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3 years ago

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A child stands between two mirrors with his arms out, perpendicular to the mirrors. One plane mirror is 5 m away from his left h

and and another plane mirror is 7 m away from his right hand. How far apart are the two images produced by the mirrors if the child has an arm span of 0.5 m?A. 2 mB. 12 mC. 12.5 mD. 24.5 m

1 answer:

Lelu [443]3 years ago

6 0

Answer:

A. 2

Explanation:

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The cheetah is considered the fastest running animal in the world. Cheetahs can accelerate to a speed of 21.7 m/s in 2.50 s and

viktelen [127]

Answer:

1) 64.2 mi/h

2) 3.31 seconds

3) 47.5 m

4) 5.26 seconds

Explanation:

t = Time taken = 2.5 s

u = Initial velocity = 0 m/s

v = Final velocity = 21.7 m/s

s = Displacement

a = Acceleration

1) Top speed = 28.7 m/s

1 mile = 1609.344 m

$1\ m=\frac{1}{1609.344}\ miles$

1 hour = 60×60 seconds

$1\ s=\frac{1}{3600}\ hours$

$28.7\ m/s=\frac{\frac{28.7}{1609.344}}{\frac{1}{3600}}=64.2\ mi/h$

Top speed of the cheetah is 64.2 mi/h

Equation of motion

$v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow t=\frac{21.7-0}{2.5}\\\Rightarrow a=8.68\ m/s^2$

Acceleration of the cheetah is 8.68 m/s²

2)

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{28.7-0}{8.68}\\\Rightarrow t=3.31\ s$

It takes a cheetah 3.31 seconds to reach its top speed.

3)

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{28.7^2-0^2}{2\times 8.68}\\\Rightarrow s=47.5\ m$

It travels 47.5 m in that time

4) When s = 120 m

$s=ut+\frac{1}{2}at^2\\\Rightarrow 120=0\times t+\frac{1}{2}\times 8.68\times t^2\\\Rightarrow t=\sqrt{\frac{120\times 2}{8.68}}\\\Rightarrow t=5.26\ s$

The time it takes the cheetah to reach a rabbit is 120 m is 5.26 seconds

8 0

3 years ago

A 2kg mass is moving at 3m/s. What is its kinetic energy?

Illusion [34]

<h2><u>KINETIC ENERGY</u></h2>

<h3>Problem:</h3>

» A 2kg mass is moving at 3m/s. What is its kinetic energy?

<h3>Answer:</h3>

$\color{hotpink} \bold{9 \: J} \\$

— — — — — — — — — —

<h3>Formula:</h3>

To calculate the velocity of a kinetic energy, we can use formula

$\underline{ \boxed{ \tt KE = \frac{1}{2} m{v}^{2} } }$

where,

v is the velocity in m/s
KE is the kinetic energy in J (joules)
m is the mass in kg

— — —

Based on the problem, the givens are:

KE (Kinetic energy) = ? (unknown)
m (mass) = 2 kg
v (velocity) = 3 m/s

<h3>Solution:</h3>

To get the velocity, substitute the givens in the formula above then solve.

$\: \: \tt KE = \frac{1}{2} m{v}^{2} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt \: KE = \frac{1}{2} \times 2 \times {(3)}^{2} \\ \tt \: \: \: \: \: \: \: \: \: \: \: \: \: \: KE = \frac{1}{2} \times(2\times 9) \\ \tt KE = \underline{ \boxed{ \blue{ \tt9 \: J}}}$

Therefore, the kinetic energy is 9 Joules.

3 0

1 year ago

Please answer these few questions. This is supposed to be a science question but I had to put it as physics.

Kaylis [27]

I don’t understand how I’m supposed to answer that if you have to watch a video and answer certain questions on certain pages which we don’t have

3 0

3 years ago

Read 2 more answers

If mechanical energy is conserved in a system the energy at any point in time can be in the form of

I am Lyosha [343]

potential, kinetic, elastc energies

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3 years ago

Suppose there are 10,000 civilizations in the Milky Way Galaxy. If the civilizations were randomly distributed throughout the di

vekshin1

Here is the full question

Suppose there are 10,000 civilizations in the Milky Way Galaxy. If the civilizations were randomly distributed throughout the disk of the galaxy, about how far (on average) would it be to the nearest civilization?

(Hint: Start by finding the area of the Milky Way's disk, assuming that it is circular and 100,000 light-years in diameter. Then find the average area per civilization, and use the distance across this area to estimate the distance between civilizations.)

Answer:

1000 light-years (ly)

Explanation:

If we go by the hint; The area of the disk can be expressed as:

$A = \pi (\frac{D}{2})^2$

where D = 100, 000 ly

Let's divide the Area by the number of civilization; if we do that ; we will be able to get 'n' disk that is randomly distributed; so ;

$d= \frac{A}{N} =\frac{\pi (\frac{D}{2})^2 }{10, 000}$

The distance between each disk is further calculated by finding the radius of the density which is shown as follows:

$d = \pi r^2 e$

$r^2_e= \frac{d}{\pi}$

$r_e = \sqrt{\frac{d}{\pi} }$

replacing d = $\frac{\pi (\frac{D}{2})^2 }{10, 000}$ in the equation above; we have:

$r_e = \sqrt{\frac{\frac{\pi (\frac{D}{2})^2 }{10, 000}}{\pi} }$

$r_e = \sqrt{\frac{(\frac{D}{2})^2 }{10, 000}}$

$r_e = \sqrt{\frac{(\frac{100,000}{2})^2 }{10, 000}}$

$r_e = 500 ly$

The distance (s) between each civilization = $2(r_e)$

= 2 (500 ly)

= 1000 light-years (ly)

4 0

3 years ago