Well if it was traveling for an hour then the answer is 8 miles.
Answer: Accelaration is 2.77 m/s*s
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Explanation:
V0=0km/h=0m/s
V1=100 km/h=27.7 m/s
t=10s=
Use equation for accelaration : a=(V1-V0)/t
a=(0m/s-27.7m/s)/10s
a=-27.7s/10s
a=2.77m/s*s
Answer:
K_{f} / K₀ =1.12
Explanation:
This problem must work using the conservation of angular momentum (L), so that the moment is conserved in the system all the forces must be internal and therefore the torque is internal and the moment is conserved.
Initial moment. With arms outstretched
L₀ = I₀ w₀
the wo value is 5.0 rad / s
final moment. After he shrugs his arms
= I_{f} w_{f}
indicate that the moment of inertia decreases by 11%
I_{f} = I₀ - 0.11 I₀ = 0.89 I₀
L_{f} = L₀
I_{f} w_{f} = I₀ w₀
w_{f} = I₀ /I_{f} w₀
let's calculate
w_{f} = I₀ / 0.89 I₀ 5.0
w_{f} = 5.62 rad / s
Having these values we can calculate the change in kinetic energy
/ K₀ = ½ I_{f} w_{f}² (½ I₀ w₀²)
K_{f} / K₀ = 0.89 I₀ / I₀ (5.62 / 5)²
K_{f} / K₀ =1.12
Answer:
42.05 m/s
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 0 m/s
Height (h) = 239 m
Acceleration due to gravity (g) = 3.7 m/s²
Final velocity (v) =?
The velocity with which the camera hits the ground can be obtained as follow:
v² = u² + 2gh
v² = 0² + 2 × 3.7 × 239
v² = 0 + 1768.6
v² = 1768.6
Take the square root of both side
v = √(1768.6)
v = 42.05 m/s
Therefore, the velocity with which the camera hits the ground is 42.05 m/s