Answer:
yes
Explanation:
because motion is relevant
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
As per the question, the mass of meteorite [ m]= 50 kg
The velocity of the meteorite [v] = 1000 m/s
When the meteorite falls on the ground, it will give whole of its kinetic energy to earth.
We are asked to calculate the gain in kinetic energy of earth.
The kinetic energy of meteorite is calculated as -
![Kinetic\ energy\ [K.E]\ =\frac{1}{2} mv^2](https://tex.z-dn.net/?f=Kinetic%5C%20energy%5C%20%5BK.E%5D%5C%20%3D%5Cfrac%7B1%7D%7B2%7D%20mv%5E2)
![=\frac{1}{2}50kg*[1000\ m/s]^2](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B2%7D50kg%2A%5B1000%5C%20m%2Fs%5D%5E2)

Here, J stands for Joule which is the S.I unit of energy.
Answer:
-15 m/s
Explanation:
The computation of the velocity of the 4.0 kg fragment is shown below:
For this question, we use the correlation of the momentum along with horizontal x axis
Given that
Weight of stationary shell = 6 kg
Other two fragments each = 1.0 kg
Angle = 60
Speed = 60 m/s
Based on the above information, the velocity = v is



= -15 m/s