Answer:
d) the amount of work is the same whether the bag is moved all at once or in two stages, provided the total height lifted is the same in either case.
Explanation:
While moving the bag to the shelf in one shot we can say that the total work done is given as
here we know that
2H = total height raised by the bag
now when we raise the bag to first shelf and then move it to next shelf
then we will have
![W = W_1 + W_2[tex][tex]W = mgH + mgH](https://tex.z-dn.net/?f=W%20%3D%20W_1%20%2B%20W_2%5Btex%5D%3C%2Fp%3E%3Cp%3E%5Btex%5DW%20%3D%20mgH%20%2B%20mgH)
so the correct answer will be
d) the amount of work is the same whether the bag is moved all at once or in two stages, provided the total height lifted is the same in either case.
Explanation:
H= mass× specific heat capacity×change in temperature
=2×380×(10-0)
=2×380×10
=7600Joules
Answer:
F = 1.2×10⁻³ N
Explanation:
From the question,
Applying newton's second law of motion,
F = m(v-u)/t................... Equation 1
Given: F = magnitude of the average force exerted on the ball, m = mass of the ball, v = final velocity, u = initial velocity, t = time of contact.
Note: let downward be negative and upward be positive.
Given: m = 48 g = 48/1000 = 0.048 kg, v = 17 m/s, u = -28 m/s (downward),
t = 1800 s
Substitute into equation 1
F = 0.048(17-[28])/1800
F = 1.2×10⁻³ N
I want to improve my speed
Answer:
Explanation:
For a body on a ramp with mass m, the forces acting on the body along the vertical component are the weight and the normal reaction.
The weight of the body acts in the negative y direction while the normal reaction acts in the positive y direction
Taking the sum of forces along the y component
Sum Fy = -W+R = ma
Since acceleration is zero
-W+R = m(0)
-W+R = 0
-W = -R
W = R
Hence the Normal reaction force acting on the on the body is equal to normal force
