Answer:
Explanation:
total weight acting downwards
= 3g + 10g
13 g
volume of lead = 10 / 11.3 = .885 cm³
Let the volume of bobber submerged in water be v in floating position . buoyant force on bobber = v x 1 x g
Buoyant force on lead = .885 x 1 x g
total buoyant force = vg + .885 g
For floating
vg + .885 g = 13 g
v = 12.115 cm³
total volume of bobber
= 4/3 x 3.14 x 2³
= 33.5 cm³
fraction of volume submerged
= 12.115 / 33.5
= .36
= 36 %
Answer:
The tension in the cable when the craft was being lowered to the seafloor is 4700 N.
Explanation:
Given that,
When the craft was stationary, the tension in the cable was 6500 N.
When the craft was lowered or raised at a steady rate, the motion through the water added an 1800 N.
The drag force of 1800 N will act in the upward direction. As it was lowered or raised at a steady rate, so its acceleration is 0. As a result, net force is 0. So,
T + F = W
Here, T is tension
F = 1800 N
W = 6500 N
Tension becomes :
So, the tension in the cable when the craft was being lowered to the seafloor is 4700 N.
Answer:
Explanation:
Let electric potential at A ,B and C be Va , Vb and Vc respectively.
Work done = charge x potential difference
Wab = q ( Va - Vb )
Wac = q ( Va - Vc )
Given
Wac = - Wab / 3
3Wac = - Wab
Now
Wbc = q ( Vb - Vc )
= q [ ( Va-Vc ) - ( Va - Vb )]
= Wac - Wab
= Wac + 3Wac
= 4Wac
Neglecting air resistance, the horizontal component remains constant. The angle doesn't matter.
Answer:
Change in potential energy of the block-spring-Earth
system between Figure 1 and Figure 2 = 1 Nm.
Explanation:
Here, spring constant, k = 50 N/m.
given block comes down eventually 0.2 m below.
here, g = 10 m/s.
let block be at a height h above the ground in figure 1.
⇒In figure 2,
potential energy of the block-spring-Earth
system = m×g×(h - 0.2) + 1/2× k × x². where, x = change in spring length.
⇒ Change in potential energy of the block-spring-Earth
system between Figure 1 and Figure 2 = (m×g×(h - 0.2)) - (1/2× k × x²)
= (1×10×0.2) - (1/2×50×0.2×0.2) = 1 Nm.
