A. True
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D ............................
The work done will be equal to the potential energy of the piano at the final position
P.E=m.g.h
.consider the plank the hypotenuse of the right triangle formed with the ground
.let x be the angle with the ground=31.6°
.h be the side opposite to the angle x (h is the final height of the piano)
.let L be the length of the plank
sinx=opposite side / hypotenuse
= h/L
then h=L.sinx=3.49×sin31.6°=0.638m
weight w=m.g
m=w/g=3858/10=385.8kg
Consider Gravity g=10m/s2
then P.E.=m.g.h=385.8kg×10×0.638=2461.404J
then Work W=P.E.=2451.404J
<span>Px = 0
Py = 2mV
second, Px = mVcosφ
Py = –mVsinφ
add the components
Rx = mVcosφ
Ry = 2mV – mVsinφ
Magnitude of R = âš(Rx² + Ry²) = âš((mVcosφ)² + (2mV – mVsinφ)²)
and speed is R/3m = (1/3m)âš((mVcosφ)² + (2mV – mVsinφ)²)
simplifying
Vf = (1/3m)âš((mVcosφ)² + (2mV – mVsinφ)²)
Vf = (1/3)âš((Vcosφ)² + (2V – Vsinφ)²)
Vf = (V/3)âš((cosφ)² + (2 – sinφ)²)
Vf = (V/3)âš((cos²φ) + (4 – 2sinφ + sin²φ))
Vf = (V/3)âš(cos²φ) + (4 – 2sinφ + sin²φ))
using the identity sin²(Ď)+cos²(Ď) = 1
Vf = (V/3)âš1 + 4 – 2sinφ)
Vf = (V/3)âš(5 – 2sinφ)</span>
Answer:’B’
Explanation:Since, the nucleus is in the middle and it carries protons which carry positive charge and neutrons which carry no charge.