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3 years ago

PLEASE ANSWER THANKS

Physics

2 answers:

Veseljchak [2.6K]3 years ago

8 0

The answers are
1.no
2. they will attract each other
3.b
4.neither direction

kolbaska11 [484]3 years ago

6 0

Answer:

They will attract each other

Neither direction

Explanation:

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ocean waves is the example of mechanical waves

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What we call the state in a human body if cerebrum is not functioning well

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3 years ago

What is the gravitational force between mars and Phobos

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Answer:

$F=5.16\times 10^{15}\ N$

Explanation:

We have,

Mass of Mars is, $m_M=6.42\times 10^{23}\ kg$

Mass of its moon Phobos, $m_P=1.06\times 10^{16}\ kg$

Distance between Mars and Phobos, d = 9378 km

It is required to find the gravitational force between Mars and Phobos. The force between two masses is given by

$F=G\dfrac{m_Mm_P}{d^2}$

Plugging all values, we get :

$F=6.67\times 10^{-11}\times \dfrac{6.42\times 10^{23}\times 1.06\times 10^{16}}{(9378\times 10^3)^2}\\\\F=5.16\times 10^{15}\ N$

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4 years ago

A tank has a gate that automatically opens if the water levelhis high enough. The gate has a squarecross section of side1m and c

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Answer:

The gate will open if the height of water is equal to or more than 0.337m.

Explanation:

From the diagram attached, (as seen from the reference question found on google)

The forces are given as

Force on OA

$F_1=P A_1\\F_1=\rho g \bar{h} A_{OA}$

Here

ρ is the density of water.
g is the gravitational acceleration constant
$\bar{h}$ is the equivalent height given as

$\bar{h}=h+\frac{0.6}{2}\\\bar{h}=h+0.3$

$A_{OA}$ is the area of the OA part of the door which is calculated as follows:

$A_{OA}=L\times W\\A_{OA}=1\times 0.6\\A_{OA}=0.6 m^2$

The Force is given as

$F_1=0.6\rho g[h+0.3]$

Force on OB

$F_2=P A_2\\F_2=\rho g \bar{h} A_{OB}$

Here

ρ is the density of water.
g is the gravitational acceleration constant
$\bar{h}$ is the equivalent height given as

$\bar{h}=h+0.6+\frac{0.4}{2}\\\bar{h}=h+0.8$

$A_{OA}$ is the area of the OB part of the door which is calculated as follows:

$A_{OB}=L\times W\\A_{OB}=1\times 0.4\\A_{OB}=0.4 m^2$

The Force is given as

$F_2=0.4\rho g[h+0.8]$

Now the moment arms are given as

$\bar{y}_a=\bar{h}+\frac{I}{A\bar{h}}\\\bar{y}_a=h+0.3+\frac{\frac{1}{12}\times 0.6^3 \times 1}{0.6 \times[h+0.3]}\\\bar{y}_a=h+0.3+\frac{0.03}{h+0.3}$

$\bar{y}_b=\bar{h}+\frac{I}{A\bar{h}}\\\bar{y}_b=h+0.8+\frac{\frac{1}{12}\times 0.4^3 \times 1}{0.4 \times[h+0.8]}\\\bar{y}_b=h+0.8+\frac{0.0133}{h+0.8}$

Taking moment about the point O as zero

$F_1(h+0.6-\bar{y}_a)=F_2(\bar{y}_b-h+0.6)\\F_1(h+0.6-h-0.3-\frac{0.03}{h+0.3})=F_2(h+0.8+\frac{0.0133}{h+0.8}-h-0.6)\\F_1(0.3-\frac{0.03}{h+0.3})=F_2(0.2+\frac{0.0133}{h+0.8})\\0.6\rho g[h+0.3](0.3-\frac{0.03}{h+0.3})=0.4\rho g[h+0.8](0.2+\frac{0.0133}{h+0.8})\\0.6[h+0.3](0.3-\frac{0.03}{h+0.3})=0.4[h+0.8](0.2+\frac{0.0133}{h+0.8})\\0.18h -0.054-0.018=0.08h+0.064+0.00533\\h=0.337 m$

So the gate will open if the height of water is equal to or more than 0.337m.

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