The four tires of an automobile are inflated to a gauge pressure of 2.2 105 Pa. Each tire has an area of 0.023 m2 in contact wit
1 answer:
Answer:
Force, F = 20240 N
Explanation:
It is given that,
Pressure exerted by the four tires of an automobile,
Area of each tire,
Area of 4 tires,
We know that the pressure exerted by an object is equal to the force per unit area. Its formula is given by :
F = 20240 N
So, the weight of the automobile is 20240 N. Hence, this is the required solution.
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Answer: A 100-lb child stands on a scale while riding in an elevator. Then, the scale reading approaches to 100lb, while the elevator slows to stop at the lowest floor
Explanation: To find the correct answer, we need to know more about the apparent weight of a body in a lift.
<h3>What is the apparent weight of a body in a lift?</h3>- Consider a body of mass m kept on a weighing machine in a lift.
- The readings on the machine is the force exerted by the body on the machine(action), which is equal to the force exerted by the machine on the body(reaction).
- The reaction we get as the weight recorded by the machine, and it is called the apparent weight.
- Here we have given with the actual weight of the body as 100lbs.
- This 100lb child is standing on the scale or the weighing machine, when it is riding .
- During this condition, the acceleration of the lift is towards downward, and thus, a force of ma .
- There is also<em> mg </em>downwards and a normal reaction in the upward direction.
- when we equate both the upward force and downward force, we get,
i.e. during riding the scale reads a weight less than that of actual weight.
- When the lift goes slow and stops the lowest floor, then the acceleration will be approaches to zero.
Thus, from the above explanation, it is clear that ,when the elevator moves to the lowest floor slowly and stops, then the apparent weight will become the actual weight.
Learn more about the apparent weight of the body in a lift here:
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Answer:
Explanation:
Given that,
B(t) = B0 cos(ωt) • k
Radius r = a
Inner radius r' = a/2 and resistance R.
Current in the loop as a function of time I(t) =?
Magnetic flux is given as
Φ = BA
And the Area is given as
A = πr², where r = a/2
A = πa²/4
Then,
Φ = ¼ Bπa²
Φ(t) = ¼πa²Bo•Cos(ωt)
Then, the EMF is given as
ε(t) = -dΦ/dt
ε(t) = -¼πa²Bo • -ωSin(ωt)
ε(t) = ¼ωπa²Bo•Sin(ωt)
From ohms law,
ε = iR
Then, i = ε/R
I(t) = ¼ωπa²Bo•Sin(ωt) /R
This is the current induced in the loop.
Check attachment for better understanding
