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3 years ago

13

A wire perpendicular to the screen carries a current in the direction shown.

2 answers:

mrs_skeptik [129]3 years ago

8 0

Answer:

A) Up

Explanation:

The direction of the current in the wire is going out of the screen. We can use the right-hand rule to find the direction of the magnetic field at point Z. In fact, we have to:

- Put the thumb of the right hand in the same direction as the current (out of the screen)

- Wrap the other fingers around the thumb. The direction of the fingers will be the same as the direction of the magnetic field.

By doing this procedure with out right hand, we see that the other fingers at point Z are directed up, so the magnetic field is also directed up.

____ [38]3 years ago

4 0

A) Up is the direction of the magnetic field at point Z.

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PE is mgh in this context.

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An artificial satellite is in a circular orbit around a planet of radius r= 2.05 x103 km at a distance d 310.0 km from the plane

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Answer:

$\rho = 12580.7 kg/m^3$

Explanation:

As we know that the satellite revolves around the planet then the centripetal force for the satellite is due to gravitational attraction force of the planet

So here we will have

$F = \frac{GMm}{(r + h)^2}$

here we have

$F =\frac {mv^2}{(r+ h)}$

$\frac{mv^2}{r + h} = \frac{GMm}{(r + h)^2}$

here we have

$v = \sqrt{\frac{GM}{(r + h)}}$

now we can find time period as

$T = \frac{2\pi (r + h)}{v}$

$T = \frac{2\pi (2.05 \times 10^6 + 310 \times 10^3)}{\sqrt{\frac{GM}{(r + h)}}}$

$1.15 \times 3600 = \frac{2\pi (2.05 \times 10^6 + 310 \times 10^3)}{\sqrt{\frac{(6.67 \times 10^{-11})(M)}{(2.05 \times 10^6 + 310 \times 10^3)}}}$

$M = 4.54 \times 10^{23} kg$

Now the density is given as

$\rho = \frac{M}{\frac{4}{3}\pi r^3}$

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3 years ago

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third law of motion

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Answer:

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Given the data in the question;

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∴ N(t) has distribution of poisson distribution with parameter (λt)

so

the mean is;

λ = 1 every month = 1/3 per month

E[N(t)] = λt

E[N(t)] = (1/3)(24)

E[N(t)] = 8

Therefore, the expected value is 8

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Var[N(t)] = Var[N(24)]

Var[N(t)] = E[N(24)]

Var[N(t)] = 8

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σ[N(24)] = 2.8284

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