Answer:
Force in the rocket will be 4166.64 N
Explanation:
We have given mass of the rocket m = 3 kg
Rocket acquires a speed of 50 m sec so final speed v = 50 m/sec
Initial speed u = 0 m/sec
Distance traveled s = 90 cm = 0.9 m
From third equation of motion we know that
From newton's law we F = ma
So force will be
So force in the rocket will be 4166.64 N
Answer:
4.8 m
Explanation:
When it hits the ground, all the kinetic energy is converted to potential energy. Potential energy, PE is given by mgh where m is the mass of the box, g is acceleration due to gravity and h is the height which is unknown for this case
PE=mgh and making h the subject of the formula then and substituting g with 9.81, PE with 1500 J and m with 32 kg then
Answer:
0.5 m
Explanation:
Two charges each of magnitude q
Let the third charge is Q is placed at a distance x from the origin so that the charge is in equilibrium.
The force on Q due to q at origin is balanced by the charge on Q due to the charge q placed at x = 1 m.
So,
1 - x = x
1 = 2x
x = 0.5 m
Thus, the third charge is placed at x = 0.5 m .