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Tresset [83]
4 years ago
8

Faraday’s law of induction deals with how a magnetic flux induces an emf in a circuit. recall that magnetic flux depends on magn

etic field strength and the effective area the field is passing through. we’ll start our investigation by looking at the field strength around a bar magnet. position the magnet around the coil so that the region labeled a in the figure below is inside the coil. move the magnet slowly back and forth and observe the effect on the brightness of the bulb and the needle of the voltmeter. repeat the same process for the other two regions.
Physics
1 answer:
Arte-miy333 [17]4 years ago
4 0
If an experiment is conducted, we place the magnet around the coil first and label it with A. Another set-up: the magnet is placed inside the coil. The magnet is moved slowly back and forth and observe the effect on the brightness of the bulb and the needle of the voltmeter. On the first set-up, the brightness of the bulb is higher than the second set-up. In both set-ups, the needle of the voltmeter, moves back and forth.  
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What earthquake damage is Texas likely to suffer
kumpel [21]
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-- Most of the upper part of the state is considered to be at risk
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A car traveling west in a straight line on a highway decreases its speed from 30.0 meters per second to 23.0 meters per second i
pychu [463]
Acceleration, a =  (v - u) / t

Initial Velocity, u = 30 m/s
Final Velocity, v  =  23 m/s
time                 t  =  2.00 seconds

a = (23 - 30) / 2
a = -7 / 2 = -3.5 m/s2

So the acceleration  is negative, which means it is a deceleration of 3.5 m/s2.
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3 years ago
A charging RC circuit controls the intermittent windshield wipers in a car. The emf is 12.0 V. The wipers are triggered when the
lilavasa [31]

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R=803k\Omega

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V_0 = 12V\\V=10V\\c= 1.25*10^{-6}F\\t=1.8s

We apply the equation for capacitor charging the voltage across it,

V=V_0 (1-e^{-t/x})\\e^{-t/x}=1-(\frac{V}{V_0})\\-\frac{t}{Rc}=ln(\frac{V}{V_0})\\R=-\frac{t}{ln(\frac{V}{V_0})*c}

Replacing values,

R=-\frac{1.8}{ln(10/12)*1.25*10^{-6}}

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3 0
3 years ago
Nellie pulls on a 10kg wagon with a constant horizontal force of 30N. If there are no other horizontal gorces what is the wagons
kicyunya [14]

Answer:

a=3m/s^2

Explanation:

If we have a net force F acting on a body of mass m it will experiment an acceleration a. Newton's 2nd Law gives us the relation between these quantities: F=ma.

In our case, we want to calculate the acceleration, so we write:

a=\frac{F}{m}

With the values we have we get:

a=\frac{30N}{10Kg}=3m/s^2

3 0
3 years ago
You and your friends are doing physics experiments on a frozen pond that serves as a frictionless, horizontal surface. Sam, with
Misha Larkins [42]

Answer:

Sam's and Abigail speeds before colliding were a. 12.34 m/s and b. 2.86 m/s, respectively. Their total kinetic energy was diminished by c. 1484.42 J, approximately

Explanation:

By conservation of momentum, we have

\Delta \vec{p} = 0\\\\m\vec{v}_i = m\vec{v}_f\\m_S v_S\hat{i} + m_A v_A\hat{j} = m_S \times 7 \cos{(40)} \hat{i} + m_S \times 7 \sin{(40)} \hat{j} + m_A \times 9.4 \cos{(18)} \hat{i} - m_A \times 9.4 \sin{(18)} \hat{j}.

Writing for each direction at a time,

m_S v_S = 7m_S \cos{(40)} +9.4 m_A \cos{(18)}\\v_S = 7 \cos{(40)} + 9.4 \frac{m_A}{m_S} \cos{(18)} = 7 \cos{(40)} + 9.4\times\frac{57}{73} \cos{(18)} \approx \mathbf{12.34}\\m_A v_A = 7m_S\sin{(40)} - 9.4m_A\sin{(18)}\\v_A = 7\frac{m_S}{m_A}\sin{(40)} - 9.4\sin{(18)} = 7\times\frac{73}{57}\sin{(40)} - 9.4\sin{(18)} \approx \mathbf{2.86}.

Their kinetic energy changed by

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3 0
3 years ago
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