Answer:
x = 11.23 m
Explanation:
For this interesting exercise, we must use angular kinematics, linear kinematics and the relationship between angular and linear quantities.
Let's reduce to SI system units
θ = 155 rev (2pi rad / rev) = 310π rad
α = 2.00rev / s2 (2pi rad / 1 rev) = 4π rad / s²
Let's look for the angular velocity at the time the piece is released, with starting from rest the initial angular velocity is zero (wo = 0)
w² = w₀² + 2 α θ
w =√ 2 α θ
w = √(2 4pi 310pi)
w = 156.45 rad / s
The relationship between angular and linear velocity
v = w r
v = 156.45 0.175
v = 27.38 m / s
In this part we have the linear speed and the height that it travels to reach the floor, so with the projectile launch equations we can find the time it takes to arrive
y =
t - ½ g t²
As it leaves the highest point its speed is horizontal
y = 0 - ½ g t²
t = √ (-2y / g)
t = √ (-2 (-0.820) /9.8)
t = 0.41 s
With this time we calculate the horizontal distance, because the constant horizontal speed
x = vox t
x = 27.38 0.41
x = 11.23 m
Answer:
d = 105 m
Explanation:
Speed of a car, v = 21 m/s
We need to find the distance traveled by the dar during those 5 s before it stops. Let the distance is d. It can be calculated as :
d = v × t
d = 21 m/s × 5 s
d = 105 m
So, it will cover 105 m before it stops.
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Answer:

Explanation:
Given that
Radius =r
Electric filed =E
Q=Charge on the ring
The electric filed at distance x given as

For maximum condition



For maximum condition




At
the electric field will be maximum.
Answer:
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