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3 years ago

How many times does the earth spin in a year

Physics

2 answers:

sp2606 [1]3 years ago

7 0

It spins one bc it takes 365 days to make a year and to make a year the earth has to mov some

Nata [24]3 years ago

3 0

It takes 365 days to a year to fully spin around once

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Why are there so few eclipses in 2014?

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Because the Moon casts a smaller shadow than Earth does, eclipses of the Sun tightly constrain where you can see them. If the Moon completely hides the Sun, even for a moment, the eclipse is considered total.

Explanation:

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Evan drew a diagram to illustrate radiation

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The answer would be D, electromagnetic waves

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Three capacitors having capacitances of 8.40, 8.40, and 4.20 μFμF, respectively, are connected in series across a 36.0-V potenti

son4ous [18]

Answer:

a)Q=71.4 μ C

b)ΔV' = 10.2 V

Explanation:

Given that

C ₁= 8.7 μF

C₂ = 8.2 μF

C₃ = 4.1 μF

The potential difference of the battery, ΔV= 34 V

When connected in series

1/C = 1/C ₁ + 1/C₂ + 1/C₃

1/ C= 1/8.4 +1 / 8.4 + 1/4.2

C=2.1 μF

As we know that when capacitor are connected in series then they have same charge,Q

Q= C ΔV

Q= 2.1 x 34 μ C

Q=71.4 μ C

As we know that when capacitor are connected in parallel then they have same voltage difference.

Q'= C' ΔV'

C'= C ₁+C₂+C₃ (For parallel connection)

C'= 8.4 + 8.4 + 4.2 μF

C'=21 μF

Q'= C' ΔV'

Q'=3 Q

3 x 71.4= 21 ΔV'

ΔV' = 10.2 V

3 0

3 years ago

A long, straight, horizontal wire carries a left-to-right current of 40 A. If the wire is placed in a uniform magnetic field of

Drupady [299]

Answer:

$4.5\times 10^{-5} T$

Explanation:

We are given that

Current in wire=40 A

Magnetic field= $B_1=3.5\times 10^{-5}$ T( vertically downward)

We have to find the resultant magnitude of the magnetic field 29 cm above the wire and 29 cm below the wire.

According to Bio-Savart law, the magnetic field exerted by the wire at distance R is given by

$B_{wire}=B_2=\frac{\mu_0I}{2\pi R}$

We have R=29 cm= $\frac{29}{100}=0.29 m$

1 m=100 cm

Substitute the values in the given formula

$B_2=\frac{4\pi\times 10^{-7}\times 40}{2\times \pi\times 0.29}=\frac{2\times 40\times 10^{-7}}{0.29}=2.76\times 10^{-5} T$

The resultant magnetic field is given by

$B=\sqrt{B^2_1+B^2_2}$

Substitute the values then we get

$B=\sqrt{(3.5\times 10^{-5})^2+(2.76\times 10^{-5})^2}$

$B=4.5\times 10^{-5} T$

The resultant magnitude of magnetic field is same above and below the wire as it is at same distance.

The resultant magnitude of the magnetic field 29 cm below the wire= $4.5\times 10^{-5} T$

Hence, the resultant magnitude of the magnetic field 29 cm above the wire= $4.5\times 10^{-5} T$

7 0

3 years ago