Answer:
The precise point is that gravitational energy is potential energy unless it makes something move, and then the energy is converted to kinetic energy, no work can be done. So a big rock at the top of a hill has no kinetic energy. Its only when it rolls down that work is done, and this can be converted to useful energy.
Explanation:
Answer:
Explanation:
As we know that current density is ratio of current and area of the crossection
now we have
so the current through the wire is given as
now we have
here we have
now plug in the values in above equation
now we have
now plug in both limits as mentioned
here R = 2.11 mm
Answer:
F=mg
F= 0.153kg x 9.8 m/s^2= 1.5 N
Answer:
20 m
Explanation:
We'll begin by calculating the kinetic energy of the mass. This can be obtained as follow:
Mass (m) = 10 kg
Velocity (v) = 20 m/s
Kinetic energy (KE) =?
KE = ½mv²
KE = ½ × 10 × 20²
KE = 5 × 400
KE = 2000 J
Finally, we shall the height to which the mass must be located in order to have potential energy that is the same as the kinetic energy. This can be obtained as follow:
Mass (m) = 10 kg
Acceleration due to gravity (g) = 10 m/s²
Potential energy (PE) = Kinetic energy (KE) = 2000 J
Height (h) =..?
PE = mgh
2000 = 10 × 10 × h
2000 = 100 × h
Divide both side by 100
h = 2000 / 100
h = 20 m
Thus, the object must be located at a height of 20 m in order to have potential energy that is the same as the kinetic energy.
Let m₁ = 3.0 kg and v₁ = + 8 m/s (so right is positive), and m₂ = 1.0 kg and v₂ = 0. The total momentum of the two balls before and after collision is conserved, so
m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
where v₁' = + 5 m/s and v₂' are the velocities of the two balls after colliding, so
(3.0 kg) (8 m/s) = (3.0 kg) (5 m/s) + (1.0 kg) v₂'
Solve for v₂' :
24 kg•m/s = 15 kg•m/s + (1.0 kg) v₂'
(1.0 kg) v₂' = 9 kg•m/s
v₂' = (9 kg•m/s) / (1.0 kg)
v₂' = + 9 m/s
which is to say, the second ball is given a speed of 9 m/s to the right after colliding with the first ball.
