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Fudgin [204]
3 years ago
6

Help thank you...............

Physics
1 answer:
Sergeu [11.5K]3 years ago
3 0

Answer:

do explain what u need help with?

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Distance and amount of charge are factors that determine which of the following quantities?
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<span>Distance, amount of charge, and polarity of charge are factors
that determine the electrostatic forces of attraction or repulsion
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The magnetic field of an electromagnetic wave in a vacuum is Bz =(2.4μT)sin((1.05×107)x−ωt), where x is in m and t is in s. You
tatiyna

Answer:

Explanation:

Given

B_z=(2.4\mu T)\sin (1.05\times 10^7x-\omega t)

Em wave is in the form of

B=B_0\sin (kx-\omega t)

where \omega =frequency\ of\ oscillation

k=wave\ constant

B_0=Maximum\ value\ of\ Magnetic\ Field

Wave constant for EM wave k is

k=1.05\times 10^7 m^{-1}

Wavelength of wave \lambda =\frac{2\pi }{k}

\lambda =\frac{2\pi }{1.05\times 10^7}

\lambda =5.98\times 10^{-7} m

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A 200-kg boulder is 1000-m above the ground.
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Answer: 1,960,000 J

Explanation:

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A thermal reservoir can be characterized as a thermal body that is large enough that when energy is dumped into it or taken out
hammer [34]

A thermal reservoir can be characterized as a thermal body that is large enough that when energy is dumped into it or taken out of it, the temperature of the reservoir does not vary considerably.

<h3>What is a thermal reservoir?</h3>

A thermal reservoir is as described, a body large enough to have a very high heat capacity. This heat capacity refers to the amount of energy needed to raise the temperature by one degree.

Therefore, we can confirm that bodies with a large enough heat capacity will be considered thermal reservoirs. This is due to the fact that when energy is dumped into it or taken out of it, the temperature of the reservoir <u>does not vary considerably</u>.

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3 years ago
Read 2 more answers
An electric motor can drive grinding wheel at two different speeds. When set to high the angular speed is 2000 rpm. The wheel tu
shutvik [7]

a) The initial angular speed is 209.3 m/s

b) The angular acceleration is -1.74 rad/s^2

c) The angular speed after 40 s is 139.7 rad/s

d) The wheel makes 1501 revolutions

Explanation:

a)

The initial angular speed of the wheel is

\omega_i = 2000 rpm

which means 2000 revolutions per minute.

We have to convert it into rad/s. Keeping in mind that:

1 rev = 2\pi rad

1 min = 60 s

We find:

\omega_i = 2000 \frac{rev}{min} \cdot \frac{2\pi rad/rev}{60 s/min}=209.3 rad/s

b)

To find the angular acceleration, we have to convert the final angular speed also from rev/min to rad/s.

Using the same procedure used in part a),

\omega_f = 1000 \frac{rev}{min} \cdot \frac{2\pi rad/rev}{60 s/min}=104.7 rad/s

Now we can find the angular acceleration, given by

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

where

\omega_i = 209.3 rad/s is the initial angular speed

\omega_f = 104.7 rad/s is the final angular speed

t = 60 s is the time interval

Substituting,

\alpha = \frac{104.7-209.3}{60}=-1.74  rad/s^2

c)

To find the angular speed 40 seconds after the initial moment, we use the equivalent of the suvat equations for circular motion:

\omega' = \omega_i + \alpha t

where we have

\omega_i = 209.3 rad/s

\alpha = -1.74 rad/s^2

And substituting t = 40 s, we find

\omega' = 209.3 + (-1.74)(40)=139.7 rad/s

d)

The angular displacement of the wheel in a certain time interval t is given by

\theta=\omega_i t + \frac{1}{2}\alpha t^2

where

\omega_i = 209.3 rad/s

\alpha = -1.74 rad/s^2

And substituting t = 60 s, we find:

\theta=(209.3)(60) + \frac{1}{2}(-1.74)(60)^2=9426 rad

So, the wheel turns 9426 radians in the 60 seconds of slowing down. Converting this value into revolutions,

\theta = \frac{9426 rad}{2\pi rad/rev}=1501 rev

Learn more about circular motion:

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