Complete Question
Suppose a NASCAR race car rounds one end of the Martinsville Speedway. This end of the track is a turn with a radius of approximately 57.0 m . If the track is completely flat and the race car is traveling at a constant 30.5 m/s (about 68 mph ) around the turn.
Required:
a. What is the race car's centripetal (radial) acceleration?
b. What is the force responsible for the centripetal acceleration in this case?
O normal
O gravity
O friction
O weight
Answer:
question a
question b
correct option is option 3
Explanation:
From the question we are told that
The radius is 
The constant speed at which the race car is travelling is 
Generally from the question we are told that the track is completely flat so the only force pulling the car to the middle is the frictional force hence the centripetal force is due to the frictional force
Generally the centripetal acceleration is mathematically represented as
=> 
=>
Answer:
1/4F
Explanation:
We already know thatThe electrostatic force is directly proportional to the product of the charge, from Coulomb's law.
So F α Qq
But if it is now half the initial charges, then
F α (1/2)Q *(1/2)q
F α (1/4)Qq
Thus the resultant charges are each halved is (1/4) and the first initial force experienced at full charge.
Thus the answer will be 1/4F
Answer:
The tensile stress on the wire is 550 MPa.
Explanation:
Given;
Radius of copper wire, R = 3.5 mm
extension of the copper wire, e = 5.0×10⁻³ L
L is the original length of the copper wire,
Young's modulus for copper, Y = 11×10¹⁰Pa.
Young's modulus, Y is given as the ratio of tensile stress to tensile strain, measured in the same unit as Young's modulus.
Therefore, the tensile stress on the wire is 550 MPa.
Answer:
Explanation:
a) Power consumption is 4100 J/min / 60 s/min = 68.3 W(atts)
work done raised the potential energy
b) 75(9.8)(1000) / (3(3600)) = 68.055555... 68.1 W
c) efficiency is 68.1 / 68.3 = 0.99593... or nearly 100%
Not a very likely scenario.
