Answer:
The final velocity of the bullet is 9 m/s.
Explanation:
We have,
Mass of a bullet is, m = 0.05 kg
Mass of wooden block is, M = 5 kg
Initial speed of bullet, v = 909 m/s
The bullet embeds itself in the block which flies off its stand. Let V is the final velocity of the bullet. The this case, momentum of the system remains conserved. So,

So, the final velocity of the bullet is 9 m/s.
Answer:
Kinetic energy, E = 133.38 Joules
Explanation:
It is given that,
Mass of the model airplane, m = 3 kg
Velocity component, v₁ = 5 m/s (due east)
Velocity component, v₂ = 8 m/s (due north)
Let v is the resultant of velocity. It is given by :


Let E is the kinetic energy of the plane. It is given by :


E = 133.38 Joules
So, the kinetic energy of the plane is 133.38 Joules. Hence, this is the required solution.
Answer:
KE= 1/2mv²
Explanation:
The kinetic energy of a body is the energy possessed by virtue of the body in motion
Given the parameters
m which is the mass of the body
v which is the velocity of the body too
K.E = kinetic energy
The expression for the kinetic energy of a body is given as
KE= 1/2mv²
k = spring constant of the spring = 85 N/m
m = mass of the box sliding towards the spring = 3.5 kg
v = speed of box just before colliding with the spring = ?
x = compression the spring = 6.5 cm = 6.5 cm (1 m /100 cm) = 0.065 m
the kinetic energy of box just before colliding with the spring converts into the spring energy of the spring when it is fully compressed.
Using conservation of energy
Kinetic energy of spring before collision = spring energy of spring after compression
(0.5) m v² = (0.5) k x²
m v² = k x²
inserting the values
(3.5 kg) v² = (85 N/m) (0.065 m)²
v = 0.32 m/s
Answer with Explanation:
We are given that


Charge on proton,q=
a.We have to find the electric potential of the proton at the position of the electron.
We know that the electric potential

Where 


B.Potential energy of electron,U=
Where
Charge on electron
=Charge on proton
Using the formula

