The farther the distance, the weaker the gravitational force
Answer:
Zeros to the left of a decimal can be insignificant place holders, such as in 0.043 (two significant figures).
They can be significant if they are between two digits who themselves are significant, such as in 101.000 (three significant figures).
In the case of a number like 1,000 we can see there is only one significant figure. The zero digits are not between sigfigs.
It really doesn't matter whether the two masses are joined by a light string, a light reinforced concrete and steel frame, a light coating of strong glue, or a layer of scotch tape that's one atom thick. They look like a single mass of 10 kg.
<u>Part a). </u>Force = (mass) x (acceleration)
The acceleration of 10 kg of mass pulled with 100N of force is
100 N = (10 kg) x (acceleration)
Divide each side by (10 kg):
Acceleration = (100 N) / (10 kg)
<em>Acceleration = 10 m/s²</em>
<u>Part b).</u> Now look at just the rear 16 kg. It stays hooked onto the front 4 kg, and everything moves together. So both pieces must have the same distance, velocity, acceleration etc. We know that the acceleration is 10 m/s² .
Force = (mass) x (acceleration)
Force on the rear 16 kg = (16 kg) x (10 m/s²)
<em>Force = 160 Newtons</em>
Deceleration due to rolling resistance (α) = coefficient*g/R = 0.1*9.81/0.5 = 1.962 rad/sec^2
Equations of motion:
ωf=ωo+αt
But ωf = 0 rad/sec; ωo= 2*pi*RPM/60 = 4.19 rad/sec
Therefore,
0=4.19-1.962t => 1.962t = 4.19 => t = 4.19/1.962 = 2.13 seconds
The answer is: Salt! :)
Have a great day!
