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vova2212 [387]
3 years ago
11

The sound level near a noisy air conditioner is 70 dB. If two such units operate side by side, the sound level near them would b

e The sound level near a noisy air conditioner is 70 . If two such units operate side by side, the sound level near them would be 70 dB. 73 dB. 105 dB. 140 dB.
Physics
1 answer:
Schach [20]3 years ago
7 0

Answer:

73 db

Explanation:

A single air conditioner is equivalent to 70 dB frequency. An extra air conditioning unit would therefore double the sound frequency. It does not, however, double the decibels to 140 dB. Instead, it adds only 3 dB to the 70 dB, making the total decibels of two air conditioning units equal to 73 db

Hence the correct option is b that is 73 db

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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
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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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The slope of the line on a speed-time graph tells the speed , true or false ?
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A mover loads a 100 kg box into the back of a moving truck by
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Answer:

2.7

Explanation:

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Mechanical advantage of a ramp is simply defined as the ratio of the length of the ramp to the height of the ramp. Mathematically, it is given by:

Mechanical Advantage = Lenght / height

MA= L/H

With the above formula, we can obtain the mechanical advantage of the ramp as follow:

Length (L) of ramp = 4 m

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Mechanical advantage (MA) of ramp =?

MA = 4/1.5

MA = 2.7

Therefore, the mechanical advantage of the ramp is 2.7

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2 years ago
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