Physics College


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....



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Homework-10-d....rtf



Answers

Answer 1
need to draw it first.....
but there is no option here to draw ..  why?
anyway ,  here the car is going with a uniform velocity..
so the acceleration of the car=0, a=0
so   resistive force on the car=component along the inclined plane of the weight of the car=mg*cos(90-15)=960*9.8*cos 75=?
Answer 2

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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Related Questions

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?

Answers

(23.98 g CO2) / (44.00964 g CO2/mol) × (1 mol C / 1 mol CO2) = 0.544881 mol C 
(0.544881 mol C) × (12.01078 g C/mol) = 6.54445 g C 

(4.91 g H2O) / (18.01532 g H2O/mol) × (2 mol H / 1 mol H2O) = 0.54509 mol H 
(0.54509 mol H) × (1.007947 g H/mol) = 0.54942 g H 

10 g total - 6.54445 g C - 0.54942 g H = 2.90613 g O 
(2.90613 g O) / (15.99943 g O/mol) = 0.18164 mol O 

Divide by the smallest number of moles: 
(0.544881 mol C) / 0.18164 mol = 2.9998 
(0.54509 mol H) / 0.18164 mol = 3.0009 
(0.18164 mol O) / 0.18164 mol = 1.0000 
Round to the nearest whole number to find the empirical formula: 
C3H3O 

A fellow student with a mathematical bent tells you that the wave function of a traveling wave on a thin rope is y(x,t)= 2.30mmcos[(6.98rad/m)x + (742 rad/s)t]. Being more practical, you measure the rope to have a length of 1.35 m and a mass of 0.00338 kg. You are then asked to determine the following:
A) Amplitude
B) Frequency
C) Wavelength
D) Wave Speed
E) Direction the wave is traveling
F) Tension in the rope
G) Average power transmitted by the wave

Answers

Final answer:

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 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

Answers

4.0 m/s2


it's 9 squared divided by 6
the answer is B 4.0 m/s2

Why is the halo of a galaxy the most important part of a galaxy?

Answers

The halo of a galaxy is the most important part of a galaxy because this preserves the signatures of both its formation, composition and evolution. These haloes are expansive and may differ in every galaxy. Even our own galaxy, which is the Milky Way, also possess this halo. Hope this is the answer that you are looking for.
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