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3 years ago

Suppose a current flows through a copper wire. Which two things occur?

Physics

2 answers:

ICE Princess25 [194]3 years ago

4 0

Answer:

A) A magnetic field forms around the wire.

C) The field is perpendicular to the direction of flow of the current.

Explanation:

A.pex

worty [1.4K]3 years ago

3 0

I’m pretty sure the answer is c and d hope this helps and good luck

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If v = 5.00 meters/second and makes an angle of 60° with the negative direction of the y–axis, calculate all possible values of

Goshia [24]

Vx = + 4.33 m/s. Hope this helps

8 0

3 years ago

Joffrey talks and moves slowly. When asked a question, he answers slowly in monotone monosyllables, if he answers at all. Joffre

lana [24]

Answer:

R.E.T.A.R.D.A.T.I.O.N

Explanation:

It won't let me spell it normal

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3 years ago

A forward honzontal force of 50 N is applied to a crate. A second horizontal force o

True [87]

Answer:

Explanation:

magnitude: 180-50=130N

Direction: in the direction same as the second horizontal force

4 0

3 years ago

During a very quick stop, a car decelerates at 7.00 m/s2 . (a) What is the angular acceleration of its 0.280-m-radius tires, ass

love history [14]

Answer:

-25 rad/s²

29 times

3.8 seconds

50.54 m

26.6 m/s

Explanation:

$\omega_f$ = Final angular velocity

$\omega_i$ = Initial angular velocity

$\alpha$ = Angular acceleration

$\theta$ = Angle of rotation

a = Acceleration = -7 m/s² (negative because of deceleration)

r = Radius of wheel = 0.28 m

Angular acceleration is given by

$\alpha=\frac{a}{r}\\\Rightarrow \alpha=\frac{-7}{0.28}\\\Rightarrow \alpha=-25\ rad/s^$

Angular acceleration of the wheel is -25 rad/s²

$\omega_f=\omega_i+\alpha t\\\Rightarrow t=\frac{\omega_f-\omega_i}{\alpha}\\\Rightarrow t=\frac{0-95}{-25}\\\Rightarrow t=3.8\ s$

It took 3.8 seconds for the car to stop

$\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow \theta=95\times 3.8+\frac{1}{2}\times -25\times 3.8^2\\\Rightarrow \theta=180.5\ rad$

$\frac{180.5}{2\pi}=28.72746\ rev$

The wheel rotated 29 times.

The

Initial velocity

$u=r\omega_i\\\Rightarrow u=0.28\times 95\\\Rightarrow u=26.6\ m/s$

The car’s initial velocity is 26.6 m/s

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=26.6\times 3.8+\frac{1}{2}\times -7\times 3.8^2\\\Rightarrow s=50.54\ m$

Distance covered in the time is 50.54 m

If the distance is found in meters we get 50.54 m which is very low. Considering the initial velocity is 95.76 km/h. Coming to a stop in that distance is very low. So, the values do not seem reasonable. At least a 100 m is required to stop from that speed.

3 0

3 years ago

A student repeats a reaction several times to test the effects of various

GREYUIT [131]

I would pick answer option A

5 0

3 years ago