Answer:
3500j
0.250 kg
14.14 m/s
220500 j
93.91 m/s
Explanation:
K.E =1/2 m v^2
first question
1/2×700×10=3500 j
second question
2×78.2/25^2= 0.250 kg
Third question
8kj=8000j
root (2×8000/80)= 14.14 m/s
Last one
P.E=MGH
so it will be
P.E=50×450×9.8=220500 j
the speed
root (2×220500/50)=93.91 m/s
To solve this problem we will apply the given concept for torque which explains the relationship between the force applied and the distance to a given point. Mathematically this relationship is given as
Where,
Torque
F = Force
d = Distance
Our values are given as,
Therefore replacing we have that the force is
F = 12.72N
Therefore the least amount of force that you must exert is 12.72N
Answer:
Hooke's law states that the applied force F equals a constant k times the displacement or change in length x, or F = kx. ... Hooke's law describes the elastic properties of materials only in the range in which the force and displacement are proportional.
Explanation:
D.a ball sitting on a shelf
