The type of potential energy is elastic energy
Given Information:
Mass of electron = m = 9x10⁻³¹ kg
initial speed of electron = v₁ = 0.92c
Force = F = 1.4x10⁻¹³ J
Distance = d = 3 m
Required Information:
Final speed of electron = v₂ = ?
Answer:
Final speed of electron = v₂ = 2.974x10⁸ m/s
Explanation:
As we know from the conservation of energy,
E₂ - E₁ = W
E₂ = E₁ + W
Where E₂ is the final energy of electron and E₁ is the initial energy of electron
The above equation can be written in the form of particle energy
γ₂mc² = γ₁mc² + W
where γ₁ and γ₂ are given by
γ₁ = 1/√1 - (v₁/c)²
γ₂ = 1/√1 - (v₂/c)²
First calculate γ₁
γ₁ = 1/√1 - (0.92c/c)²
γ₁ = 2.55 m
Now calculate γ₂
γ₂ = (γ₁mc² + W)/mc²
First we need to find the work done
W = F*d
W = 1.4x10⁻¹³*3
W = 4.2x10⁻¹³ J
so γ₂ is
γ₂ = (2.55*9x10⁻³¹*(3x10⁸)² + 4.2x10⁻¹³)/9x10⁻³¹*(3x10⁸)²
γ₂ = 7.73
Now we can find the new speed of the electron
γ₂ = 1/√1 - (v₂/c)²
Re-arranging the above equation results in
v₂ = c*√(1 - 1/γ₂²)
v₂ = 3x10⁸*√(1 - 1/7.73²)
v₂ = 2.974x10⁸ m/s
<span> velocity increases by √3</span>
Answer:
Net force ( F ) = 208 N.
Speed of car ( v ) = 68.58 m/s.
Explanation:-
The radius ( r ) = 480 m.
Mass of passenger ( m ) = 65 kg.
Horizontal force ( 
) = 208 N.
Vertical force ( 
) = 637 N.
Net vertical component of force ( ) = m*g - 637 = 65 * 9.8 - 637 = 0 N.
a ) magnitude of net force .
= 208 N.
b) The speed of car.
where v is the velocity.
v² = 4704
v = 68.58 m/s.
Answer:
1. Collision- the sudden, forceful coming together in direct contact of two bodies
2. Newton's third law of motion is<em> naturally applied to collisions between two objects</em>. In a collision between two objects, both objects experience forces that are equal in magnitude and opposite in direction.
3. Momentum is the product of a moving object's mass and velocity. When two objects collide the total momentum before the collision is equal to the total momentum after the collision
