Answer:
yes you can find a motor in a circuit
It will take a shorter amount of time for the cylinder to go down the plane down off the plane Because more pressure is applied one going up then going down there’s no pressure at all it’s the gravity is helping
Answer:
Explanation:
Before it hits the ground:
The initial potential energy = the final potential energy + the kinetic energy
mgH = mgh + 1/2 mv²
gH = gh + 1/2 v²
v = √(2g (H - h))
v = √(2 * 9.81 m/s² * (0.42 m - 0.21 m))
v ≈ 2.0 m/s
When it hits the ground:
Initial potential energy = final kinetic energy
mgH = 1/2 mv²
v = √(2gH)
v = √(2 * 9.81 m/s² * 0.42 m)
v ≈ 2.9 m/s
Using a kinematic equation to check our answer:
v² = v₀² + 2a(x - x₀)
v² = (0 m/s)² + 2(9.8 m/s²)(0.42 m)
v ≈ 2.9 m/s
Answer:
The magnitude of the torque due to gravity if it is supported at the 28-cm mark is 0.5 N-m.
Explanation:
Given that,
Mass of the meter stick, m = 0.3 kg
Center of mass is located at its 45 cm mark.
We need to find the magnitude of the torque due to gravity if it is supported at the 28-cm mark. Torque acting on the object is given by :
or
So, the magnitude of the torque due to gravity if it is supported at the 28-cm mark is 0.5 N-m.
