Answer:
9. A 1500kg car traveling +6m/s with a 2000kg truck at rest. The vehicles collide, but do not stick together. The car has a velocity -3m/s after the collision. What is the velocity of the truck? a. What type of collision occurred above?
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Answer:
A) Emin = eV
B) Vo = (E_light - Φ) ÷ e
Explanation:
A)
Energy of electron is the product of electron charge and the applied potential difference.
The energy of an electron in this electric field with potential difference V will be eV. Since this is the least energy that the electron must reach to break out, then the minimum energy required by this electron will be;
Emin = eV
B)
The maximum stopping potential energy is eVo,
The energy of the electron due to the light is E_light.
If the minimum energy electron must posses is Φ, then the minimum energy electron must have to reach the detectors will be equal to the energy of the light minus the maximum stopping potential energy
Φ = E_light - eVo
Therefore,
eVo = E_light - Φ
Vo = (E_light - Φ) ÷ e
Answer:



Explanation:
= Torque = 36.5 Nm
= Initial angular velocity = 0
= Final angular velocity = 10.3 rad/s
t = Time = 6.1 s
I = Moment of inertia
From the kinematic equations of linear motion we have

Torque is given by

The wheel's moment of inertia is 
t = 60.6 s
= 10.3 rad/s
= 0

Frictional torque is given by

The magnitude of the torque caused by friction is 
Speeding up

Slowing down

Total number of revolutions


The total number of revolutions the wheel goes through is
.
Answer:
V = 20.5 m/s
Explanation:
Given,
The mass of the cart, m = 6 Kg
The initial speed of the cart, u = 4 m/s
The acceleration of the cart, a = 0.5 m/s²
The time interval of the cart, t = 30 s
The final velocity of the cart is given by the first equation of motion
v = u + at
= 4 + (0.5 x 30)
= 19 m/s
Hence the final velocity of cart at 30 seconds is, v = 19 m/s
The speed of the cart at the end of 3 seconds
V = 19 + (0.5 x 3)
= 20.5 m/s
Hence, the final velocity of the cart at the end of this 3.0 second interval is, V = 20.5 m/s