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Advocard [28]
3 years ago
13

An electric field can be created by a single charge or a distribution of charges. The electric field a distance from a point cha

rge has magnitude E = k|q'|/r^2.The electric field points away from positive charges and toward negative charges. A distribution of charges creates an electric field that can be found by taking the vector sum of the fields created by individual point charges. Note that if a charge is placed in an electric field created by q', q will not significantly affect the electric field if it is small compared to q'.Imagine an isolated positive point charge with a charge Q (many times larger than the charge on a single electron).There is a single electron at a distance from the point charge. On which of the following quantities does the force on the electron depend?Check all that apply.A the distance between the positive charge and the electronB the charge on the electronC the mass of the electronD the charge of the positive chargeE the mass of the positive chargeF the radius of the positive chargeG the radius of the electron
Physics
1 answer:
joja [24]3 years ago
8 0

Answer:

A the distance between the positive charge and the electron

B the charge on the electron

D the charge of the positive charge

Explanation:

The electric field produced by the positive charge Q at the location of the electron is given by

E=k\frac{Q}{r^2}

where

k is the Coulomb constant

Q is the charge

r is the distance between the charge Q and the electron

The force exerted on a charged particle by an electric field is given by

F=qE

where q is the magnitude of the charged particle. So, the force exerted on the electron in this problem is

F=eE = k\frac{eQ}{r^2}

where e is the charge of the electron. As we see from the equation, the force depends only the following quantities:

A the distance between the positive charge and the electron (r)

B the charge on the electron (e)

D the charge of the positive charge (Q)

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An electric motor rotating a workshop grinding wheel at a rate of 1.31 ✕ 102 rev/min is switched off. Assume the wheel has a con
antoniya [11.8K]

(a) 4.03 s

The initial angular velocity of the wheel is

\omega_i = 1.31 \cdot 10^2 \frac{rev}{min} \cdot \frac{2\pi rad/rev}{60 s/min}=13.7 rad/s

The angular acceleration of the wheel is

\alpha = -3.40 rad/s^2

negative since it is a deceleration.

The angular acceleration can be also written as

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

where

\omega_f = 0 is the final angular velocity (the wheel comes to a stop)

t is the time it takes for the wheel to stop

Solving for t, we find

t=\frac{\omega_f - \omega_i }{\alpha}=\frac{0-13.7 rad/s}{-3.40 rad/s^2}=4.03 s

(b) 27.6 rad

The angular displacement of the wheel in angular accelerated motion is given by

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

where we have

\omega_i=13.7 rad/s is the initial angular velocity

\alpha = -3.40 rad/s^2 is the angular acceleration

t = 4.03 s is the total time of the motion

Substituting numbers, we find

\theta= (13.7 rad/s)(4.03 s) + \frac{1}{2}(-3.40 rad/s^2)(4.03 s)^2=27.6 rad

6 0
3 years ago
Elsa has three metal blocks that look the same.
9966 [12]

Answer:

Aluminium is lightweight

Iron is magnetic

and one is not magnetic

4 0
3 years ago
What is true when an object is moved farther from a plane mirror?
musickatia [10]

Answer:

For a plane mirror, the image distance equals the object distance, so the image distance will increase as the object distance increases

The height of the image stays the same and the image distance increases.)

Explanation:

For plane mirrors, the object distance (is equal to the image distance. That is the image is the same distance behind the mirror as the object is in front of the mirror. If you stand a distance of 2 meters from a plane mirror, you must look at a location 2 meters behind the mirror in order to view your image

8 0
3 years ago
Describe and give an example of the doppler effect
Alexxandr [17]
The Doppler effect is when the frequency of a sound changes as the position of you and the sound changes. For example, if a fire truck comes at you, the siren sound will increase, then, after it passes you, then the sound will decrease. The thing that just happened is the Doppler effect!

Have a nice day! :)
4 0
3 years ago
Read 2 more answers
A closed flat loop conductor with radius 2mm is located in a changing uniform magnetic field. If the emf induced in the loop is
BARSIC [14]

Answer:

Rate of magnetic field is 15923.668 T/sec

Explanation:

We have given radius of loop conductor r = 2 mm = 0.002 m

Induced emf = 2 volt

As the magnetic field and plane are perpendicular to each other so angle between magnetic field and area is 0°

Cross sectional area of the conductor is equal to A=\pi r^2

A=3.14\times 0.002^2=1.256\times 10^{-6}m^2

Induced emf is given by e=\frac{-d\Phi }{dt}=-A\frac{dB}{dt}

2=1.256\times 10^{-6}\times \frac{dB}{dt}

\frac{dB}{dt}=159235.668T/sec

6 0
3 years ago
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