Answer:
1.045 m from 120 kg
Explanation:
m1 = 120 kg
m2 = 420 kg
m = 51 kg
d = 3 m
Let m is placed at a distance y from 120 kg so that the net force on 51 kg is zero.
By use of the gravitational force
Force on m due to m1 is equal to the force on m due to m2.
3 - y = 1.87 y
3 = 2.87 y
y = 1.045 m
Thus, the net force on 51 kg is zero if it is placed at a distance of 1.045 m from 120 kg.
Complete question:
The coordinate of a particle in meters is given by x(t)=1 6t- 3.0t³ , where the time tis in seconds. The
particle is momentarily at rest at t is:
Select one:
a. 9.3s
b. 1.3s
C. 0.75s
d.5.3s
e. 7.3s
Answer:
b. 1.3 s
Explanation:
Given;
position of the particle, x(t)=1 6t- 3.0t³
when the particle is at rest, the velocity is zero.
velocity = dx/dt
dx /dt = 16 - 9t²
16 - 9t² = 0
9t² = 16
t² = 16 /9
t = √(16 / 9)
t = 4/3
t = 1.3 s
Therefore, the particle is momentarily at rest at t = 1.3 s
Newton's first law of motion is an object in motion stays in motion until acted upon by another force. Driving at 30 mph in a car is going to stay constant until you crash the car into a wall, stopping the car.
I believe the answer is Ca
