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pashok25 [27]
3 years ago
14

English or metric What is 16in to cm

Physics
1 answer:
il63 [147K]3 years ago
6 0
40.64 cm is the answer to your question.
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If 20 beats are produce with one second which of the following frequency could possibly be held by two sound waves traveling thr
musickatia [10]

ANY two frequencies whose difference is 20 Hz.

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3 years ago
How does the wavelength of a wave change when frequency decreases? when frequency increases?
KonstantinChe [14]

Wavelength increases as Frequency decreases and vice versa.

This is because Wavelength and frequency are inversely related.


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3 years ago
Calculate the location xcm of the center of mass of the Earth-Moon system. Use a coordinate system in which the center of the Ea
Jet001 [13]

Answer:

The center of mass of the Earth-Moon system is 4.673 kilometers away from center of Earth.

Explanation:

Let suppose that planet and satellite can be treated as particles. The masses of Earth and Moon (m_{E}, m_{M}) are 5.972\times 10^{24}\,kg and 7.349\times 10^{22}\,kg, respectively. The distance between centers is 384,403 kilometers. The location of the center of mass can be found by using weighted averages:

\bar x = \frac{x_{E}\cdot m_{E}+x_{M}\cdot m_{M}}{m_{E}+m_{M}}

If x_{E} = 0\,km and x_{M} = 384,403\,km, then:

\bar x = \frac{(0\,km)\cdot (5.972\times 10^{24}\,kg)+(384,403\,km)\cdot (7.349\times 10^{22}\,kg)}{5.972\times 10^{24}\,kg+7.349\times 10^{22}\,kg}

\bar x = 4.673\,km

The center of mass of the Earth-Moon system is 4.673 kilometers away from center of Earth.

8 0
3 years ago
Llus
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Answer:

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8 0
3 years ago
Read 2 more answers
A 120-kg object and a 420-kg object are separated by 3.00 m At what position (other than an infinitely remote one) can the 51.0-
djverab [1.8K]

Answer:

1.045 m from 120 kg

Explanation:

m1 = 120 kg

m2 = 420 kg

m = 51 kg

d = 3 m

Let m is placed at a distance y from 120 kg so that the net force on 51 kg is zero.

By use of the gravitational force

Force on m due to m1 is equal to the force on m due to m2.

\frac{Gm_{1}m}{y^{2}}=\frac{Gm_{2}m}{\left ( d-y \right )^{2}}

\frac{m_{1}}{y^{2}}=\frac{m_{2}}{\left ( d-y \right )^{2}}

\frac{3-y}{y}=\sqrt{\frac{7}{2}}

3 - y = 1.87 y

3 = 2.87 y

y = 1.045 m

Thus, the net force on 51 kg is zero if it is placed at a distance of 1.045 m from 120 kg.

6 0
3 years ago
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