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zhannawk [14.2K]

3 years ago

7

hich best explains why no current is induced? The wire needs to be coiled less tightly. The wire needs to be straight, not coile

d. The magnet needs to be moved through the coils of wire. The magnet needs to be held above the coils of wire.

1 answer:

Dahasolnce [82]3 years ago

7 0

The magnet needs to be held above the coils of wires

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A cyclist maintains a constant velocity of

tensa zangetsu [6.8K]

Hi there!

We can use the equation:

d = x₀ + vt, where:

x₀ = initial distance from the reference point

v = velocity (m/s)

t = time (sec)

Plug in the given values:

d = 248 + 5(49)

d = 493m

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3 years ago

HELP ASAP PLS<br><br> Describe what characterizes science

r-ruslan [8.4K]

Science is characterized by empirical observations, testable questions, formation of hypotheses and experiments that result in stable and replicable results, logic reasoning and theoretical constructs

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3 years ago

He was a priest who issued the Grito de Dolores, a call for a peasant rebellion in Mexico.

uranmaximum [27]

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3 years ago

Coulomb's law for the magnitude of the force FFF between two particles with charges QQQ and Q′Q′Q^\prime separated by a distance

SSSSS [86.1K]

Answer:

${F_{tot} = -1.092*10^{-2}N$

Explanation:

The question: What is the net force exerted by these two charges on a third charge q_3 = 47.0 nC placed between q_1 and q_2 at x_3 = -1.240 mm ?

<u>Your answer may be positive or negative, depending on the direction of the force.</u>

Solution:

The coulomb force is given by the equation

$F =k\dfrac{q_1 q_2}{d^2}.$

where $d$ is the separation between the charges $q_1$ and $q_2$ .

Now, in our case

$q_1=-13.5nC=-13.5*10^{-9}C$

$q_2=-1.735nC=-1.735*10^{-9}C$

$q_3= 47*10^{-9}C$

The separation between charges $q_3$ and $q_1$ is

$d_{31}= (-1.240mm)-(-1.735mm)=0.495mm=4.95*10^{-4} m$

Therefore, the force between them is

$F_{31} = (9*10^9NmC^{-2})\dfrac{47*10^{-9}*(-13.5*10^{-9})}{4.95*10^{-4}} =-0.01152N$

and it is directed in the negative x-direction.

The separation between charges $q_2$ and $q_3$ is

$d_{23} =-1.240mm-0=-1.240*10^{-3}m$

therefore, the force between them is

$F_{23} =(9*10^9Nm^2C^{-2})\dfrac{47*10^{-9}C*(-1.735*10^{-9}C)}{-1.240*10^{-3}m}=5.94*10^{-4}N$

Therefore the total force on charge $q_3$ is

$F_{tot} =5.94*10^{-4}-0.01152N = -0.010926N\\\\\boxed{F_{tot} = -1.092*10^{-2}N}$

8 0

3 years ago

A 100 sprinter starts from rest and final velocity of 18.8m/s what is the sprinter average acceleration if takes him 10.6 s to r

AlladinOne [14]

Answer: 1.77 m/s²

Explanation: acceleration is expressed in:

a= ∆v / t = vf-vi/t

a= 18.8 m/s - 0 / 10.6 s

a= 1.77 m/s²

5 0

3 years ago