When the spring is extended by 44.5 cm - 34.0 cm = 10.5 cm = 0.105 m, it exerts a restoring force with magnitude R such that the net force on the mass is
∑ F = R - mg = 0
where mg = weight of the mass = (7.00 kg) g = 68.6 N.
It follows that R = 68.6 N, and by Hooke's law, the spring constant is k such that
k (0.105 m) = 68.6 N ⇒ k = (68.6 N) / (0.105 m) ≈ 653 N/m
Explanation:
When m=<em>mass</em>
G=<em>a</em><em>c</em><em>c</em><em>e</em><em>l</em><em>e</em><em>r</em><em>a</em><em>t</em><em>i</em><em>o</em><em>n</em><em> </em><em>d</em><em>u</em><em>e</em><em> </em><em>t</em><em>o</em><em> </em><em>gravity</em>
<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>H</em><em>=</em><em>h</em><em>e</em><em>i</em><em>g</em><em>h</em><em>t</em>
<em>U</em><em>s</em><em>i</em><em>n</em><em>g</em><em> </em><em>f</em><em>o</em><em>r</em><em>m</em><em>u</em><em>l</em><em>a</em>
<em>M</em><em>g</em><em>h</em>
<em>(</em><em>M</em><em>=</em><em>6</em><em>, </em><em>g</em><em>=</em><em>10</em><em>,</em><em>h</em><em>=</em><em>?</em><em>) </em>
6×10×h
=60joules
Answer:
Fc = 19.2 N
Explanation:
In this case, the force of the Honda over the rock, is a centripetal force. Then, you have:

m: mass of the rock = 600g = 0.6 kg
v: tangential velocity of the Honda = 4m/s
r: radius of the Honda = 50cm = 0.5m
You replace the values of m, r and v in the equation for Fc:

hence, the force has a magnitude of 19.2 N
If the rock would have more mass the centripetal force would be higher
Answer:
7.04 m
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity = 0
s = Displacement on Earth = 1.2 m
a = Acceleration due to gravity on Moon = 1.67 m/s²
a = Acceleration due to gravity Earth= 9.81 m/s²
Accelration going up is considered as negetive
Initial Velocity of the ball

Assuming that the ball is thrown with the same velocity on the Moon, displacement of the ball is

The displacement of the ball on the moon is 7.04 m