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3 years ago

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A 75 kg man and a 100. Kg woman are playing tug of war on a frictionless surface. The man accelerates toward the woman at 2.0 m/

s2.

1 answer:

Artemon [7]3 years ago

6 0

Complete Question

A 75 kg man and a 100. Kg woman are playing tug of war on a frictionless surface. The man accelerates toward the woman at 2.0 m/s2.How much force does she use to pull on the rope

Answer:

The value is $F = 150 \ N$

Explanation:

From the question we are told that

The mass of the man is $m_m = 75 \ kg$

The mass of the woman is $m_f = 100 \ kg$

The acceleration of the man is $a_m = 2.0 \ m/s ^2$

Generally the force which she used to pull the rope is mathematically represented as

$F = m * a$

=> $F = 75 * 2$

=> $F = 150 \ N$

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Six artificial satellites complete one circular orbit around a space station in the same amount of time. Each satellite has mass

oee [108]

Answer:

The ranking of the net force acting on different satellite from largest to smallest is ${F_E} > {F_F} > {F_A} = {F_B} = {F_D} > {F_C}$

Explanation:

In order to get a good understanding of this solution we need to understand that the main concepts used to solve this problem are centripetal force and velocity of satellite.

Initially, use the expression of the velocity of satellite and find out its dependence on the radius of orbit. Use the dependency in the centripetal force expression.

Finally, we find out the velocity of the six satellites and use that expression to find out the force experienced by the satellite. Find out the force in terms of mass (m) and radius of orbit (L) and at last compare the values of force experienced by six satellites.

Fundamentals

The centripetal force is necessary for the satellite to remain in an orbit. The centripetal force is the force that is directed towards the center of the curvature of the curved path. When a body moves in a circular path then the centripetal force acts on the body.

The expression of the centripetal force experienced by the satellite is given as follows:

${F_{\rm{c}}} = \frac{{m{v^2}}}{L}$

Here, m is the mass of satellite, v is the velocity, and L is the radius of orbit.

The velocity of the satellite with which the satellite is orbiting in circular path is given as follows:

$v = \frac{{2\pi L}}{T}$

Here, T is the time taken by the satellite.

The velocity of the satellite with which the satellite is orbiting in circular path is given as follows;

$v = \frac{{2\pi L}}{T}$

Since, all the satellites complete the circular orbit in the same amount of time. The factor of $\frac{{2\pi }}{T}$ is not affected the velocity value for the six satellites. Therefore, we can write the expression of v given as follows:

Substitute $v = \frac{{2\pi L}}{T}$ in the force expression ${F_{\rm{c}}} = \frac{{m{v^2}}}{L}$ as follows:

$\begin{array}{c}\\{F_c} = \frac{{m{{\left( {\frac{{2\pi L}}{T}} \right)}^2}}}{L}\\\\ = \frac{{4{\pi ^2}}}{{{T^2}}}mL\\\end{array}$

Since, all the satellites complete the circular orbit in the same amount of time. The factor of $\frac{{4{\pi ^2}}}{{{T^2}}}$ not affect the force value for six satellites.Therefore, we can write the expression of ${F_c}$ given as follows:

${F_c} = kmL$

Here, k refers to constant value and equal to $\frac{{4{\pi ^2}}}{{{T^2}}}$

${F_A} = k{m_A}{L_A}$

Substitute 200 kg for ${m_A}$ and 5000 m for LA in the expression ${F_A} = k{m_A}{L_A}$

$\begin{array}{c}\\{F_A} = k\left( {200{\rm{ kg}}} \right)\left( {5000{\rm{ m}}} \right)\\\\ = {10^6}k{\rm{ N}}\\\end{array}$

The force acting on satellite B from their rocket is given as follows: ${F_B} = k{m_B}{L_B}$

Substitute 400 kg for ${m_B}$ and 2500 m for in the expression ${F_B} = k{m_B}{L_B}$

$\begin{array}{c}\\{F_B} = k\left( {400{\rm{ kg}}} \right)\left( {2500{\rm{ m}}} \right)\\\\ = {10^6}k{\rm{ N}}\\\end{array}$

The force acting on satellite C from their rocket is given as follows: ${F_C} = k{m_C}{L_C}$

