You throw a ball upwards at 22.0 m/s how high will it go?
1 answer:
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Answer:
x = 6.94 m
Explanation:
For this exercise we can find the speed at the bottom of the ramp using energy conservation
Starting point. Higher
Em₀ = K + U = ½ m v₀² + m g h
Final point. Lower
= K = ½ m v²
Em₀ = Em_{f}
½ m v₀² + m g h = ½ m v²
v² = v₀² + 2 g h
Let's calculate
v = √(1.23² + 2 9.8 1.69)
v = 5.89 m / s
In the horizontal part we can use the relationship between work and the variation of kinetic energy
W = ΔK
-fr x = 0- ½ m v²
Newton's second law
N- W = 0
The equation for the friction is
fr = μ N
fr = μ m g
We replace
μ m g x = ½ m v²
x = v² / 2μ g
Let's calculate
x = 5.89² / (2 0.255 9.8)
x = 6.94 m
Answer:
Explanation:
The artificial gravity generated by the rotating space station is the same centripetal acceleration due to the rotational motion of the station, which is given by:
Here, r is the radius and v is the tangential speed, which is given by:
Here is the angular velocity, we replace (2) in (1):
Recall that .
Solving for :
Answer : south
Explanation :
Given that,
Force = 200 N
Length = 2.00 m
Mass = 5.00 kg
We know that , the formula of the angular acceleration
We know the moment of inertia of rod
Now, the torque is
The angular acceleration
Negative sign represent of south direction.
Hence, this is the required solution.