The formula is=1/2(m x v^2)
so = 1/2*(0.05)*(310)^2
ans is =2402.5 joules
Thermal- transfer of heat thru space
Radiation- the average amount of energy of motion in the molecules of a substance
Thermometer- a thin glass tube with a bulb on one end that contains a liquid, usually mercury or colored alcohol
Brainly?
GPE I am assuming is gravitational potential energy. I'll denote it as U for simplicity.
U = mgy
U = (70kg)(9.81m/s^2)(1m) = 686.7J
U = 686.7J
Hope this helps!
Answer:
Magnitude of static friction force is 70 sin40° = 44.99 N.
No, it is not necessary that it is maximum static friction.
Normal force is equal to 70 cos40° = 53.62 N.
Explanation:
We apply newton law of moton equation along the plane and perpendicular to plane;
Along the plane,
70 sin 40° = ---------------(1)
70 cos 40° = N --------------(2)
= μN -----------------(3)
So, it depends on the value of μ that the friction is maximum or not .