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storchak [24]
2 years ago
6

Two rocks are thrown directly upward with the same initial speeds, one on earth and one on our moon, where the acceleration due

to gravity is one-sixth what it is on earth. Part A If the rock on the moon rises to a height H, how high will the rock rise on the earth
Physics
1 answer:
emmasim [6.3K]2 years ago
8 0

Answer:

The rock will rise H/6 units high on earth

Explanation:

In order to find the height to which rock rises, we use 3rd equation of motion. The 3rd equation of motion is as follows:

2gh = Vf² - Vi²

h = (Vf² - Vi²)/2g

where,

h = height

Vf = Final Velocity

Vi = Initial Velocity

g = acceleration due to gravity

<u>ON MOON</u>:

On moon:

h = H

Vf = 0 m/s (rock stops at highest point for a moment)

Vi = Vi

g = -(1/6) g (negative sign due to upward motion)

Therefore,

H = (0² - Vi²)/[-(2)(1/6)(g)]

H = 3Vi²/g

H/3 = Vi²/g  ------ equation (1)

<u>ON EARTH</u>:

On earth:

h = ?

Vf = 0 m/s (rock stops at highest point for a moment)

Vi = Vi (same initial velocity)

g = - g (negative sign due to upward motion)

Therefore,

h = (0² - Vi²)/(-2g)

h = Vi²/2g

h = (1/2)(Vi²/g)

using equation (1), we get:

h = (1/2)(H/3)

<u>h = H/6</u>

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Explanation:

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What is a reasonable estimate of the average kinetic energy of an athlete during a 100 m race that takes 10s?
arlik [135]
The average weight of an athlete should be around 60kg so from the information that the athlete can run 100m in 10s, we can calculate that their average speed is 10m/s. Using the kinetic energy formula, Ek = 1/2mv^2 we can calculate the kinetic energy using 60kg as the mass.

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A 0.350 kg block at -27.5°C is added to 0.217 kg of water at 25.0°C. They come to equilibrium at 16.4°C. What is the specific he
igor_vitrenko [27]

Answer:

508 J/kg/C

Explanation:

Energy Lost by water = Energy gained by block

mcT = <em>m</em><em>c</em><em>T</em><em> </em><em> </em>[bolded is for water, <em>i</em><em>t</em><em>a</em><em>l</em><em>i</em><em>c</em><em>i</em><em>s</em><em>e</em><em>d</em><em> </em>is for block]

(0.217)(4186)(25 - 16.4) = <em>(</em><em>0</em><em>.</em><em>3</em><em>5</em><em>0</em><em>)</em><em>(</em><em>c</em><em>)</em><em>(</em><em>1</em><em>6</em><em>.</em><em>4</em><em> </em><em>+</em><em> </em><em>2</em><em>7</em><em>.</em><em>5</em><em>)</em>

<em>1</em><em>5</em><em>.</em><em>3</em><em>6</em><em>5</em><em>c</em><em> </em>= 7811.9132

c = <u>5</u><u>0</u><u>8</u><u> </u><u>J</u><u>/</u><u>k</u><u>g</u><u>/</u><u>C</u><u> </u><u>(</u><u>3</u><u> </u><u>s</u><u>f</u><u>)</u>

5 0
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Suppose we replace the mass in the video with one that is four times heavier. How far from the free end must we place the pivot
Llana [10]

We must place the pivot to keep the meter stick in balance at 90 cm (10 cm from the weight) from the free end.

Answer: Option B

<u>Explanation:</u>

In initial stage, the meter stick’s mass and mass hanged in meter stick at one end are same. Refer figure 1, the mater stick’s weight acts at the stick’s mid-point.

If in case, the meter stick is to be at balanced form, then the acting torques sum would be zero. So,

                  m \times g \times(x)+((m \times g)(x-50 \mathrm{cm}))=0

                  (m \times g \times x)-(50 \times m \times g)+(m \times g \times x)=0

Taking out ‘mg’ as common and we get

                  2 x-50=0

                  2 x=50

                  x=\frac{50}{2}=25 \mathrm{cm}

Hence, the stick should be pivoted at a distance of,

                 x^{\prime}=100 \mathrm{cm}-25 \mathrm{cm}=75 \mathrm{cm}

So, the stick should be pivoted at a distance of 75 cm at the free end

Now, replace mass with another mass. i.e., four times the initial mass (as given)

If in case, the meter stick is to be at balanced form, then the acting torques sum would be zero. So,

                   4 m g(x)+(m g)(x-50 c m)=0

                   4 m g x+m g x-50 m g=0

Taking out ‘mg’ as common and we get

                   5 x=50

                   x=\frac{50}{5}=10 \mathrm{cm}

Hence, the stick should be pivoted at a distance of,

                   x^{\prime}=100 \mathrm{cm}-10 \mathrm{cm}=10 \mathrm{cm}

So, the stick should be pivoted at a distance of 10 cm from the free end.

Therefore, the option B is correct 90 cm (10 cm from the weight).

3 0
2 years ago
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