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3 years ago

What is neccessary for a magnetic field to create electric current in a copper coil?

Physics

1 answer:

Viktor [21]3 years ago

6 0

A, Lenz' Law. There need to be a difference of flux, so if you use AC you will get a current too.

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O((>ω< ))oo((>ω< ))oo((>ω< ))oo((>ω< ))oo((>ω< ))o

Paladinen [302]

Answer:

(•_•)

Explanation:

7 0

2 years ago

An advertisement for an all-terrain vehicle (ATV) claims that the ATV can climb inclined slopes of 35°. What is the minimum coef

navik [9.2K]

An advertisement for an all-terrain vehicle (ATV) claims that the ATV can climb inclined slopes of 35°. The minimum coefficient of static friction needed for this claim to be possible is 0.7

In an inclined plane, the coefficient of static friction is the angle at which an object slide over another.

As the angle rises, the gravitational force component surpasses the static friction force, as such, the object begins to slide.

Using the Newton second law;

$\sum F_x = \sum F_y = 0$

$\mathbf{mg sin \theta -f_s= N-mgcos \theta = 0 }$

So; On the L.H.S

$\mathbf{mg sin \theta =f_s}$

$\mathbf{mg sin \theta =\mu_s N}$

On the R.H.S

N = mg cos θ

Equating both force component together, we have:

$\mathbf{mg sin \theta =\mu_s \ mg \ cos \theta}$

$\mathbf{sin \theta =\mu_s \ \ cos \theta}$

$\mathbf{\mu_s = \dfrac{sin \theta }{ cos \theta}}$

From trigonometry rule:

$\mathbf{tan \theta= \dfrac{sin \theta }{ cos \theta}}$

∴

$\mathbf{\mu_s =\tan \theta}}$

$\mathbf{\mu_s =\tan 35^0}}$

$\mathbf{\mu_s = 0.700}}$

Therefore, we can conclude that the minimum coefficient of static friction needed for this claim to be possible is 0.7

Learn more about static friction here:

brainly.com/question/24882156?referrer=searchResults

8 0

2 years ago

g Two mirrors are touching so they have an angle of 35.4 degrees with one another. A light ray is incident on the first at an an

Mamont248 [21]

Answer:

add the angles

HOPE you understand

5 0

2 years ago

If an object is thrown in an upward direction from the top of a building 1.6 x 10^2 ft. high at an initial velocity of 21.82 mi/

Goshia [24]

If an object is thrown in an upward direction from the top of a building 1.60 x 102 ft. high at an initial velocity of 21.82 mi/h, what is its final velocity when it hits the ground? (Disregard wind resistance. Round answer to nearest whole number and do not reflect negative direction in your answer.)

this question is troubling me i guessed 96 ft/s

can someone help me out and explain it thanks so much!!!!!!

7 0

3 years ago

Read 2 more answers

A silver wire has a cross sectional area a = 2.0 mm2. a total of 9.4 × 1018 electrons pass through the wire in 3.0 s. the conduc

marta [7]

This problem uses the relationships among current I, current density J, and drift speed vd. We are given the total of electrons that pass through the wire in t = 3s and the area A, so we use the following equation to to find vd, from J and the known electron density n, so:

$v_{d} = \frac{J}{n\left | q \right |}$

<span>The current I is any motion of charge from one region to another, so this is given by:

</span> $I = \frac{\Delta Q}{\Delta t} = \frac{9.4x1018electrons}{3s} = 3189.73(A)$

The magnitude of the current density is:

$J = \frac{I}{A} = \frac{3189.73}{2x10^{-6}} = 1594.86(A/m^{2})$

Being:

$A=2mm^{2} = 2x10^{-6}m^{2}$
<span>
Finally, for the drift velocity magnitude vd, we find:

</span> $v_{d} = \frac{1594.86}{5.8x1028\left |1.60x10^{-19}|\right } = 1.67x10^{18}(m/s)$

Notice: The current I is very high for this wire. The given values of the variables are a little bit odd

6 0

2 years ago