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3 years ago

The volume of the moon is about 33% of the volume of the earth.true or false

Physics

2 answers:

stealth61 [152]3 years ago

4 0

The moons volume is that of 2 percent of the earth.

Montano1993 [528]3 years ago

3 0

Answer: False

Explanation: The volume of the Moon is approximately 21.9 billion cubic kilometers. If we compare the volume of the Moon with the volume of the Earth, which is about 1 trillion cubic kilometers, then it is possible to obscure the size and the volume of the Earth in relation to the Moon. In other words, when calculating, the Moon's volume is about 2% of the Earth's volume.

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When a football is kicked, the action and reaction forces do NOT cancel each other out because

AfilCa [17]

Answer:

Explanation:

The kicker exerts more force of the football which is why the football moves when its kicked.

6 0

3 years ago

A tug-of-war game is played by five c/hildren: three on one team and two on the other. How much force will the two child team ha

dem82 [27]

Answer:

no, 3 porces is more tha 2 so the power between the 3 should be more than 2

Explanation:

4 0

3 years ago

The specific weight of sea water is 10.1 kN/m^3. Convert to lbs/in^3.

Viktor [21]

Answer:

0.03719 lbs/in³

Explanation:

Specific weight is given by multiplying the density of an object to the acceleration due to gravity.

$\gamma =\rho g\\\Rightarrow \rho=\frac{gamma}{g}\\\Rightarrow \rho=\frac{10.1\times 10^3}{9.81}\\\Rightarrow \rho=1029.562\ kg/m^3$

$1\ kg=2.20462\ lb$

$1\ m=39.3701\ in$

$\\\Rightarrow 1029.562\ kg/m^3=\frac{1029.562\times 2.20462}{39.3701^3}=0.03719\ lbs/in^3$

So,

$10.1\ kN/m^3=0.03719\ lbs/in^3$

8 0

3 years ago

I stretch a rubber band and "plunk" it to make it vibrate in its fundamental frequency. I then stretch it to twice its length an

Nikitich [7]

Answer:

The new frequency (F₂ ) will be related to the old frequency by a factor of one (1)

Explanation:

Fundamental frequency = wave velocity/2L

where;

L is the length of the stretched rubber

Wave velocity = $\sqrt{\frac{T}{\frac{M}{L}}}$

Frequency (F₁) = $\frac{\sqrt{\frac{T}{\frac{M}{L}}}}{2*L}$

To obtain the new frequency with respect to the old frequency, we consider the conditions stated in the question.

Given:

L₂ =2L₁ = 2L

T₂ = 2T₁ = 2T

(M/L)₂ = 0.5(M/L)₁ = 0.5(M/L)

F₂ = $\frac{\sqrt{\frac{2T}{0.5(\frac{M}{L})}}}{4*L} = \frac{\sqrt{4(\frac{T}{\frac{M}{L}}})}{4*L} = \frac{2}{2} [\frac{\sqrt{\frac{T}{\frac{M}{L}}}}{2*L}] = F_1$

Therefore, the new frequency (F₂ ) will be related to the old frequency by a factor of one (1).

7 0

3 years ago

X<br> x<br> 2.5 N<br> 1.7 N<br> 2.5 N

Olin [163]

Answer:

2.5*1.7/2.5

Explanation:

6 0

3 years ago