Answer:
Thus, the initial velocity of the bullet is 507.5 m/s.
Explanation:
mass of bullet, m = 0.0085 kg
mass of block, M = 0.99 kg
Height raised, h = 0.95 m
Let the initial velocity of bullet is u and the final velocity of block and bullet system is v.
Use conservation of energy
Potential energy of bullet bock system = kinetic energy of bullet block system
Now use conservation of linear momentum
mu = (M+m) v
0.0085 x u = (0.99 + 0.0085) x 4.32
0.0085 u = 4.314
u = 507.5 m/s
Thus, the initial velocity of the bullet is 507.5 m/s.
Explanation:
Below is an attachment containing the solution.
Answer
given,
L(t) = 10 - 3.5 t
mass of particle = 2 Kg
radius of the circle = 3.1 m
a) torque
τ = 
τ = 
τ = -3.5 N.m
Particle rotates clockwise as i look down the plane. Hence, its angular velocity is downward.
L decreases the angular acceleration upward. so, net torque is upward.
b) Moment of inertia of the particle
I = m R^2
I = 2 x 3.1²
I = 19.22 kg.m²
L = I ω
ω = 
ω = 
ω = 
A = 0.52 rad/s B = -0.182 rad/s²
<h2>
Answer: 34.78 m/s</h2>
Explanation:
The momentum 
is given by the following equation:
(1)
Where:
is the mass of the object
is the velocity of the object
Finding the velocity from (1):
(2)
<u>Finally:</u>
>>>This is the velocity of the object
The carpal bones in the hands are an example of __________.
Answer: gliding joints
<span>A gliding joint means a freely moving joint in which the articulations allow only gliding motions</span>
