D. it is thickest in the middle
Answer:
1,050 Joules
Explanation:
<u>Step 1:</u> work done in moving the box 30 meters
work done = force X distance
= 25N X 30 = 750 Joules
<u>Step 2: </u>calculate total internal energy
Total internal energy = work done + kinetic energy
= 750 Joules + 300 Joules
= 1,050 Joules = 1.05 KJ
Answer:
According to the data given in the question, experiment on table two pulling and falling masses are arranged in the fig. 250 g is pulling right side and 100 g pulling down. The gravitational force is common to both the masses, so we cannot say that the block moves towards heavier mass, also the block does not move towards the lighter mass.
Obviously, the effect of heavier mass of 250 g is more on the block, so the block moves towards right bottom corner. i.e., diagonally between two masses
please find the attachment.
Answer:
The equation of equilibrium at the top of the vertical circle is:
\Sigma F = - N - m\cdot g = - m \cdot \frac{v^{2}}{R}
The speed experimented by the car is:
\frac{N}{m}+g=\frac{v^{2}}{R}
v = \sqrt{R\cdot (\frac{N}{m}+g) }
v = \sqrt{(5\,m)\cdot (\frac{6\,N}{0.8\,kg} +9.807\,\frac{kg}{m^{2}} )}
v\approx 9.302\,\frac{m}{s}
The equation of equilibrium at the bottom of the vertical circle is:
\Sigma F = N - m\cdot g = m \cdot \frac{v^{2}}{R}
The normal force on the car when it is at the bottom of the track is:
N=m\cdot (\frac{v^{2}}{R}+g )
N = (0.8\,kg)\cdot \left(\frac{(9.302\,\frac{m}{s} )^{2}}{5\,m}+ 9.807\,\frac{m}{s^{2}} \right)
N=21.690\,N
Answer:
22 N applied force
Explanation: Since they are both pushing the wagon in the same direction the force adds up.
