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2 years ago

5. What is the velocity of a Nolan Catholic football player if he runs 200.0 m in 24.0 s?

Physics

1 answer:

Degger [83]2 years ago

6 0

Answer:

$\boxed{ \bold{ \huge{ \boxed {\sf{8.33 \: m \:/s \: }}}}}$

Explanation:

Distance travelled = 200 metre

Time taken = 24 second

Velocity = ?

Finding the velocity 

$\boxed{ \sf{velocity = \frac{distance \: travelled \: }{time \: taken}}}$

$\dashrightarrow{ \sf{velocity = \frac{200 \: m}{24 \: s} }}$

$\dashrightarrow{ \sf{velocity = 8.33 \: m/s}}$

Hope I helped!

Best regards!

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What is the relationship between the masses of the objects and the gravitational force between them

irga5000 [103]

The relationship between the masses of the object and the gravitational force between them is a direct relationship

Explanation:

The gravitational force between two objects is given by the equation:

$F=G\frac{m_1 m_2}{r^2}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between them

We observe that:

- The gravitational force is proportional to the masses of the two objects, m1 and m2, so if the masses increase, the force will increase as well (so, this is a direct relationship)

- The gravitational force is inversely proportional to the square of the separation between the objects, so if the distance is increased, the force will decrease (so, this is an inverse relationship)

Learn more about gravitational force here:

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2 years ago

An open organ pipe of length 0.47328 m and another pipe closed at one end of length 0.702821 m are sounded together. What beat f

sineoko [7]

Answer:

fb = 240.35 Hz

Explanation:

In order to calculate the beat frequency generated by the first modes of each, organ and tube, you use the following formulas for the fundamental frequencies.

Open tube:

$f=\frac{v_s}{2L}$ (1)

vs: speed of sound = 343m/s

L: length of the open tube = 0.47328m

You replace in the equation (1):

$f=\frac{343m/s}{2(0.47228m)}=362.36Hz$

Closed tube:

$f'=\frac{v_s}{4L'}$

L': length of the closed tube = 0.702821m

$f'=\frac{343m/s}{4(0.702821m)}=122.00Hz$

Next, you use the following formula for the beat frequency:

$f_b=|f-f'|=|362.36Hz-122.00Hz|=240.35Hz$

The beat frequency generated by the first overtone pf the closed pipe and the fundamental of the open pipe is 240.35Hz

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2 years ago

An object is made of glass and has the shape of a cube 0.13 m on a side, according to an observer at rest relative to it. Howeve

Vlad1618 [11]

Answer:

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The density changes means that the length in the direction of the motion is changed.

Therefore,

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Given :

Side, b = h = 0.13 m

Mass, m = 3.3 kg

Density = 8100 $kg/m^3$

So,

$8100=\frac{3.3}{l \times 0.13 \times 0.13}$

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$0.024= 0.13 \sqrt{1-\frac{v^2}{c^2}}$

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$0.033= 1-\frac{v^2}{c^2}}$

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Therefore, the speed of the observer relative to the cube is 0.9833 c (in the units of c).

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2 years ago

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bearhunter [10]

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2 years ago

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Tatiana [17]

Complete Question

The complete question is shown on the first uploaded image

Answer:

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