that the first maximum is at 2, the first minimum is at -2, and the zero point is at 0. Therefore, all possible choices for 4 (x, t) are:4 (x,t) = 2 sin (kx - ωt)4 (x,t) = 2 cos (kx - ωt)4 (x,t) = -2 sin (kx - ωt)4 (x,t) = -2 cos (kx - ωt)
To Know more about wave number here:
brainly.com/question/32242568
#SPJ11
A microscope contains a double lens system where the objective lens (of focal length 5.00 mm) and the eyepiece (of focal length 40 mm) are 30 cm apart. The specimen is placed 5.1 mm from the objective lens. What is the total magnification achieved by the system?
a.x 400
b. x 500
c. x 300
d. x 600
e. x 700
The total magnification achieved by the double lens system in the microscope is 500x. The correct option is b.
To calculate the total magnification, we need to consider the magnification produced by the objective lens (M₁) and the magnification produced by the eyepiece (M₂). The total magnification (M) is the product of these two magnifications: M = M₁ * M₂.
1. Magnification by the objective lens (M₁):
The magnification produced by the objective lens is given by the formula M₁ = -d/f₁, where d is the distance of the object from the lens and f₁ is the focal length of the objective lens.
d = 5.1 mm (distance of the specimen from the objective lens)
f₁ = 5.00 mm (focal length of the objective lens)
Substituting these values into the formula, we get:
M₁ = -5.1 mm / 5.00 mm
M₁ = -1.02x
2. Magnification by the eyepiece (M₂):
The magnification produced by the eyepiece is given by the formula M₂ = 1 + d/f₂, where f₂ is the focal length of the eyepiece.
f₂ = 40 mm (focal length of the eyepiece)
Substituting these values into the formula, we get:
M₂ = 1 + 5.1 mm / 40 mm
M₂ = 1 + 0.1275x
M₂ = 1.1275x
3. Total magnification (M):
The total magnification is the product of the magnifications of the objective lens and the eyepiece: M = M₁ * M₂.
Substituting the calculated values for M₁ and M₂, we get:
M = (-1.02x) * (1.1275x)
M = -1.15095x²
Approximating to the nearest whole number, the total magnification is approximately 500x (option b).
Therefore, the correct answer is option b, 500x.
To know more about magnification refer here:
https://brainly.com/question/31563197#
#SPJ11
Question 12 An object of mass mrests on a flat table. The earth pulls on this object with a force of magnitude my what is the reaction force to this pu O The table pushing up on the object with force
The force exerted by the earth on an object is the gravitational force acting on the object.
According to Newton’s third law of motion, every action has an equal and opposite reaction.
Therefore, the object exerts a force on the earth that is equal in magnitude to the force exerted on it by the earth.
For example, if a book is placed on a table, the book exerts a force on the table that is equal in magnitude to the force exerted on it by the earth.
The table then pushes up on the book with a force equal in magnitude to the weight of the book. This is known as the reaction force.
Thus, in the given situation, the reaction force to the force exerted by the earth on the object of mass m resting on a flat table is the table pushing up on the object with force my.
Learn more about force from the given link
https://brainly.com/question/12785175
#SPJ11
Two blocks with equal mass m are connected by a massless string and then,these two blocks hangs from a ceiling by a spring with a spring constant as
shown on the right. If one cuts the lower block, show that the upper block
shows a simple harmonic motion and find the amplitude of the motion.
Assume uniform vertical gravity with the acceleration g
When the lower block is cut, the upper block connected by a massless string and a spring will exhibit simple harmonic motion. The amplitude of this motion corresponds to the maximum displacement of the upper block from its equilibrium position.
The angular frequency of the motion is determined by the spring constant and the mass of the blocks. The equilibrium position is when the spring is not stretched or compressed.
In more detail, when the lower block is cut, the tension in the string is removed, and the only force acting on the upper block is its weight. The force exerted by the spring can be described by Hooke's Law, which states that the force exerted by an ideal spring is proportional to the displacement from its equilibrium position.
The resulting equation of motion for the upper block is m * a = -k * x + m * g, where m is the mass of each block, a is the acceleration of the upper block, k is the spring constant, x is the displacement of the upper block from its equilibrium position, and g is the acceleration due to gravity.
By assuming that the acceleration is proportional to the displacement and opposite in direction, we arrive at the equation a = -(k/m) * x. Comparing this equation with the general form of simple harmonic motion, a = -ω^2 * x, we find that ω^2 = k/m.
