Answer:

Explanation:
From the question we are told that:
Distance 
Tangential speed 
Distance 2 
Generally the equation for Angular velocity is mathematically given by

Therefore



Answer: A 2.00-kilogram object weighs 19.6 newtons on Earth.
Explanation:
Answer:
1. The car which is twice as massive as the other will have twice potential energy.
Explanation:
The potential energy of an object is given by:

where
m is the mass of the object
g is the gravitational acceleration
h is the height of the object measured to some reference level
In this problem, we have two cars at same elevation (= same h in the formula) but one car has twice the mass of the second car. Calling m the mass of one car and 2m the mass of the second car, we have:
- Potential energy of the first car:

- potential energy of the second car:

So, the car which is twice as massive as the other will have twice potential energy.
Before coming into conclusion first we have to understand both scalar and vector .
A scalar quantity is a physical quantity which has only magnitude for it's complete specification.
A vector quantity is that physical quantity which not only requires magnitude but also possesses direction for it's complete specification.
So the most important factor that differentiate vector from scalar is the direction.
As per the question the student is doing an experiment where he is recording the data obtained during the process.
In order to arrange them in data table, he should ask about the direction of the quantity under consideration.
Hence the correct option is the third option(C)i.e does the measurement include direction?
Answer:
Part A) the angular acceleration is α= 44.347 rad/s²
Part B) the angular velocity is 195.13 rad/s
Part C) the angular velocity is 345.913 rad/s
Part D ) the time is t= 7.652 s
Explanation:
Part A) since angular acceleration is related with angular acceleration through:
α = a/R = 10.2 m/s² / 0.23 m = 44.347 rad/s²
Part B) since angular acceleration is related
since
v = v0 + a*(t-t0) = 51.0 m/s + (-10.2 m/s²)*(3.4 s - 2.8 s) = 44.88 m/s
since
ω = v/R = 44.88 m/s/ 0.230 m = 195.13 rad/s
Part C) at t=0
v = v0 + a*(t-t0) = 51.0 m/s + (-10.2 m/s²)*(0 s - 2.8 s) = 79.56 m/s
ω = v/R = 79.56 m/s/ 0.230 m = 345.913 rad/s
Part D ) since the radial acceleration is related with the velocity through
ar = v² / R → v= √(R * ar) = √(0.23 m * 9.81 m/s²)= 1.5 m/s
therefore
v = v0 + a*(t-t0) → t =(v - v0) /a + t0 = ( 1.5 m/s - 51.0 m/s) / (-10.2 m/s²) + 2.8 s = 7.652 s
t= 7.652 s