The tangential velocity of the car's tire is the product of the angular velocity and radius of the car's tire which is 11(r) m/s.
<h3>
Angular velocity of the tire</h3>
The angular velocity of the tire is the rate of change of angular displacement of the tire with time.
The magnitude of the angular velocity of the tire is calculated as follows;
ω = 2πN
where;
- N is the number of revolutions per second
ω = 2π x (5.25 / 3)
ω = 11 rad/s
<h3>Tangential velocity of the tire</h3>
The tangential velocity of the car's tire is the product of the angular velocity and radius of the car's tire.
The magnitude of the tangential velocity is caculated as follows;
v = ωr
where;
- r is the radius of the car's tire
v = 11r m/s
Learn more about tangential velocity here: brainly.com/question/25780931
Answer:
Mass of the aluminium chunk = 278.51 g
Explanation:
For an isolated system as given the energy lost and gains in the system will be zero therefore sum of all transfer of energy will be zero,as the temperature will also remain same
A specific heat formula is given as
Energy Change = Mass of liquid x Specific Heat Capacity x Change in temperature
Q = m×c×ΔT
Heat gain by aluminium + heat lost by copper = 0 (1)
For Aluminium:
Q = 
Q = m x 17.94 joule
For Copper:

Q= 4996.53 Joule
from eq 1
m x 17.94 = 4996.53

Mass of the aluminium chunk = 278.51 g
Answer:
1923 N
Explanation:
From the question given above, the following data were obtained:
Mass (m) = 65 Kg
Radius (r) = 2.5 m
Velocity (v) = 8.6 m/s
Centripetal force (F) =?
The centripetal force, F, can be obtained by using the following formula:
F = mv²/r
F = 65 × 8.6² / 2.5
F = 65 × 73.96 / 2.5
F = 4807.4 / 2.5
F = 1922.96 ≈ 1923 N
Thus, the magnitude of the centripetal's force acting on the student is approximately 1923 N
okay this is kinda easy
<u>What is the gravitational field strength on the moon?</u>
The Moon has a gravitational field strength of 1.6 N/kg.
Answer:
6 m/s
Explanation:
Given that :
mass of the block m = 200.0 g = 200 × 10⁻³ kg
the horizontal spring constant k = 4500.0 N/m
position of the block (distance x) = 4.00 cm = 0.04 m
To determine the speed the block will be traveling when it leaves the spring; we applying the work done on the spring as it is stretched (or compressed) with the kinetic energy.
i.e 





v = 6 m/s
Hence,the speed the block will be traveling when it leaves the spring is 6 m/s