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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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When placed in direct sunlight, which an object will absorb the most visible light energy
Vikentia [17]
I would guess anything that is black? black absorbs there most light energy. that's is why a black car in the sun will always be hotter inside than a white car.
3 0
3 years ago
A man-made satellite of mass 6105 kg is in orbit around the earth, making one revolution in 430 minutes. What is the magnitude o
blondinia [14]

Answer:

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

Explanation:

At first we assume that Earth is represented by an uniform sphere, such that the man-made satellite rotates in a circular orbit around the planet. Hence, the following condition must be satisfied:

\left(\frac{4\pi^{2}}{T^{2}} \right)\cdot r = \frac{G\cdot M}{r^{2}} (1)

Where:

T - Period of rotation of the satellite, measured in seconds.

r - Distance of the satellite with respect to the center of the planet, measured in meters.

G - Gravitational constant, measured in newton-square meters per square kilogram.

M - Mass of the Earth, measured in kilograms.

Now we clear the distance of the satellite with respect to the center of the planet:

r^{3} = \frac{G\cdot M\cdot T^{2}}{4\pi^{2}}

r = \sqrt[3]{\frac{G\cdot M\cdot T^{2}}{4\pi^{2}} } (2)

If we know that G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}, M = 6.0\times 10^{24}\,kg and T = 25800\,s, then the distance of the satellite is:

r = \sqrt[3]{\frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6.0\times 10^{24}\,kg)\cdot (25800\,s)^{2}}{4\pi^{2}} }

r \approx 18.897\times 10^{6}\,m

The gravitational force exerted on the satellite by the Earth is determined by the Newton's Law of Gravitation:

F = \frac{G\cdot m\cdot M}{r^{2}} (3)

Where:

m - Mass of the satellite, measured in kilograms.

F - Force exerted on the satellite by the Earth, measured in newtons.

If we know that G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}, M = 6.0\times 10^{24}\,kg, m = 6105\,kg and r \approx 18.897\times 10^{6}\,m, then the gravitational force is:

F = \frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6105\,kg)\cdot (6\times 10^{24}\,kg)}{(18.897\times 10^{6}\,m)^{2}}

F = 6841.905\,N

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

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In an open system such as a campfire matter can
eimsori [14]
In an open system such as a campfire, matter can lose particles, gain particles or exchange particles.
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Akio draws the ray diagram shown.
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Answer: Move the small car so it appears on the left side of the lens.

Explanation:

Because the lens is reflective the small car would apear on the same side as the normal car.

Hope this helps :)

3 0
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. A car starts to move from rest. If its velocity becomes 90km/hr after 3s, calculate its acceleration​
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Answer:

32.5/s squared

Explanation:

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