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3 years ago

The spring stretches from 10cm to 22cm

Physics

1 answer:

icang [17]3 years ago

4 0

Supposed to find extension? Extension= 22-10= 12cm .....

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A hockey player strikes a puck, giving it an initial velocity of 14.0 m/s in the positive x-direction. The puck slows uniformly

Advocard [28]

a) $-1.71 m/s^2$

b) 7.7 m/s

c) 4.39 s

Explanation:

The acceleration of an object is the rate of change of velocity of an object.

In this problem, the acceleration of the puck can be found using the following suvat equation:

$v^2-u^2=2as$

where:

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance travelled

For the puck in this problem:

u = 14.0 m/s

v = 6.50 m/s

s = 45.0 m

So, the acceleration is:

$a=\frac{v^2-u^2}{2s}=\frac{6.50^2-14.0^2}{2(45.0)}=-1.71 m/s^2$

The velocity of the puck at time t can be found by using another suvat equation:

$v=u+at$

where

u is the initial velocity

a is the acceleration

t is the time elapsed

v is the final velocity

Here, we have:

u = 14.0 m/s

$a=-1.71 m/s^2$ (found in part a)

Therefore, the velocity of the puck after t = 3.70 s is:

$v=14.0+(-1.71)(3.70)=7.7 m/s$

Here we want to find the time taken for the puck to travel a distance of

s = 45.0 m

To solve this part, we can use again the suvat equation:

$v=u+at$

Where in this case, we use:

u = 14.0 m/s is the initial velocity

v = 6.50 m/s is the final velocity when the puck has travelled 45.0 m (this information is given in the question)

$a=-1.71 m/s^2$ is the acceleration (found in part a)

Therefore, by re-arranging the equation, we find the time taken to cover 45.0 m:

$t=\frac{v-u}{a}=\frac{6.50-14.0}{-1.71}=4.39 s$

7 0

3 years ago

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 ${F_E} > {F_F} > {F_A} = {F_B} = {F_D} > {F_C}$

7 0

4 years ago

A 12 g plastic ball is dropped from a height of 2.5 m and is moving at 3.2 m/s just before it hits the floor. How much mechanica

Nataly [62]

Answer:

0.24

Explanation:

Mass of ball= 12g=0.012Kg

height of ball= 2.5m

velocity of ball before falling= 3.2m/s

potential energy of the ball=mgh= 0.012*10*2.5=0.3J

kinetic energy of the ball=0.5*m $v^{2}$ =0.5*0.012*3.2*3.2=0.6J

Loss in mechanical energy during the fall= potential energy- Kinetic energy= 0.3-0.06=0.24J

note: During the fall, the potential energy of the ball is converted to kinetic energy. the loss in energy is due to air resistance.

4 0

3 years ago

__________ explains what happens in an event and __________ applies knowledge to create something or change an existing process.

Yuliya22 [10]

An hypothesis is a sort of "untested idea" as to what might be happening. I guess it becomes a theory if experiments back it up, and if the combination leads to other things.So, I'd go for The top pair of answers.

3 0

3 years ago

What is magnestism? Explain

Lisa [10]

Answer:

a physical phenomenon produced by the motion of electric charge, which results in attractive and repulsive forces between objects.
the ability to attract and charm people.

3 0

2 years ago

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