<h3>Question 1</h3>
Answer
option C) velocity
Explanation
acceleration = Δv ÷ Δt
<h3>Question 2</h3>
Answer
option C) m/s²
Explanation
Δv ÷ Δt
= m/s ÷ s
= m/s x 1/s
= m/s²
<h3>Question 3</h3>
Answer
option B) velocity has both direction and speed.
That is why velocity can be negative but speed can not and velocity is rate of change of displacement where as speed is rate of change of distance.
<span>The correct option is C. Energy cannot be created or destroyed. This statement is known as law of conservation of energy, and it implies that whenever a certain form of energy does change, the loss of this form of energy must have converted into an another type of energy. A typical example is an object falling to the ground: initially, the object has gravitational potential energy. As the object falls down, it loses potential energy (since its altitude from the grounf decreases), but it acquires kinetic energy (because its velocity increases). In this example, potential energy has converted into kinetic energy, but the total energy of the object has remained constant.</span>
Explanation:
It is given that,
Radius of circular particle accelerator, r = 1 m
The distance covered by the particle is equal to the circumference of the circular path, d = 2πr
d = 2π × 1 m
(a) The speed of satellite is given by total distance divided by total time taken as :

Let t is the period of the particle.

d = distance covered
s = speed of particle
It is given that the charged particle is moving nearly with the speed of light



(b) On the circular path, the centripetal acceleration is given by :



Hence, this is the required solution.
I’m so sorry I been trying to figure this out for sun