The amplitude is 2.30mm, frequency is 118 Hz, wavelength is 0.899 m, wave speed is 106 m/s, the wave is moving in the direction of decreasing x, tension in the rope is 28 N and the average power transmitted by the wave is approximately 0.38W.
Explanation:The wave function of a travelling wave on a thin rope is given by y(x,t) = 2.30mm * cos[(6.98rad/m) * x + (742 rad/s) * t]. From this equation, we can derive the following information:
Amplitude - This is the maximum displacement of a point on the wave and is determined by the coefficient before the cosine in the equation. Therefore, the amplitude is 2.30mm.
Frequency - This can be obtained by dividing the wave speed by the wavelength. The coefficient in front of the 't' in the equation is equivalent to 2π times the frequency (since there are 2π rad in one wave cycle). Therefore, the frequency f = (742 rad/s)/(2π) = 118 Hz.
Wavelength - This is represented by the reciprocal of the coefficient in front of 'x' in the equation. Therefore, the wavelength λ = 2π/(6.98 rad/m) = 0.899 m.
Wave Speed - This can be found by multiplying frequency and wavelength together. Hence, wave speed v = f * λ = 118 Hz * 0.899 m = 106 m/s.
Direction the wave is travelling - The symbol '+' in front of 't' in the equation tells us that the wave is moving in the direction of decreasing x.
Tension in the rope - This can be found using the wave speed that we have already calculated and the rope's mass per unit length µ (mass/length = 0.00338 kg / 1.35 m = 0.00250 kg/m). The tension T can be calculated using the formula v = sqrt(T/µ), which gives T = µ * v² = 0.00250 kg/m * (106 m/s)² = 28 N.
Average power transmitted by the wave - The power P carried by wave is given by the formula P = 0.5 * µ * v * A² * ω², where A is the amplitude, ω is the angular frequency (which is the coefficient in front of 't' in the equation). Substituting the values we find: P = 0.5 * 0.00250 kg/m * 106 m/s * (2.30 mm)² * (742 rad/s)² ≈ 0.38 W.
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A 10.00 g sample of a compound containing only carbon, hydrogen, and oxygen forms 23.98 g CO2 and 4.91 g H2O upon complete combustion. What is the empirical formula of the compound?
What is the approximate weight of a 20-kg cannonball on earth?
a. 2 n
b. 20 n
c. 200 n
d. 2000 n?
Answer: The weight of the cannonball on Earth is 200 N
Explanation:
Weight is defined as the force exerted by the body on any surface. It is also defined as the product of mass of the body multiplied by the acceleration due to gravity.
Mathematically,
[tex]W=mg[/tex]
where,
W = weight of the cannonball
m = mass of the cannonball = 20 kg
g = acceleration due to gravity on Earth = [tex]10m/s^2[/tex]
Putting values in above equation, we get:
[tex]W=20kg\times 10m/s^2=200N[/tex]
Hence, the weight of the cannonball on Earth is 200 N
A ride at an amusement park moves the riders in a circle at a rate of 6.0 m/s. If the radius of the ride is 9.0 meters, what is the acceleration of the ride?
4.0 m/s2
0.67 m/s2
0.075 m/s2
54 m/s2
Why is the halo of a galaxy the most important part of a galaxy?
A car of mass 960 kg is free-wheeling down an incline at a constant speed of 9.0 m s–1.
The slope makes an angle of 15° with the horizontal.
Deduce that the average resistive force acting on the car is 2.4×103N
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1. This question is about the breaking distance of a car and specific heat capacity. (a) A car of mass 960 kg is free-wheeling down an incline at a constant speed of 9.0 m s–1. The slope makes an angle of 15° with the horizontal. (i) Deduce that the average resistive force acting on the car is 2.4×103N. ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... (2) (ii) Calculate the kinetic energy of the car. ........................................................................................................................... ........................................................................................................................... (1) (b) The driver now applies the brakes and the car comes to rest in 15 m. Use your answer to (a)(ii) to calculate the average braking force exerted on the car in coming to rest. ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... (2) (c) The same braking force is applied to each rear wheel of the car. The effective mass of each brake is 5.2 kg with a specific heat capacity of 900 J kg–1 K–1....
Attachments:
Homework-10-d....rtf
i) The average resistive force acting on the car is approximately 2.4 × 10³ N. ii) The kinetic energy of the car is 3,888 J. b) The average braking force exerted on the car is coming to rest is approximately 259.2 N. ii) The average braking force exerted on the car is coming to rest is approximately 259.2 N. 2) c) The change in temperature of each rear brake is approximately 0.907 Kelvin.
i) Given information:
Mass of the car (m) = 960 kg
Speed of the car (v) = 9.0 m/s
The angle of the slope (θ) = 15°
Resistive force (Fr) = m × g × sin(θ)
Fr = 960 × 9.8 × sin(15°)
Fr = 2.4 × 10³ N
Therefore, the average resistive force acting on the car is approximately 2.4 × 10³ N.
(ii) To calculate the kinetic energy of the car, we can use the formula:
Kinetic energy (KE) = (1/2) × m × v²
KE = (1/2) × 960 × (9.0)²
KE = 3,888 J
Therefore, the kinetic energy of the car is 3,888 J.
(b) The average braking force exerted on the car is:
Average braking force = Change in kinetic energy/braking distance
Average braking force = KE / braking distance
Average braking force = 3,888 / 15
Average braking force = 259.2 N
Therefore, the average braking force exerted on the car in coming to rest is approximately 259.2 N.
(ii) The work done by a force can be calculated using the formula:
Work = Force × Distance × cos(θ)
3,888 = Force × 15 × cos(0°)
Since cos(0°) = 1, the equation simplifies to:
3,888 = Force × 15
Force = 3,888 ÷ 15
Force = 259.2 N
Therefore, the average braking force exerted on the car in coming to rest is approximately 259.2 N.
2) (c) To calculate the change in temperature of each rear brake, we can use the formula:
Change in heat energy = mass × specific heat capacity × change in temperature
Change in heat energy = Work
The work done by the braking force is equal to the product of the force and the braking distance (15 m),
Therefore, the change in heat energy for each brake is:
Change in heat energy = Work = Force × braking distance
Change in heat energy = 259.2 × 15 = 3,888 J
Now, we can use the formula to find the change in temperature:
Change in heat energy = mass × specific heat capacity × change in temperature
3,888 = 5.2 × 900 × change in temperature
change in temperature ≈ 0.907 K
Therefore, the change in temperature of each rear brake is approximately 0.907 Kelvin.
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