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amm1812
3 years ago
11

Kevin decides to soup up his car by replacing the car's wheels with ones that have 1.4 times the diameter of the original wheels

. Note that the speedometer in a car is calibrated based on the tire's diameter and on the distance the tire covers in each revolution. (a) Will the reading of the speedometer change
Physics
1 answer:
True [87]3 years ago
7 0

Answer:

No.

Explanation:

Given that Kevin decides to soup up his car by replacing the car's wheels with ones that have 1.4 times the diameter of the original wheels. Note that the speedometer in a car is calibrated based on the tire's diameter and on the distance the tire covers in each revolution. (a) Will the reading of the speedometer change ?

Considering the formula

V = wr

Where

V = linear speed

W = angular speed

r = radius of the wheel.

But W = 2πrf

Where the the 2 and pi are constant. The radius of the first wheel will be small but counter balance with the larger frequency.

While the radius of the second wheel may be large but it will be of a small frequency.

We can therefore conclude that the reading on the speedometer will not change. Because speedometer will read the linear speed V.

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A parallel-plate capacitor with plates of area 360 cm2 is charged to a potential difference V and is then disconnected from the
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Answer:

Q=3.9825\times 10^{-9} C

Explanation:

We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.

1 m =100 cm

Surface area =S=\frac{360}{10000}=0.036 m^2

\Delta d=0.8 cm=0.008 m

\Delta V=100 V

We have to find the charge Q on the positive plates of the capacitor.

V=Initial voltage between plates

d=Initial distance between plates

Initial Capacitance of capacitor

C=\frac{\epsilon_0 S}{d}

Capacitance of capacitor after moving plates

C_1=\frac{\epsilon_0 S}{(d+\Delta d)}

V=\frac{Q}{C}

Potential difference between plates after moving

V=\frac{Q}{C_1}

V+\Delta V=\frac{Q}{C_1}

\frac{Qd}{\epsilon_0S}+100=\frac{Q(d+\Delta d)}{\epsilon_0S}

\frac{Q(d+\Delta d)}{\epsilon_0 S}-\frac{Qd}{\epsilon_0S}=100

\frac{Q\Delta d}{\epsilon_0 S}=100

\epsilon_0=8.85\times 10^{-12}

Q=\frac{100\times 8.85\times 10^{-12}\times 0.036}{0.008}

Q=3.9825\times 10^{-9} C

Hence, the charge on positive plate of capacitor=Q=3.9825\times 10^{-9} C

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Answer:

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Explanation:

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