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patriot [66]
3 years ago
6

7. An automobile with a radio antenna 1.0 m long travels at 100.0 km/h in a location where the Earth’s horizontal magnetic field

is 5.5×10−5T. What is the maximum possible emf induced in the antenna due to this motion?
Physics
1 answer:
VashaNatasha [74]3 years ago
3 0

Answer:

0.00479 volts

Explanation:

From Faraday's law of electromagnetic induction, the induced emf equals the change in magnitude flux (magnetic field strength multiplied by the area = BA) divided by the time change

Therefore we have the equation

EMF = BA÷t

Since area A = 2πr

EMF = Bπr²÷t

B = 5.5×10^(-5)

Velocity = 100km/h = 27.7778m/s

r = 1m, t = r÷V = 0.036

EMF = Bπr²÷t = (5.5×10^(-5) x π x (1)²)÷0.036 = 0.00479 Volts

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Rubber rods charged by rubbing with cat fur repel each other. Glass rods charged by rubbing with silk repel each other. A rubber
puteri [66]

Answer:

C. A rubber rod and a glass rod charged this way have opposite charges on them.

Explanation:

When a rubber rod is rubbed against cat fur, it acquires a negative charge, it becomes negatively charged.

When you then try to bring two rubber rod's together, they repel because like charges repel.

Meanwhile, when you rub a glass rod against silk, it loses electrons to the silk material and becomes positively charged.

When you bring two positively charged glass rod's together, they repel, because like charges repel.

However, when you bring the rubber rod and a glass rod together, the attract each other because unlike/opposite charges attract.

5 0
3 years ago
PLS HELP
qwelly [4]

Answer:

B

Explanation:

6 0
2 years ago
Why is the coefficient of friction independent of the mass?
Svet_ta [14]
Because the coefficient of friction depends on the surface
3 0
3 years ago
True or False? 13. All living things are made of cells 14. All cells have DNA within their nucleus 15. The cell is the basic uni
Leni [432]

Answer:

1: True

2: False

3: True

4: False

5: True

3 0
2 years ago
One end of a 34-m unstretchable rope is tied to a tree; the other end is tied to a car stuck in the mud. The motorist pulls side
anyanavicka [17]

Answer:

Fc = 89.67N

Explanation:

Since the rope is unstretchable, the total length will always be 34m.

From the attached diagram, you can see that we can calculate the new separation distance from the tree and the stucked car H as follows:

L1+L2=34m

L1^2=L2^2=L^2=2^2+(H/2)^2  Replacing this value in the previous equation:

\sqrt{2^2+H^2/4}+ \sqrt{2^2+H^2/4}=34  Solving for H:

H=\sqrt{52}

We can now, calculate the angle between L1 and the 2m segment:

\alpha = atan(\frac{H/2}{2})=60.98°

If we make a sum of forces in the midpoint of the rope we get:

-2*T*cos(\alpha ) + F = 0  where T is the tension on the rope and F is the exerted force of 87N.

Solving for T, we get the tension on the rope which is equal to the force exerted on the car:

T=Fc=\frac{F}{2*cos(\alpha) } = 89.67N

7 0
3 years ago
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