Answer:
Hello your question is incomplete attached is the missing diagram and solution
One elevator arrangement includes the passenger car, a counterweight, and two large pulleys, as shown in (Figure 1). Each pulley has a radius of 1.2 m and a moment of inertia of 410 kg⋅m2. The top pulley is driven by a motor. The elevator car plus passengers has a mass of 3000 kg, and the counterweight has a mass of 2600 kg.
a) If the motor is to accelerate the elevator car upward at 1.8 m/s2, how much torque must it generate? Express your answer to two significant figures and include appropriate units.
Answer : 6030 N.m ≈ 6000 N.m
Explanation:
To determine how much torque the motor accelerating at 1.8 m/s^s will generate we will have to determine T1 ( tension generated for upward acceleration in the rope ) and Tg ( tension generated in the rope when there is a counterweight accelerating downwards)
Torque in the pulley (T2) = (T1 - Tg ) r
torque generated by motor = T1 + T2 = 1230 + 4800 = 6030 N.m to two significant figures = 6000 N.m
r = 1.2
a = 1.8
attached below is the remaining part of the solution
Answer:
The speed of q₂ is
Explanation:
Given that,
Distance = 0.4 m apart
Suppose, A small metal sphere, carrying a net charge q₁ = −2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q₂ = −8μC and mass 1.50g, is projected toward q₁. When the two spheres are 0.800m apart, q₂ is moving toward q₁ with speed 20m/s.
We need to calculate the speed of q₂
Using conservation of energy
Put the value into the formula
Hence, The speed of q₂ is
Answer:
<em>The velocity of the two cars is 10 m/s after the collision.</em>
Explanation:
<u>Law Of Conservation Of Linear Momentum </u>
The total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and velocity v is
P=m.v
If we have a system of bodies, then the total momentum is the sum of them all
If some collision occurs, the velocities change to v' and the final momentum is:
In a system of two masses, the law of conservation of linear momentum takes the form:
If both masses stick together after the collision at a common speed v', then:
The car of mass m1=1000 Kg travels at v1=25 m/s and collides with another car of m2=1500 Kg which is at rest (v2=0).
Knowing both cars stick and move together after the collision, their velocity is found solving for v':
v' = 10 m/s
The velocity of the two cars is 10 m/s after the collision.