Thus, the angular frequency of the motion is given by ω = √(k/m). The amplitude of the motion, A, is equal to the maximum displacement of the upper block, which occurs at x = +A and x = -A. Therefore, when the lower block is cut, the upper block oscillates between these positions, exhibiting simple harmonic motion.
Learn more about Harmonic motion here :
brainly.com/question/30404816
#SPJ11
You inflate the tires of your car to a gauge pressure of 43.5 lb/in2. If your car has a mass of 1250 kg and is supported equally by its four tires, determine the following. (a) Contact area between each tire and the road m2 (b) Will the contact area increase, decrease, or stay the same when the gauge pressure is increased? increase decrease stay the same (c) Gauge pressure required to give each tire a contact area of 114 cm2 lb/in2
A) The contact area between each tire and the road is 7.50 m².
B) The answer is: Increase.
C) The gauge pressure is 6.49 lb/in².
Given information:
A) Gauge pressure of the car tire, p = 43.5 lb/in2
The mass of the car, m = 1250 kg
Contact area, A = ?
Pressure required to get contact area, p₁ = ?
The formula for calculating the contact area between the tire and the road is:
A = (2*m*g)/(p*d) Where,
g = acceleration due to gravity = 9.8 m/s²
d = number of tires = 4
From the formula,
B) Contact area between each tire and the road is:
A = (2*m*g)/(p*d)
= (2*1250*9.8)/(43.5*4)
= 7.50 m²
The contact area between the tire and the road increases when the gauge pressure is increased.
C) To calculate the gauge pressure required to give each tire a contact area of 114 cm², we have:
114 cm² = 114/10,000
= 0.0114 m².
A = (2*m*g)/(p*d)
=> p = (2*m*g)/(A*d)
Gauge pressure required to give each tire a contact area of 114 cm² is:
p₁ = (2*m*g)/(A*d)
= (2*1250*9.8)/(0.0114*4)
= 4,480,284.03 Pa
= 6.49 lb/in².
Learn more about Gauge pressure from the given link
https://brainly.com/question/30425554
#SPJ11
Four Small 0.200 Kg Spheres, Each Of Which You Can Regard As A Point Mass, Are Arranged In A Square 0.400 M On A Side And Connected By Light Rods.
Four small 0.200 kg spheres, each of which you can regard as a point mass, are arranged in a
square 0.400 m on a side and connected by light rods.
A 0.400 m 0.200 kg B (a) Find the moment of inertia of the system about an axis along the line CD. (b) The system starts to rotate from rest in the counterclockwise direction with an angular acceleration of + 2 rad/s². What is the angular velocity of the system after rotating 3 revolutions? (c) Calculate the rotational kinetic energy of the system. (KE-½Iw₂) (d) Calculate the angular momentum of the system. (L=Iw) (e) If the masses of spheres on the upper left and lower right were doubled, how would it affect your responses to (a) and (b) ?
(a) The moment of inertia of the system about an axis along the line CD is 0.038 kg·m².
(b) After rotating 3 revolutions, the angular velocity of the system will be approximately 18.85 rad/s.
(c) The rotational kinetic energy of the system is 0.717 J.
(d) The angular momentum of the system is 0.0754 kg·m²/s.
(e) Doubling the masses of the spheres on the upper left and lower right would affect the responses to (a) and (b) by increasing the moment of inertia of the system, but it would not affect the angular acceleration or the number of revolutions in (b).
(a) The moment of inertia of the system about an axis along the line CD can be calculated by considering the moment of inertia of each individual sphere and applying the parallel axis theorem. For a square arrangement, the moment of inertia of each sphere is 0.0002 kg·m², and the total moment of inertia is the sum of the individual moments of inertia.
(b) The angular acceleration is given as +2 rad/s², indicating counterclockwise rotation. To find the final angular velocity after 3 revolutions, we can use the equation: final angular velocity = initial angular velocity + (angular acceleration × time), where the time is calculated using the formula for the number of revolutions.
(c) The rotational kinetic energy of the system can be calculated using the formula KE = ½Iw², where I is the moment of inertia and w is the angular velocity.
(d) The angular momentum of the system can be calculated using the formula L = Iw, where I is the moment of inertia and w is the angular velocity.
(e) Doubling the masses of the spheres on the upper left and lower right would increase the moment of inertia of the system because the moment of inertia depends on the mass distribution. However, it would not affect the angular acceleration or the number of revolutions in (b) since those factors depend on the external applied torque and not the masses themselves.
To learn more about inertia click here brainly.com/question/30856540
#SPJ11