Answer:
The total energy of the object does not change is the true statement.
Explanation:
1. The force applied to the object is perpendicular to the movement so doesn't work on it, assuming the table is at the same height in all of its points the only energy we have to analyze is the kinetic energy , because the table top is frictionless by Newton's first law the velocity (v) doesn't change so the kinetic energy is constant and the total energy too.
2. There's no friction in this situation
3. As stated on 1. because the applied force is perpendicular to the movement of the object, there force doesn't work on it.
4. There are over one force on the object, the weight and the normal force table top exerts to the object.
So the only true statement is 1.
What is a projectile give example?
The motion of such a particle is called Projectile Motion. In the above diagram, where a particle is projected at an angle θ, with an initial velocity u.
...
Few Examples of Two – Dimensional Projectiles.
Sure !
Start with Newton's second law of motion:
Net Force = (mass) x (acceleration) .
This formula is so useful, and so easy, that you really
should memorize it.
Now, watch:
The mass of the box is 5.25 kilograms, and the box is
accelerating at the rate of 2.5 m/s² .
What's the net force on the box ?
Net Force = (mass) x (acceleration)
= (5.25 kilograms) x (2.5 m/s²)
Net force = 13.125 newtons .
But hold up, hee haw, whoa ! Wait a second !
Bella is pushing with a force of 15.75 newtons, but the box
is accelerating as if the force on it is only 13.125 newtons.
What happened to the rest of Bella's force ? ?
==> Friction is pushing the box in the opposite direction,
and cancelling some of Bella's force.
How much ?
(Bella's 15.75 newtons) minus (13.125 that the box feels)
= 2.625 newtons backwards, applied by friction.
Answer:
The net force exerted by the two charges is 10.97 x 10⁻⁵ N along negative x-direction.
Explanation:
K=1/4πϵ0, where ϵ0=8.854×10−12C2
K = 9x10⁹
The electric force on point charge q₃ due to charge q₂ is
F₃₂ = kq₃q₂ / (1.245)²
= (9x10⁹ * 49.5x10⁻⁹ * 30.5x10⁻⁹) / (1.245)²
= 13,587.75 x 10⁻⁹ / 1.55
= 8.76629 x 10⁻⁵ N
The electric force on point charge q₃ due to charge q₁ is
F₃₁ = kq₃q₁ / (1.695 - 1.245)²
= (9x10⁹ * 49.5x10⁻⁹ * 10.0x10⁻⁹) / 0.2025
= 2.2000 x 10⁻⁵ N
The net electric force on point charge q₃ is
F₃ = -F₃₁ - F₃₂
= - 8.76629 x 10⁻⁵ N - 2.2000 x 10⁻⁵ N
= 10.97 x 10⁻⁵ N along negative x-direction