Substitute 100 kg for ${m_C}$ and 2500 m for in the above expression ${F_C} = k{m_C}{L_C}$

$\begin{array}{c}\\{F_C} = k\left( {100{\rm{ kg}}} \right)\left( {2500{\rm{ m}}} \right)\\\\ = 0.25 \times {10^6}k{\rm{ N}}\\\end{array}$

The force acting on satellite D from their rocket is given as follows: ${F_D} = k{m_D}{L_D}$

Substitute 100 kg for ${m_D}$ and 10000 m for ${L_D}$ in the expression ${F_D} = k{m_D}{L_D}$

$\begin{array}{c}\\{F_D} = k\left( {100{\rm{ kg}}} \right)\left( {10000{\rm{ m}}} \right)\\\\ = {10^6}k{\rm{ N}}\\\end{array}$

The force acting on satellite E from their rocket is given as follows: ${F_E} = k{m_E}{L_E}$

Substitute 800 kg for ${m_E}$ and 5000 m for ${L_E}$ in the expression ${F_E} = k{m_E}{L_E}$

$\begin{array}{c}\\{F_E} = k\left( {800{\rm{ kg}}} \right)\left( {5000{\rm{ m}}} \right)\\\\ = 4.0 \times {10^6}k{\rm{ N}}\\\end{array}$

The force acting on satellite F from their rocket is given as follows: ${F_F} = k{m_F}{L_F}$

Substitute 300 kg for ${m_F}$ and 7500 m for ${L_F}$ in the expression ${F_F} = k{m_F}{L_F}$

$\begin{array}{c}\\{F_F} = k\left( {300{\rm{ kg}}} \right)\left( {7500{\rm{ m}}} \right)\\\\ = 2.25 \times {10^6}k{\rm{ N}}\\\end{array}$

The value of forces obtained for the six-different satellite are as follows.

$\begin{array}{l}\\{F_A} = {10^6}k{\rm{ N}}\\\\{F_B} = {10^6}k{\rm{ N}}\\\\{F_C} = 0.25 \times {10^6}k{\rm{ N}}\\\\{F_D} = {10^6}k{\rm{ N}}\\\\{F_E} = 4.0 \times {10^6}k{\rm{ N}}\\\\{F_F} = 2.25 \times {10^6}k{\rm{ N}}\\\end{array}$

The ranking of the net force acting on different satellite from largest to smallest is ${F_E} > {F_F} > {F_A} = {F_B} = {F_D} > {F_C}$

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An object of mass M oscillates on the end of a spring. To double the period, replace the object with one of mass: (a) 2M. (b)M/2

harkovskaia [24]

Answer:

(c) 4M

Explanation:

The system is a loaded spring. The period of a loaded spring is given by

$T = 2\pi\sqrt{\dfrac{m}{k}}$

<em>m</em> is the mass and <em>k</em> is the spring constant.

It follows that, since <em>k</em> is constant,

$T\propto\sqrt{m}$

$\dfrac{T}{\sqrt{m}} = C$ where <em>C</em> represents a constant.

$\dfrac{T_1}{\sqrt{m_1}} = \dfrac{T_2}{\sqrt{m_2}}$

$m_2 = m_1\left(\dfrac{T_2}{T_1}\right)^2$

When the period is doubled, $T_2 = 2T_1$ .

$m_2 = m_1\left(\dfrac{2T_1}{T_1}\right)^2 = 4m_1$

Hence, the mass is replaced by 4M.

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4 years ago

Under what conditions might a stream's volume increase, and under what conditions might it decrease? How would the size of the s

JulijaS [17]

The volume of the stream would increase when it was raining and would decrease when there hadn't been rain, not necessarily a drought but just no rain for a while. As far as the sediment particles part they would be slightly bigger because there would be less water for them to mix into causing them to join together.

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3 years ago

A box with the mass of 20 kg at 5 m is lifted to 20 m. How much work was 7 points<br> done?

kirza4 [7]

Answer:

$2,900\: \mathrm{J}$

Explanation:

Work is given by the equation $W=F\Delta x$ where $F$ is force and $\Delta x$ is displacement.

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The force acting on the box is the force of gravity, given as $F_g=mg$ .

Therefore, the total work done is:

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