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 
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, 
= 110 km/h = 110 x 1000/3600 = 30.6 m/s
initial height, 
= 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 = 
gh = 
h = 
= 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 - )
= 750 x 9.8 x (47.6 - 22)
= 188160 Joule
= 1.88 x 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 /(22/sin 2.5°)
= 373 N
The answer is A.
Explanation:
We know that the average acceleration a for an interval of time Δt is expressed as:
a = Δv
Δt
where Δv is the change in velocity that occurs during Δt.
e formula for the instantaneous acceleration a is almost the same, except that we need to indicate that we're interested in knowing what the ratio of Δv to Δt approaches as Δt approaches zero.
We can indicate that by using the limit notation.
So, the formula for the instantaneous acceleration is:
a = lim Δv
Δt→0 Δt
Answer:
Solution
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(a) The labelled diagram is shown.
(b) The refractive index of diamond is 2.42. Refractive index of diamond is the ratio of the speed of light in air to the speed of light in diamond.i.e.,
μ=
Speedoflightindiamond
Speedoflightinair
and, the ratio of these velocities is 2.42. i.e., This means that the speed of light in diamond will reduce by a factor of 2.42 as compared to its speed in air. In other words, the speed of light in diamond is
1/2.42
times the speed of light in vacuum.
Explanation:
a) Draw and label the diagram given :
(i) Incident ray
(ii) Refracted ray
(iii) Emergent ray
(iv) Angle of reflection
(v) Angle of deviation
(v) Angle of emergence
(b) The refractive index of diamond is 2.42. What is the meaning of this statement in relation to speed of light?
No because there must be an even # if their is an even amount one of the forces isn’t being cancelled
