The work done by force on a spring hung from the ceiling will be 1.67 J
Any two things with mass are drawn together by the gravitational pull. We refer to the gravitational force as attractive because it consistently seeks to draw masses together rather than pushing them apart.
Given that a spring is hung from the ceiling with a 2.0-kg mass suspended hung from the spring extends it by 6.0 cm and a downward external force applied to the mass extends the spring an additional 10 cm.
We need to find the work done by the force
Given mass is of 2 kg
So let,
F = 2 kg
x = 0.1 m
Stiffness of spring = k = F/x
k = 20/0.006 = 333 n/m
Now the formula to find the work done by force will be as follow:
Workdone = W = 0.5kx²
W = 0.5 x 333 x 0.1²
W = 1.67 J
Hence the work done by force on a spring hung from the ceiling will be 1.67 J
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Since D=M/V, the answer would be 2.7
a) It is absolute, so it does not change.
b) Inertial ones.
c) Inside the train the time will slow down relatively to the outside clock. So if one travel at nearly the speed if light for 2 hours on his clock, for outdoor observers it will look like 3 hours.
We take the derivative of Ohm's law with respect to time: V = IR
Using the product rule:
dV/dt = I(dR/dt) + R(dI/dt)
We are given that voltage is decreasing at 0.03 V/s, resistance is increasing at 0.04 ohm/s, resistance itself is 200 ohms, and current is 0.04 A. Substituting:
-0.03 V/s = (0.04 A)(0.04 ohm/s) + (200 ohms)(dI/dt)
dI/dt = -0.000158 = -1.58 x 10^-4 A/s
