Answer:
21.21 m/s
Explanation:
Let KE₁ represent the initial kinetic energy.
Let v₁ represent the initial velocity.
Let KE₂ represent the final kinetic energy.
Let v₂ represent the final velocity.
Next, the data obtained from the question:
Initial velocity (v₁) = 15 m/s
Initial kinetic Energy (KE₁) = E
Final final energy (KE₂) = double the initial kinetic energy = 2E
Final velocity (v₂) =?
Thus, the velocity (v₂) with which the car we travel in order to double it's kinetic energy can be obtained as follow:
KE = ½mv²
NOTE: Mass (m) = constant (since we are considering the same car)
KE₁/v₁² = KE₂/v₂²
E /15² = 2E/v₂²
E/225 = 2E/v₂²
Cross multiply
E × v₂² = 225 × 2E
E × v₂² = 450E
Divide both side by E
v₂² = 450E /E
v₂² = 450
Take the square root of both side.
v₂ = √450
v₂ = 21.21 m/s
Therefore, the car will travel at 21.21 m/s in order to double it's kinetic energy.
Answer:
Neither.
Explanation:
When an electron is released from rest, in an uniform electric field, it will accelerate moving in a direction opposite to the field (as the field has the direction that it would take a positive test charge, and the electron carries a negative charge).
It will move towards a point with a higher potential, so its kinetic energy will increase, while its potential energy will decrease:
⇒ ΔK + ΔU = 0 ⇒ ΔK = -ΔU = - (-e*ΔV)
As ΔV>0, we conclude that the electric potential energy decreases while the kinetic energy increases in the same proportion, in order to energy be conserved, in absence of non-conservative forces.
Answer:
Explanation:
- The expression for acceleration of the rolling body on an inclined plane is given as a = gsinФ/1 + k²/R²
- where Ф is the angle of inclination, R is the radius, k is the radius of gyration.
- The potential energy of the system is given as ; PE = mgh
- The potential energy will be constant for ring, cylinder, solid sphere, and hollow sphere.
- The total kinetic energy of the rolling body is ; KE = mv²/2 + Iw²/2
- Hence, the total kinetic energy of the ring, cylinder, solid sphere and hollow sphere will be constant.
2. The moment of inertia of the ring is given as ;
I = mR²
The moment of inertia of the ring is maximum and therefore reaches the bottom last.
