The answer depends on the context:
You can find the acceleration if you know any three of : initial velocity, final velocity, time, distance travelled.
You can find it if you know the mass and force.
You know the two masses and the distance between them (gravitational acceleration).
weight
Find out the time using speed and acceleration, (time=speed/acceleration) and then use it to find out uniform velocity. From that find out uniform acceleration. (as uniform acceleration is equal changes of velocity over equal intervals of time)
Use the formula Acceleration = (final velosity - initial velocity)/ time.
F = M aa = F / M = 65 / 10 = 6.5 meters per second2
If by N you mean Newtons, then the formula to use would be Force=mass x acceleration. Newtons is the unit of measurement for force
You ignore the acceleration, and just give them the mass. Now, if they give you the acceleration and the applied force, you could use m = F/a.
Not enough information. One equation you can often use is Newton's Second Law: force = mass x acceleration Which, when solved for acceleration, gives you: acceleration = force / mass
If you are talking about problems involving Newton's second law of motion, F = ma, you would need to define two of the three variables of force, mass, and acceleration in order to find the third variable. If you have force and mass, you can find acceleration. If you have force and acceleration, you can find mass. If you have mass and acceleration you can find force.
Then the acceleration would also double.Then the acceleration would also double.Then the acceleration would also double.Then the acceleration would also double.
To find the theoretical acceleration, you must use the following formula: Acceleration Theory = Applied Net Force / Total Mass Applied Net force (F): F=Ma
Use the equation a=(v-u)/t, whereby v stands for final velocity, u for initial velocity and t for time.
The acceleration would be increased if the person pushing started to use a greater force. This would increase the force but keep the mass constant.