Answer:
26.82m/s
Explanation:
Given
Mass = m= 0.4kg
Initial Velocity = u = 0
Charge = 4.0E-5C
Distance= d = 0.5m
Object Charge = 2E-4C
First, we'll calculate the initial energy (E)
E = Potential Energy
PE = kQq / d
Where k = coulomb constant = 8.99E9Nm²/C²
Energy is then calculated by;
PE = 8.99E9 * 4E-5 * 2E-4 / 0.5
PE = 143.84J
Energy = Potential Energy = Kinetic Energy
K.E = ½mv² = 143.84J
½mv² = ½ * 0.40 * v² = 143.85
0.2v² = 143.85
v² = 143.85/0.2
v² = 719.25
v = √719.25
v = 26.81883666380777
v = 26.82m/s
Hence, the object is 26.82m/s fast when the cart moving is very far (infinity) from the fixed charge
A ball falling through the air has a mass, a density, a volume...it is facing air resistance and is being acted on by gravity...it is accelerating and gaining velocity...and it is increasing in kinetic energy.
I suppose out of all those the biggest thing the ball has in this case is ENERGY. There are two main types to focus on...
Kinetic Energy - The further the ball fall the more KE it has...until terminal velocity is reach, then KE would become constant.
Potential Energy - Conversely to that of KE, the further the ball falls the less PE it will have.
<em>Heat/Thermal Energy is technically also present due to the friction from the air resistance, but the transfer of energy between the air and ball is quite complex and not necessary important for basic physics.
</em>
The question itself seem kind of vague and open ended, but I could just be viewing it the wrong way.
Comment if you need more help!
Answer:
this is just a guess bc i only looked at it for 5 seconds but i think 150 m/s
I think it's C, longer wave length.
Answer:
0.167m/s
Explanation:
According to law of conservation of momentum which States that the sum of momentum of bodies before collision is equal to the sum of the bodies after collision. The bodies move with a common velocity after collision.
Given momentum = Maas × velocity.
Momentum of glider A = 1kg×1m/s
Momentum of glider = 1kgm/s
Momentum of glider B = 5kg × 0m/s
The initial velocity of glider B is zero since it is at rest.
Momentum of glider B = 0kgm/s
Momentum of the bodies after collision = (mA+mB)v where;
mA and mB are the masses of the gliders
v is their common velocity after collision.
Momentum = (1+5)v
Momentum after collision = 6v
According to the law of conservation of momentum;
1kgm/s + 0kgm/s = 6v
1 =6v
V =1/6m/s
Their speed after collision will be 0.167m/s