The coefficient of friction between the road and the car's tire is determined as 0.78.
<h3>Acceleration of the car</h3>
The acceleration of the car is calculated as follows;
v² = u² - 2as
0 = u² - 2as
a = u²/2s
where;
- u is the initial velocity = 97 km/h = 26.94 m/s
a = (26.94)²/(2 x 47)
a = 7.72 m/s²
<h3>Coefficient of friction</h3>
μ = a/g
μ = (7.72)/9.8
μ = 0.78
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Answer:
s
Explanation:
From the question we are told that
The outer ring with a radius of 30 m
inner Gravity Approximately 9.80 m/s'
Outer Gravity Approximately 5.35 m/s.
Generally the equation for centripetal force is given mathematically as
Centripetal acceleration enables Rotation therefore?

Considering the outer ring,




Therefore solving for Period T
Generally the equation for solving Period T is mathematically given as


s
Atomic number=Proton count
Atomic mass=Proton count+ neutron count
Neuton Count=Atomic mass-Proton count
Proton count=Atomic number=27
Mass number=74
Neuton count= 74-27=47
Answer:
The value of resistance when power is 1100 watts =
= 50 ohms
Explanation:
Power
= 2200 Watts
Resistance
= 25 ohms
Power
= 1100 Watts
Resistance
= we have to calculate
Given that the power in an electric circuit varies inversely with the resistance
⇒ P ∝ 
⇒
= 
⇒
= 
⇒
= 50 ohms
This is the value of resistance when power is 1100 watts.