Equations:
F=m*a
d=vi*t+1/2*a*t²
Given Variables
m=.5kg
vi=0m/s
d=8m
t=2s
8=0m/s*2s+1/2*a*2s²
a=4m/s²
F=.5kg*4m/s
F=2N
8/2=4
Force = mass * acceleration you do the math
If, somehow miraculously, the block is on a frictionless slide, and there isno frictional force opposing its horizontal motion, thenF = M AA = F/M = 583/44 = 13.25 m/s2 = about 1.35 G's
90kg
I think so. I'm not a physics expert but see if this makes sense to you.Force = mass * accelerationdouble the mass and you getForce = 2 * mass * accelerationdouble the force with double the mass and you get2 * Force = 2 * mass * accelerationusing simple division math you cancel the 2's which results inForce = mass * accelerationSo I think it would be the same acceleration.
Remember Newton's Second Law of Universal Dynamics. Put into algebraic form it is F = ma To answer your question a = F/m Acceleration is equal to force(F) divided by mass(m) of the object being pushed/[pulled.
Well it's obviously not flying.
the force acting down the slope = sin 30 * 25 (kg) = 0.5 * 25 = 12.5 kg = (12.5 * 9.806 ) 122.575 newtons = resultant force of (122.575-120 ) 2.575 newtons downhill giving downhill acceleration of (using f=ma) 0.103 (m/s)/s
weight is a function of (mass * acceleration due to gravity) and is a force acting toward the earths centre (vertically down) . if your pulling force is horizontal, then it wont affect the weight
Force=mass*acceleration 80N=10kg*acceleration 80N/10kg=acceleration 8m/s2=acceleration The acceleration is 8m/s2.
The acceleration in the block will be 4.59 m/s2
Force = mass * acceleration you do the math
No, but in a sleigh pulled by reindeer.
If, somehow miraculously, the block is on a frictionless slide, and there isno frictional force opposing its horizontal motion, thenF = M AA = F/M = 583/44 = 13.25 m/s2 = about 1.35 G's
F=mass * acceleration 60kg m/s^2=10kg * acceleration 6m/s^2 = acceleration
true
Santa travels on his sleigh, pulled by reindeer.
F = m aa = F/m = (800) / (1,000) = 0.8 m/sec2