The equation for work (W) done by an electric field is:
W = qΔV
where q is the magnitude of the charge and ΔV is the potential difference. The question gives you W and q, so plug n' play to find ΔV:
10 = 2ΔV
ΔV = 5
The correct answer is 1.2 m/s
: mv+mv=mv+mv
(0.5kg)(2m/s)+(0.4kg)(0m/s)=(0.5kg)v+(0.4kg)(1m/s)
= 1kg*m/s=(0.5kg)v+0.4kg*m/s
=1kg*m/s-0.4kg*m/s=(0.5kg)v
=0.6kg*m/s=(0.5kg)v
to solve for v we divide both side by 0.5kg
v=1.2m/s
Answer:
(a) the work done by the student is 110.1 J
(b) The gravitational force that acts on the amplifier is 102.9 N
Explanation:
Given;
mass of the amplifier, m = 10.5 kg
initial position of the amplifier, x₀ = 1.82 m
final position of the amplifier, x₁ =0.75 m
The dispalcement of the amplifier Δx = x₁ - x₀ = 1.82 m - 0.75 m = 1.07 m
(b) The gravitational force that acts on the amplifier;
F = mg
F = 10.5 x 9.8
F = 102.9 N
(a) the work done by the student is calculated as;
W = FΔx
W = 102.9 x 1.07
W = 110.1 J
Answer:
we need the graph to answer the question.
