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2 years ago

11

Give two examples of where a permanent magnet might used. <3

1 answer:

N76 [4]2 years ago

5 0

Two examples of where you can find permanent magnets are
1.compass
2.speakers

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A 545-kg satellite is in a circular orbit about Earth at a height above Earth equal to Earth's mean radius. (a) Find the satelli

Art [367]

Answer

given,

mass of satellite = 545 Kg

R = 6.4 x 10⁶ m

H = 2 x 6.4 x 10⁶ m

Mass of earth = 5.972 x 10²⁴ Kg

height above earth is equal to earth's mean radius

a) satellite's orbital velocity

centripetal force acting on satellite = $\dfrac{mv^2}{r}$

gravitational force = $\dfrac{GMm}{r^2}$

equating both the above equation

$\dfrac{mv^2}{r} = \dfrac{GMm}{r^2}$

$v = \sqrt{\dfrac{GM}{r}}$

$v = \sqrt{\dfrac{6.67 \times 10^{-11}\times 5.972 \times 10^{24}}{2 \times 6.4 \times 10^6}}$

v = 5578.5 m/s

b) $T= \dfrac{2\pi\ r}{v}$

$T= \dfrac{2\pi\times 2\times 6.4 \times 10^6}{5578.5}$

$T= \dfrac{2\pi\times 2\times 6.4 \times 10^6}{5578.5}$

T = 14416.92 s

$T = \dfrac{14416.92}{3600}\ hr$

T = 4 hr

c) gravitational force acting

$F = \dfrac{GMm}{r^2}$

$F = \dfrac{6.67 \times 10^{-11}\times 545 \times 5.972 \times 10^{24} }{(6.46 \times 10^6)^2}$

F = 5202 N

4 0

3 years ago

Linus builds an electrical circuit with a battery and with wires that carry the current. As the battery weakens, the current als

pantera1 [17]

Answer:

The field gets weaker

Explanation:

I’m taking the test right now, hope this helps!!

7 0

3 years ago

Read 2 more answers

As part of a daring rescue attempt, the Millennium Eagle coasts between a pair of twin asteroids, as shown in the figure below w

Luden [163]

The correct answer is 3 wiv 2=6

4 0

3 years ago

What is the atomic mass of Jupiter?

lana [24]

The mass of Jupiter is 1.9 x 1027 kg.

4 0

3 years ago

A) An automobile light has a 1.0-A current when it is connected to a 12-V battery. Determine the resistance of the light.

kirill [66]

Answer:

The resistance in first case is 12 Ω, power delivered is 12 W, and potential difference is 0.01 V

Explanation:

Given:

(A)

Current $I = 1$ A

Voltage $V = 12$ V

For finding the resistance,

$V = IR$

$R = \frac{V}{I}$

$R = \frac{12}{1}$

$R =$ 12Ω

(B)

For finding power delivered,

$P = I^{2} R$

$P = (1) ^{2} \times 12$

$P = 12$ Watt

(C)

For finding the potential difference,

$V = IR$

$V = 5 \times 10^{-3} \times 2$

$V = 10 \times 10^{-3}$

$V = 0.01$ V

Therefore, the resistance in first case is 12 Ω, power delivered is 12 W, and potential difference is 0.01 V

4 0

3 years ago