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matrenka [14]
2 years ago
9

What happens if you move a bar magnet back and forth along the axis of the

Physics
2 answers:
nadezda [96]2 years ago
8 0

C) A current is induced in the coiled wire, which lights the light bulb

The moving magnetic field creates electricity which lights the light bulb

Hope it helps!

iogann1982 [59]2 years ago
5 0

Answer:

C) A current is induced in the coiled wire, which lights the light bulb

Explanation:

You might be interested in
You lift a 10-kg box to a height of 1m. How much work do you do on the box when you lift it from the ground? (g= 9.81 m/s2)
Andreyy89

Answer:

98.1 Joule

Explanation:

Solution,

⇒Mass(m)=10kg

⇒Weight(F)=Mg=10×9.81=98.1N

⇒Distance(d)=1m

Now,

Work done=F×d

                  =98.1×1

                  =98.1J

5 0
2 years ago
What part of the atom takes up most of the atoms space?
Ahat [919]

Answer:

The electrons

Explanation:

3 0
3 years ago
Read 2 more answers
A basketball referee tosses the ball straight up for the starting tipoff. at what velocity must a basketball player leave the gr
sladkih [1.3K]
Mgh=mv²/2

v=√2gh=√2·9.8·1.25=4.95m/s
5 0
3 years ago
Two identical tiny spheres of mass m =2g and charge q hang from a non-conducting strings, each of length L = 10cm. At equilibriu
Citrus2011 [14]

Answer:

0.247 μC

Explanation:

As both sphere will be at the same level at wquilibrium, the direction of the electric force will be on the x axis. As you can see in the picture below, the x component of the tension of the string of any of the spheres should be equal to the electric force of repulsion. And its y component will be equal to the weight of one sphere. We can use trigonometry to find the components of the tensions:

F_y:  T_y - W = 0\\T_y = m*g = 0.002 kg *9.81m/s^2 = 0.01962 N

T_y = T_*cos(50)\\T = \frac{T_y}{cos(50)} = 0.0305 N

T_x = T*sin(50) = 0.0234 N

The electric force is given by the expression:

F = k*\frac{q_1*q_2}{r^2}

In equilibrium, the distance between the spheres will be equal to 2 times the length of the string times sin(50):

r = 2*L*sin(50) = 2 * 0.1m * sin(50) 0.1532 m

And k is the coulomb constan equal to 9 *10^9 N*m^2/C^2. q1 y q2 is the charge of each particle, in this case, they are equal.

F_x = T_x - F_e = 0\\T_x = F_e = k*\frac{q^2}{r^2}

q = \sqrt{T_x *\frac{r^2}{k}} = \sqrt{0.0234 N * \frac{(0.1532m)^2}{9*10^9 N*m^2/C^2} } = 2.4704 * 10^-7 C

O 0.247 μC

8 0
3 years ago
An initially uncharged 3.47-μF capacitor and a 6.43-kΩ resistor are connected in series to a 1.50-V battery that has negligible
harkovskaia [24]

Answer: a) io=233.28 A ( initial current); b) τ=R*C= 22.31 ms; c) 81.7 ms

Explanation:  In order to explain this problem we have to use, the formule for the variation of the current in a RC circuit:

I(t)=io*Exp(-t/τ)

and also we consider that io=V/R=(1.5/6.43*10^3)

=233.28 A

then the time constant for the RC circuit is τ=R*C=6.43*10^3*3.47*10^-6

=22.31 ms

Finally the time to reduce the current to 2.57% of its initial value is obtained from:

I(t)=io*Exp(-t/τ)  for I(t)/io=0.0257=Exp(-t/τ) then

ln(0.0257)*τ =-t

t=-ln(0.0257)*τ=81.68 ms

3 0
3 years ago
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