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Aleksandr [31]
2 years ago
14

you are flying a kite that has two strings attached to either side of it you are holding those strings in your left and right ha

nd if you pull down with right hand which direction will it move
Physics
2 answers:
Svetach [21]2 years ago
8 0
The kite would move to the right as well
lys-0071 [83]2 years ago
5 0

Answer:

Suppose that originally the kite is parallel to the ground, there is no force of the wind pushing it to neither side.

Now, if you pull the right side of the kite down, the left side will rise a little bit, and now there will be a surface where the wind cand apply pressure, and in this case, the pressure will be applied to the side with the lower part (in this case the right direction)

so the direction in which the kite will move is also right.

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If Sienna is using the gymnasium floor to cool off, which position will lower her skin temperature the quickest?
maksim [4K]

Answer:

Flat on her back

Explanation:

So c

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A shift in one fringe in the Michelson-Morley experiment corresponds to a change in the round-trip travel time along one arm of
olya-2409 [2.1K]

Explanation:

When Michelson-Morley apparatus is turned through 90^{o} then position of two mirrors will be changed. The resultant path difference will be as follows.

      \frac{lv^{2}}{\lambda c^{2}} - (-\frac{lv^{2}}{\lambda c^{2}}) = \frac{2lv^{2}}{\lambda c^{2}}

Formula for change in fringe shift is as follows.

          n = \frac{2lv^{2}}{\lambda c^{2}}

       v^{2} = \frac{n \lambda c^{2}}{2l}

             v = \sqrt{\frac{n \lambda c^{2}}{2l}}

According to the given data change in fringe is n = 1. The data is Michelson and Morley experiment is as follows.

             l = 11 m

    \lambda = 5.9 \times 10^{-7} m

           c = 3.0 \times 10^{8} m/s

Hence, putting the given values into the above formula as follows.

            v = \sqrt{\frac{n \lambda c^{2}}{2l}}

               = \sqrt{\frac{1 \times (5.9 \times 10^{-7} m) \times (3.0 \times 10^{8})^{2}}{2 \times 11 m}}

               = 2.41363 \times 10^{9} m/s

Thus, we can conclude that velocity deduced is 2.41363 \times 10^{9} m/s.

3 0
1 year ago
To make the jump, Neo and Morpheus have pushed against their respective launch points with their legs applying a _____ to the la
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Force, newtons 3rd law of motion stated for every action there is an equal and opposite reaction
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2 years ago
A turntable rotates counterclockwise at 76 rpm . A speck of dust on the turntable is at 0.47 rad at t=0. What is the angle of th
icang [17]

To solve this exercise it is necessary to apply the kinematic equations of angular motion.

By definition we know that the displacement when there is constant angular velocity is

\theta= \theta_0 +\omega t

From our given data we know that,

\omega = 76\frac{rev}{min}

\omega = 76\frac{rev}{min}(\frac{2\pi rad}{1rev})(\frac{1 min}{60s})

\omega = 7.958rad/s

Moreover we know that

\theta_0 = 0.47 rad

Therefore for time t=8.1s we have,

\theta= \theta_0+ \omega t

\theta= 0.47+(7.958)(8.1)

\theta = 64.9298rad

That number in revolution is:

\theta = 64.9298rad(\frac{1rev}{2\pi})

\theta = 15.108 Revolutions

Here, we see that there are 15 complete revolutions

And 0.108 revolutions i not complete, so the tunable rotation is

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4 0
2 years ago
Two large non-conducting plates of surface area A = 0.25 m 2 carry equal but opposite charges What is the energy density of the
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Answer:

5.1*10^3 J/m^3

Explanation:

Using E = q/A*eo

And

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Energy density = 1/2*eo*(E^2) = 1/2*eo*(q/A*eo)^2 = [q^2] / [2*(A^2)*eo]

= [(75*10^-6)^2] / [2*(0.25)^2*8.85*10^-12]

= 5.1*10^3 J/m^3

8 0
2 years ago
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