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4 years ago

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An object with charge 4.3x10-5 C pushes another object 0.31 micrometers away with a force of 7 N. What is the total charge of th

e second object?

1 answer:

ad-work [718]4 years ago

5 0

Answer:

Charge on the other particle is given as

$q_2 = 1.74 \times 10^{-18} C$

Explanation:

As we know that the force between two charges is given as

$F = \frac{kq_1q_2}{r^2}$

here we know that

$F = 7 N$

$r = 0.31 \mu m$

$q_1 = 4.3 \times 10^{-5} C$

now we have

$7 = \frac{9 \times 10^9 (4.3 \times 10^{-5}q_2}{(0.31\times 10^{-6})^2}$

now we have

$q_2 = 1.74 \times 10^{-18} C$

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A reciprocating compressor is a device that compresses air by a back-and-forth straight-line motion, like a piston in a cylinder

Stella [2.4K]

Answer:

The temperature change per compression stroke is 32.48°.

Explanation:

Given that,

Angular frequency = 150 rpm

Stroke = 2.00 mol

Initial temperature = 390 K

Supplied power = -7.9 kW

Rate of heat = -1.1 kW

We need to calculate the time for compressor

Using formula of compression

$\terxt{time for compression}=\text{time for half revolution}$

$\terxt{time for compression}=\dfrac{1}{2}\times T$

$\terxt{time for compression}=\dfrac{1}{2}\times \dfrac{1}{f}$

Put the value into the formula

$\terxt{time for compression}=\dfrac{1}{2}\times \dfrac{1}{150}\times60$

$\terxt{time for compression}=0.2\ sec$

We need to calculate the rate of internal energy

Using first law of thermodynamics

$U=Q-W$

$\dfrac{\Delta U}{\Delta t}=\dfrac{\Delta Q}{\Delta t}-\dfrac{\Delta W}{\Delta t}$

Put the value into the formula

$\dfrac{\Delta U}{\Delta t}=(-1.1)-(7.9)$

$\dfrac{\Delta U}{\Delta t}=6.8\ kW$

We need to calculate the temperature change per compression stroke

Using formula of rate of internal energy

$\dfrac{\Delta U}{\Delta t}=\dfrac{nc_{v}\Delta \theta}{\Delta t}$

$\Delta\theta=\dfrac{\Delta U}{\Delta t}\times\dfrac{\Delta t}{n\times c_{c}}$

Put the value into the formula

$\Delta \theta=6.8\times10^{3}\dfrac{0.2}{2.0\times20.93}$

$\Delta\theta=32.48^{\circ}$

Hence, The temperature change per compression stroke is 32.48°.

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4 years ago

If you decrease the distance between successive crests of a wave, this changesA. The frequency.B. The wavelength.C. The speed.D.

Feliz [49]

Answer:

option E

Explanation:

The correct answer is option E

Writing the relation between wavelength and frequency

$f = \dfrac{v}{\lambda}$

f is the frequency

v is the velocity of wave

λ is the wavelength

From the above expression we can clearly see that frequency is inversely proportional to wavelength.

When the distance between the two successive crest is decreased then wave length of the wave also decrease.

If wavelength of the wave decreases then frequency of the wave increase.

hence, we can say that both wavelength and the frequency changes.

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3 years ago

Akhtar, Kiran and Rahul were riding in a motorocar that was moving with a high velocity on

VikaD [51]

Answer:

According to the law of conservation of momentum:

Momentum of the car and insect system before collision = Momentum of the car and insect

system after collision

Hence, the change in momentum of the car and insect system is zero.

The insect gets stuck on the windscreen. This means that the direction of the insect is

reversed. As a result, the velocity of the insect changes to a great amount. On the other hand,

the car continues moving with a constant velocity. Hence, Kiran’s suggestion that the insect

suffers a greater change in momentum as compared to the car is correct. The momentum of

the insect after collision becomes very high because the car is moving at a high speed.

Therefore, the momentum gained by the insect is equal to the momentum lost by the car.

Akhtar made a correct conclusion because the mass of the car is very large as compared to

the mass of the insect.

Rahul gave a correct explanation as both the car and the insect experienced equal forces

caused by the Newton’s action-reaction law. But, he made an incorrect statement as the

system suffers a change in momentum because the momentum before the collision is equal to

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3 years ago

A 5 kg block moves with a constant speed of 10 ms to the right on a smooth surface where frictional forces are considered to be

nirvana33 [79]

Answer:

Work done, W = 19.6 J

Explanation:

It is given that,

Mass of the block, m = 5 kg

Speed of the block, v = 10 m/s

The coefficient of kinetic friction between the block and the rough section is 0.2

Distance covered by the block, d = 2 m

As the block passes through the rough part, some of the energy gets lost and this energy is equal to the work done by the kinetic energy.

$W=\mu_kmgd$

$W=0.2\times 5\times 9.8\times 2$

W = 19.6 J

So, the change in the kinetic energy of the block as it passes through the rough section is 19.6 J. Hence, this is the required solution.

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3 years ago

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