Answer:
The total momentum before and after collision is 72000 kg-m/s.
Explanation:
Given that,
Mass of car = 1200 kg
Velocity of car = 10 m/s
Mass of truck = 2000 kg
Velocity of truck = 30 m/s
Using conservation of momentum
The total momentum before the collision is equal to the total momentum after collision.
Where, 
=mass of car
=velocity of car
=mass of truck
=velocity of truck
Put the value into the formula
Now, The total momentum before collision is
The total momentum after collision is
Hence, The total momentum before and after collision is 72000 kg-m/s.
Answer:
The value is 
Explanation:
The radius is 
The mass of the ordinary matter is 
Generally the speed of the star is mathematically represented as
Here G is the gravitational constant with a value
So
=>
efficiency= [useful energy transferred ÷ total energy supply]×100%
So, [5500÷10000]×100%=0.55×100
=55%
eat do work screen time eat dinner and have fun and seep:)
have a good day
Answer:
v_2 = 2*v
Explanation:
Given:
- Mass of both charges = m
- Charge 1 = Q_1
- Speed of particle 1 = v
- Charge 2 = 4*Q_1
- Potential difference p.d = 10 V
Find:
What speed does particle #2 attain?
Solution:
- The force on a charged particle in an electric field is given by:
F = Q*V / r
Where, r is the distance from one end to another.
- The Net force acting on a charge accelerates it according to the Newton's second equation of motion:
F_net = m*a
- Equate the two expressions:
a = Q*V / m*r
- The speed of the particle in an electric field is given by third kinetic equation of motion.
v_f^2 - v_i^2 = 2*a*r
Where, v_f is the final velocity,
v_i is the initial velocity = 0
v_f^2 - 0 = 2*a*r
Substitute the expression for acceleration in equation of motion:
v_f^2 = 2*(Q*V / m*r)*r
v_f^2 = 2*Q*V / m
v_f = sqrt (2*Q*V / m)
- The velocity of first particle is v:
v = sqrt (20*Q / m)
- The velocity of second particle Q = 4Q
v_2 = sqrt (20*4*Q / m)
v_2 = 2*sqrt (20*Q / m)
v_2 = 2*v
