Answer:
A ball moving through the air.
Explanation:
The ball has momentum which is a form of kinetic energy.
I don't know if that is correct, but I hope it helps!!!!
Answer:
(a). The charge on the outer surface is −2.43 μC.
(b). The charge on the inner surface is 4.00 μC.
(c). The electric field outside the shell is 
Explanation:
Given that,
Charge q₁ = -4.00 μC
Inner radius = 3.13 m
Outer radius = 4.13 cm
Net charge q₂ = -6.43 μC
We need to calculate the charge on the outer surface
Using formula of charge
The charge on the inner surface is q.
We need to calculate the electric field outside the shell
Using formula of electric field
Put the value into the formula
Hence, (a). The charge on the outer surface is −2.43 μC.
(b). The charge on the inner surface is 4.00 μC.
(c). The electric field outside the shell is
Answer:
The approximate terminal velocity of a sky diver before the parachute opens is 320 km/h.
Explanation:
- The terminal velocity is the maximum magnitude of velocity that is attained by the diver when he or she falls in the air.
- The terminal velocity of the person diving in air before opening parachute is 320 km/h that means the velocity when the person is experiencing free fall is 320 km/h.
- During terminal velocity, we can represent mathematical equation as;
Buoyancy force + drag force = Gravity
Answer:
a) the elastic force of the pole directed upwards and the force of gravity with dissects downwards
Explanation:
The forces on the athlete are
a) at this moment the athlete presses the garrolla against the floor, therefore it acquires a lot of elastic energy, which is absorbed by the athlete to rise and gain potential energy,
therefore the forces are the elastic force of the pole directed upwards and the force of gravity with dissects downwards
b) when it falls, in this case the only force to act is batrachium by the planet, this is a projectile movement for very high angles
c) When it reaches the floor, it receives an impulse that opposes the movement created by the mat. The attractive force is the attraction of gravity.
Answer:
r = 58.44 [m]
Explanation:
To solve this problem we must use the following equation that relates the centripetal acceleration with the tangential velocity and the radius of rotation.
a = v²/r
where:
a = centripetal acceleration = 15.4 [m/s²]
v = tangential speed = 30 [m/s]
r = radius or distance [m]
r = v²/a
r = 30²/15.4
r = 58.44 [m]
