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tigry1 [53]
3 years ago
13

He back window of this car contains a heating element.The heating element is part of an electrical circuit connected to the batt

ery of the car.
Physics
1 answer:
Strike441 [17]3 years ago
7 0
Um okay then????????????
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A very long straight current-carrying wire produces a magnetic field of 25 µT at a distance d from the wire. How far will the ma
daser333 [38]

The magnetic field strength of a very long current-carrying wire is proportional to the inverse of the distance from the wire. The farther you go from the wire, the weaker the magnetic field becomes.

B ∝ 1/d

B = magnetic field strength, d = distance from wire

Calculate the scaling factor for d required to change B from 25μT to 2.8μT:

2.8μT/25μT = 1/k

k = 8.9

You must go to a distance of 8.9d to observe a magnetic field strength of 2.8μT

6 0
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Jackson ran in a marathon for 3 hours. The marathon was 185 miles long. How fast was the runner running?
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Dinosaurs’ skeletons can be distinguished from those of other reptiles by the structure of the hips and legs.
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5 0
3 years ago
What is used as evidence for sea-floor spreading?
raketka [301]

Answer:

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4 0
3 years ago
Read 2 more answers
The sun emits electromagnetic waves with a power of 4.0 × 10²⁶ W. Determine the intensity of electromagnetic waves from the sun
Sphinxa [80]

Answer:

I_v = 2,700 W / m^2

I_m = 610 W / m^2

I_s = 16 W / m^2

Explanation:

Given:

- The Power of EM waves emitted by Sun P_s = 4.0*10^26 W

- Radius of Venus r_v = 1.08 * 10^11 m

- Radius of Mars r_m = 2.28 * 10^11 m

- Radius of Saturn r_s = 1.43 * 10^12 m

Find:

Determine the intensity of electromagnetic waves from the sun just outside the atmospheres of (a) Venus, (b) Mars, and (c) Saturn.

Solution:

- We know that Power is related to intensity and surface area of an object follows:

                                        I = P / 4*pi*r^2

Where, A is the surface area of a sphere models the atmosphere around the planets.

a)

- The intensity at the surface of Venus is calculated as:

                                       I_v = P_s / 4*pi*r^2_v

                                       I_v = 4.0*10^26 / 4*pi*(1.08*10^11)^2

                                       I_v = 2,700 W / m^2

b)

- The intensity at the surface of Mars is calculated as:

                                       I_m = P_s / 4*pi*r^2_m

                                       I_m = 4.0*10^26 / 4*pi*(2.28*10^11)^2

                                      I_m = 610 W / m^2

c)

- The intensity at the surface of Saturn is calculated as:

                                       I_s = P_s / 4*pi*r^2_s

                                       I_s = 4.0*10^26 / 4*pi*(1.43*10^12)^2

                                      I_s = 16 W / m^2

7 0
3 years ago
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