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3 years ago

5

What must be the distance in meters between point charge q1 = 23.5 µc and point charge q2 = -64.2 µc for the electrostatic force

between them to have a magnitude of 4.67 n?

1 answer:

svlad2 [7]3 years ago

7 0

The answer for your problem is shown on the picture.

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522 J of work are done on a system in a process that decreases the system's thermal energy by 240 J. How much heat energy is tra

ELEN [110]

Answer:

ΔQ = - 762 J (Negative Sign shows heat transfer from the system)

Explanation:

The heat transferred to or from the system can be given by using First law of Thermodynamics as follows:

$\Delta Q = \Delta U + W$

where,

ΔQ = Heat Transferred to or from system = ?

ΔU = Change in Internal Energy of System = Change in Thermal Energy

ΔU = - 240 J (negative sign due to decrease in energy)

W = Work done on system = -522 J (negative sign due to work done "on" system)

Therefore,

$\Delta Q = - 240 J - 522 J$

<u>ΔQ = - 762 J (Negative Sign shows heat transfer from the system)</u>

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Answer:

The work efficiency formula is efficiency = output / input, and you can multiply the result by 100 to get work efficiency as a percentage. This is used across different methods of measuring energy and work, whether it's energy production or machine efficiency.

Explanation:

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As a tractor pulls a plow across a field, it exerts a force on the plow in the forward direction. What is the "equal and opposit

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Answer:

the force of the plow pulling backward on the tractor

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Which is a characteristic of all ions? They are made of one type of atom. They have one overall charge. They are made of two or

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3 years ago

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A compact disc (CD) stores music in a coded pattern of tiny pits 10−7m deep. The pits are arranged in a track that spirals outwa

andreev551 [17]

(a) 50 rad/s

The angular speed of the CD is related to the linear speed by:

$\omega=\frac{v}{r}$

where

$\omega$ is the angular speed

v is the linear speed

r is the distance from the centre of the CD

When scanning the innermost part of the track, we have

v = 1.25 m/s

r = 25.0 mm = 0.025 m

Therefore, the angular speed is

$\omega=\frac{1.25 m/s}{0.025 m}=50 rad/s$

(b) 21.6 rad/s

As in part a, the angular speed of the CD is given by

$\omega=\frac{v}{r}$

When scanning the outermost part of the track, we have

v = 1.25 m/s

r = 58.0 mm = 0.058 m

Therefore, the angular speed is

$\omega=\frac{1.25 m/s}{0.058 m}=21.6 rad/s$

(c) 5550 m

The maximum playing time of the CD is

$t =74.0 min \cdot 60 s/min = 4,440 s$

And we know that the linear speed of the track is

v = 1.25 m/s

If the track were stretched out in a straight line, then we would have a uniform motion, therefore the total length of the track would be:

$d=vt=(1.25 m/s)(4,440 s)=5,550 m$

(d) $-6.4\cdot 10^{-3} rad/s^2$

The angular acceleration of the CD is given by

$\alpha = \frac{\omega_f - \omega_i}{t}$

where

$\omega_f = 21.6 rad/s$ is the final angular speed (when the CD is scanned at the outermost part)

$\omega_i = 50.0 rad/s$ is the initial angular speed (when the CD is scanned at the innermost part)

$t=4440 s$ is the time elapsed

Substituting into the equation, we find

$\alpha=\frac{21.6 rad/s-50.0 rad/s}{4440 s}=-6.4\cdot 10^{-3} rad/s^2$

5 0

4 years ago

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