Which statement accurately describes the ultraviolet catastrophe

. (a) How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110

7nadin3 [17]

Answer:

(a) the high of a hill that car can coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h is 47.6 m

(b) thermal energy was generated by friction is 1.88 x $10^{5}$ J

(C) the average force of friction if the hill has a slope 2.5º above the horizontal is 373 N

Explanation:

given information:

m = 750 kg

initial velocity, $v_{0}$ = 110 km/h = 110 x 1000/3600 = 30.6 m/s $\frac{30.6^{2} }{2x9.8}$

initial height, $h_{0}$ = 22 m

slope, θ = 2.5°

(a) How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h?

according to conservation-energy

EP = EK

mgh = $\frac{1}{2} mv_{0} ^{2}$

gh = $\frac{1}{2} v_{0} ^{2}$

h = $\frac{v_{0} ^{2} }{2g}$

= 47.6 m

(b) If, in actuality, a 750-kg car with an initial speed of 110 km/h is observed to coast up a hill to a height 22.0 m above its starting point, how much thermal energy was generated by friction?

thermal energy = mgΔh

= mg (h - $h_{0}$ )

= 750 x 9.8 x (47.6 - 22)

= 188160 Joule

= 1.88 x $10^{5}$ J

(c) What is the average force of friction if the hill has a slope 2.5º above the horizontal?

f d = mgΔh

f = mgΔh / d,

where h = d sin θ, d = h/sinθ

therefore

f = (mgΔh) / (h/sinθ)

= 1.88 x $10^{5}$ /(22/sin 2.5°)

= 373 N

8 0

3 years ago

You pick up a 3.4-kg can of paint from the ground and lift it to a height of 1.8 m. (a) how much work do you do on the can of pa

MariettaO [177]

(a) For the work-energy theorem, the work done to lift the can of paint is equal to the gravitational potential energy gained by it, therefore it is equal to

$W=mg\Delta h$

where m=3.4 kg is the mass of the can, g=9.81 m/s^2 is the gravitational acceleration and $\Delta h=1.8 m$ is the variation of height. Substituting the numbers into the formula, we find

$W=(3.4 kg)(9.81 m/s^2)(1.8 m)=60.0 J$

(b) In this case, the work done is zero. In fact, we know from its definition that the work done on an object is equal to the product between the force applied F and the displacement:

$W=Fd$

However, in this case there is no displacement, so d=0 and W=0, therefore the work done to hold the can stationary is zero.

(c) In this case, the work done is negative, because the work to lower the can back to the ground is done by the force of gravity, which pushes downward. Its value is given by the same formula used in part (a):

$W=mg \Delta h=(3.4 kg)(9.81 m/s^2)(-1.8 m)=-60.0 J$