I would help but I’m also taking this
Answer:
Mass of the pull is 77 kg
Explanation:
Here we have for
Since the rope moves along with pulley, we have
For the first block we have
T₁ - m₁g = -m₁a = -m₁g/4
T₁ = 3/4(m₁g) = 323.4 N
Similarly, as the acceleration of the second block is the same as the first block but in opposite direction, we have
T₂ - m₂g = m₂a = m₂g/4
T₂ = 5/4(m₂g) = 134.75 N
T₂r - T₁r = I·∝ = 0.5·M·r²(-α/r)
∴ 
Mass of the pull = 77 kg.
Answer:
450
Explanation:
Given,
Mass= 100kg
Velocity= 3 m/s
Kinetic Energy= ?
Kinetic Energy= 1/2 mv^2
= 1/2× 100× 3^2
= 1/2× 900
= 450.
<em>HOPE</em><em> </em><em>IT</em><em> </em><em>HELPED</em><em> </em><em>:</em><em>)</em>
<span>Greek philosophers had a basic approach to studying the world. They like to question the world and incite debates but they never really bothered to gather any real information, just discussions. Due to this, many ideas about matters were put out to be discussed, but they were never resolved.</span>
<span>3.92 m/s^2
Assuming that the local gravitational acceleration is 9.8 m/s^2, then the maximum acceleration that the truck can have is the coefficient of static friction multiplied by the local gravitational acceleration, so
0.4 * 9.8 m/s^2 = 3.92 m/s^2
If you want the more complicated answer, the normal force that the crate exerts is it's mass times the local gravitational acceleration, so
20.0 kg * 9.8 m/s^2 = 196 kg*m/s^2 = 196 N
Multiply by the coefficient of static friction, giving
196 N * 0.4 = 78.4 N
So we need to apply 78.4 N of force to start the crate moving. Let's divide by the crate's mass
78.4 N / 20.0 kg
= 78.4 kg*m/s^2 / 20.0 kg
= 3.92 m/s^2
And you get the same result.</span>
