. (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
1 answer:
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
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Answer:
v_{1fy} = - 0.4549 m / s
Explanation:
This is an exercise of conservation of the momentum, for this we must define a system formed by the two balls, so that the forces during the collision have internal and the momentum is conserved
initial. Before the crash
p₀ = m v₁₀
final. After the crash
= m
+ m v_{2f}
Recall that velocities are a vector so it has x and y components
p₀ = p_{f}
we write this equation for each axis
X axis
m v₁₀ = m v_{1fx} + m v_{2fx}
Y Axis
0 = -m v_{1fy} + m v_{2fy}
the exercise tells us the initial velocity v₁₀ = 1.83 m / s, the final velocity v_{2f} = 1.15, let's use trigonometry to find its components
sin 23.3 = v_{2fy} / v_{2f}
cos 23.3 = v_{2fx} / v_{2f}
v_{2fy} = v_{2f} sin 23.3
v_{2fx} = v_{2f} cos 23.3
we substitute in the momentum conservation equation
m v₁₀ = m v_{1f} cos θ + m v_{2f} cos 23.3
0 = - m v_{1f} sin θ + m v_{2f} sin 23.3
1.83 = v_{1f} cos θ + 1.15 cos 23.3
0 = - v_{1f} sin θ + 1.15 sin 23.3
1.83 = v_{1f} cos θ + 1.0562
0 = - v_{1f} sin θ + 0.4549
v_{1f} sin θ = 0.4549
v_{1f} cos θ = -0.7738
we divide these two equations
tan θ = - 0.5878
θ = tan-1 (-0.5878)
θ = -30.45º
we substitute in one of the two and find the final velocity of the incident ball
v_{1f} cos (-30.45) = - 0.7738
v_{1f} = -0.7738 / cos 30.45
v_{1f} = -0.8976 m / s
the component and this speed is
v_{1fy} = v1f sin θ
v_{1fy} = 0.8976 sin (30.45)
v_{1fy} = - 0.4549 m / s
Answer:
The Resultant Induced Emf in coil is 4∈.
Explanation:
Given that,
A coil of wire containing having N turns in an External magnetic Field that is perpendicular to the plane of the coil which is steadily changing. An Emf (∈) is induced in the coil.
To find :-
find the induced Emf if rate of change of the magnetic field and the number of turns in the coil are Doubled (but nothing else changes).
So,
Emf induced in the coil represented by formula
∈ = ...................(1)
Where:
. { B is magnetic field }
{A is cross-sectional area}
. No. of turns in coil.
. Rate change of induced Emf.
Here,
Considering the case :-
&
Putting these value in the equation (1) and finding the new emf induced (∈1)
∈1 =
∈1 =
∈1 =
∈1 = 4∈ ...............{from Equation (1)}
Hence,
The Resultant Induced Emf in coil is 4∈.
Answer:
P = 12000 W
Explanation:
General data:
- F = 15000 N
- d = 40 m
- t = 50 s
- P = ?
Work is force times unit of distance. So in order to calculate the power we must first calculate the work.
Formula:
Replace and solve
once the work is found, we proceed to find the power according to the formula:
The power of the crane is <u>12000 Watts.</u>
Greetings.