The potential energy of the car when it let go is 20,000 J.
The speed of the car at the bottom of the ramp is 20 m/s.
The given parameters;
- <em>mass of the car, m = 100 kg</em>
- <em>height of the car, h = 20 m</em>
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The potential energy of the car is calculated as follows;
P.E = mgh
P.E = 100 x 10 x 20
P.E = 20,000 J
The speed of the car at the bottom of the ramp is calculated as follows;
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Answer:
vₓ = 0.566 m / s, W_total = 9.1 J
Explanation:
This exercise is a parabolic type movement, for the x axis where there is no acceleration
x = v t
vₓ = x / t
vₓ = 0.34 / 0.6
vₓ = 0.566 m / s
the work done is
X axis
In this axis there is no acceleration, therefore the sum of the forces is zero and since the work is the force times the distance, we conclude that the lock in this axis is zero.
W₁ = 0
Y axis
in this axis the force that exists is the force of gravity, that is, the weight of the body
W₂ = Fy y
W₂ = mg and
W₂ = m 9.8 0.70
W₂ = m 9.1
the work is a scalar for which we have to add the quantities obtained
W_total = W₁ + W₂
W_total = 0 + 9.1 m
W_total = 9.1 m
In order to complete the calculation, the mass of the body is needed if we assume that the mass is m = 1
W_total = 9.1 J
Answer:
The energy stored in the capacitor is 333.3 J
Explanation:
Energy stored in a capacitor = CV²/2
C = capacitance = 14 × 10⁻⁶F
V = 6.90 × 10³V
Energy = (14 × 10⁻⁶)(6.90 × 10³)²/2 = 333.27 = 333.3 J
Hope this Helps!!!
If your ref points are each other and not the surrounding environment, at the same speed you would both appear to be stationary
Answer:
A. 490
Explanation:
soln
mass = m = 5kg
Height = h = 10m
Acceleration due to gravity = g = 9.8ms²
K.E = 1/2 × mass × (velocity)²
Recall from equations of motion
v² = u² + 2gh
Therefore,
K.E = 1/2 × mass × ( u² + 2gh)
K.E = 1/2 × 5 × ( 0² + 2×10×9.8)
K.E = 1/2 × 5 × 196
K.E = 1/2 × 980
K.E = 490 Joules
