<span>49N is the force needed to give a .25 kg arrow an acceleration of 196m/s2. F =ma ⇒ =( 0.25kg)(196m/s2) = 49N if the arrow is shot horizontally where the applied force is entirely in the x-direction.</span>
Answer:
physical feature of a wave is related to the depth of the wave base is The circular orbital motion
B. The wave base is the depth, and the still water level is the horizontal level
Answer:
<h2>1567.09 N/m</h2>
Explanation:
Step one:
given data
mass m=5kg
compression x= 3.13cm to m= 0.0313m
<em>According to Hooke's law, provided the elastic limit of an elastic material is not exceeded the extension e is directly proportional to the applied force</em>
F=ke
where
k= spring constant in N/m
e= extension/compression in
Step two:
assume g= 9.81m/s^2
F=mg
F=5*9.81
F=49.05N
substitute in the expression F=ke
49.05=k*0.0313
k=49.05/0.0313
k=1567.09 N/m
<u>The force constant (in N/m) of the spring is 1567.09 N/m</u>
