<h2><u>Question</u><u>:</u><u>-</u></h2>
Ryan applied a force of 10N and moved a book 30 cm in the direction of the force. How much was the work done by Ryan?
<h2><u>Answer:</u><u>-</u></h2>
<h3>Given,</h3>
=> Force applied by Ryan = 10N
=> Distance covered by the book after applying force = 30 cm
<h3>And,</h3>
30 cm = 0.3 m (distance)
<h3>So,</h3>
=> Work done = Force × Distance
=> 10 × 0.3
=> 3 Joules

Answer:
d) -4.0
Explanation:
The magnification of a lens is given by

where
M is the magnification
q is the distance of the image from the lens
p is the distance of the object from the lens
In this problem, we have
p = 50 cm is the distance of the object from the lens
q = 250 cm - 50 cm is the distance of the image from the lens (because the image is 250 cm from the obejct
Also, q is positive since the image is real
So, the magnification is

The rate of acceleration of the crate would be 1 m/s^2 because the equation for force is F=ma and when you plug in your numbers you get 10=10a so a=1
Answer:
∆PE = 749.7 J
At 0.9 m high, PE = 793.8 J
At 1.75 m high, PE = 1543.5 J