The speed of the second mass after it has moved ℎ=2.47 meters will be 1.09 m/s approximately
<h3>
What are we to consider in equilibrium ?</h3>
Whenever the friction in the pulley is negligible, the two blocks will accelerate at the same magnitude. Also, the tension at both sides will be the same.
Given that a large mass m1=5.75 kg and is attached to a smaller mass m2=3.53 kg by a string and the mass of the pulley and string are negligible compared to the other two masses. Mass 1 is started with an initial downward speed of 2.13 m/s.
The acceleration at which they will both move will be;
a = ( - ) / ( + )
a = (5.75 - 3.53) / (5.75 + 3.53)
a = 2.22 / 9.28
a = 0.24 m/s²
Let us assume that the second mass starts from rest, and the distance covered is the h = 2.47 m
We can use third equation of motion to calculate the speed of mass 2 after it has moved ℎ=2.47 meters.
v² = u² + 2as
since u =0
v² = 2 × 0.24 × 2.47
v² = 1.1856
v = √1.19
v = 1.0888 m/s
Therefore, the speed of mass 2 after it has moved ℎ=2.47 meters will be 1.09 m/s approximately
Learn more about Equilibrium here: brainly.com/question/517289
#SPJ1
Answer:
Explanation:
Work done to move a charge through a given potential difference is
here
q = magnitude of charge
= potential difference
here it is given that charge of the particle is same as charge of proton
now we have
Answer:
Mass, m = 32 kg
Explanation:
Given that,
Force, F = 2000 N
The change in velocity, (v-u) = 50 m/s
Time, t = 0.8 s
We need to find the mass of the ball. The net force acting on an object is given by :
F = ma
Where a is acceleration
So, the mass of the ball is 32 kg.
Answer:
(A) Torque required is 21.205 N-m
(b) Wok done will be equal to 1199.1286 j
Explanation:
We have given moment of inertia
Wheel deaccelerate from 135 rpm to 0 rpm
135 rpm =
Time t = 8 sec
So angular speed and
Angular acceleration is given by
Torque is given by torque
Work done to accelerate the vehicle is
The change in momentum of a 3kg object is 12