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enyata [817]
2 years ago
9

Check my work please

Physics
1 answer:
Amiraneli [1.4K]2 years ago
5 0

We can use the ideal gas equation which is expressed as PV = nRT. At a constant volume and number of moles of the gas the ratio of T and P is equal to some constant. At another set of condition, the constant is still the same. Calculations are as follows:

T1/P1 = T2/P2

P2 = T2 x P1 / T1

P2 = 273 x 340 / 713

<span>P2 = 130 kPa</span>

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Suppose you have a dipole that's free to move in any way (including rotate - imagine it floating in space). And there's an objec
Setler [38]

Complete Question

The complete question shown on the first uploaded image

Answer:

a)

The force on Q due to dipole is Attractive

b)

The charge Q exerts attractive force on the dipole

c)

Yes from the above parts, force depends on the sign of charge

d)

   F = kQq[\frac{d^{2}+2rd}{r^{2}(d+r)^{2}} ]

e)

The magnitude o force decrease by a factor of 8.0 times

Explanation:

The explanation is shown on the second uploaded image

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2 years ago
A block attached to a spring with an unknown spring constant oscillates with a period of 2.0 s. What is the period if
Zigmanuir [339]

Answer:

a) If the mass is doubled, then the period is increased by \sqrt{2}. Hence, the period of the system is 2.828 seconds.

b) If the mass is halved, then the period is reduced by \frac{\sqrt{2}}{2}. Hence, the period of the system is 1.414 seconds.

c) The period of the system does not depend on amplitude. Hence, the period of the system is 2 seconds.

d) If the spring constant is doubled, then the period is reduced by \frac{\sqrt{2}}{2}. Hence, the period of the system is 1.414 seconds.

Explanation:

The statement is incomplete. We proceed to present the complete statement: <em>A block attached to a spring with unknown spring constant oscillates with a period of 2.00 s. What is the period if </em><em>a. </em><em>The mass is doubled? </em><em>b.</em><em> The mass is halved? </em><em>c.</em><em> The amplitude is doubled? </em><em>d.</em><em> The spring constant is doubled? </em>

We have a block-spring system, whose angular frequency (\omega) is defined by the following formula:

\omega = \sqrt{\frac{k}{m} } (1)

Where:

k - Spring constant, measured in newtons per meter.

m - Mass, measured in kilograms.

And the period (T), measured in seconds, is determined by the following expression:

T = \frac{2\pi}{\omega} (2)

By applying (1) in (2), we get the following formula:

T = 2\pi\cdot \sqrt{\frac{m}{k} }

a) If the mass is doubled, then the period is increased by \sqrt{2}. Hence, the period of the system is 2.828 seconds.

b) If the mass is halved, then the period is reduced by \frac{\sqrt{2}}{2}. Hence, the period of the system is 1.414 seconds.

c) The period of the system does not depend on amplitude. Hence, the period of the system is 2 seconds.

d) If the spring constant is doubled, then the period is reduced by \frac{\sqrt{2}}{2}. Hence, the period of the system is 1.414 seconds.

8 0
3 years ago
A monkey has a bit of a heavy for on the gas pedal. As soon as the light turns green the monkey pushes the gas pedal to the floo
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Answer:

s=6.86m/s^2

Explanation:

Hello,

In this case, considering that the acceleration is computed as follows:

a=\frac{v_{final}-v_{initial}}{t}

Whereas the final velocity is 28.82 m/s, the initial one is 0 m/s and the time is 4.2 s. Thus, the acceleration turns out:

a=\frac{28.82m/s-0m/s}{4.2s}\\ \\s=6.86m/s^2

Regards.

3 0
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A runner accelerates toward the finish line.
Brut [27]
C the runners feet pushing against the ground describes the acceleration toward the finish line
6 0
3 years ago
How do i solve this?
kenny6666 [7]

Answer:

hmmm i dont know....

Explanation:

i just wanted free point. TANKS YOU SIR!!

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