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lana [24]
3 years ago
6

Suppose a pendulum clock has been calibrated to be accurate in San Francisco, where g = 9.800 m/s2 . In Camrose, g = 9.811 m/s2

is slightly larger due to the effect of Earth’s rotation at a higher latitude. Explain why the clock will either run perfectly, run too quickly, or run too slowly in Camrose.
Physics
1 answer:
tamaranim1 [39]3 years ago
8 0

The clock would run too slowly in Camrose.

Since the period of the pendulum T = 2π√(L/g) where L = length of clock and g = acceleration due to gravity.

Now since g = 9.800 m/s² in San Francisco and g = 9.811 m/s² in Camrose, we see that g increases.

From the expression for the period T, since L is constant, we find that

T ∝ 1/√g

Since g increases, so, T would decrease.

Thus, the clock would run too slowly in Camrose.

Learn more about pendulum clock here:

brainly.com/question/12405819

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<h2>Answer: 745.59 nm</h2>

Explanation:

The diffraction angles \theta_{n} when we have a slit divided into n parts are obtained by the following equation:

dsin\theta_{n}=n\lambda (1)

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\lambda is the wavelength of the light  

n is an integer different from zero

Now, the first-order diffraction angle is given when n=1, hence equation (1) becomes:

dsin\theta_{1}=\lambda (2)

We know:

\theta_{1}=34\°

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d=\frac{1mm}{750}

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61. A physics student has a single-occupancy dorm room. The student has a small refrigerator that runs with a current of 3.00 A
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Answer:

Part a)

percentage = 21.3%

Part b)

percentage = 2.13 \times 10^{-5}%

Explanation:

As we know that total power used in the room is given as

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Since power supply is at 110 Volt so the current obtained from this supply is given as

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R = \frac{(2.8 \times 10^{-8})(10\times 10^3)}{\pi(4.126\times 10^{-3})^2}

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P = (4.48)^2(5.23)

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Part b)

now same power must have been supplied from the supply station at 110 kV, so we have

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P = 1.05 \times 10^{-4} W

Now percentage loss is given as

percentage = \frac{loss}{supply} \times 100

percentage = \frac{1.05 \times 10^{-4}}{493} \times 100

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