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4 years ago

The picture below shows the positions of the Earth, Sun, and Moon during an eclipse.

Physics

2 answers:

Sonja [21]4 years ago

8 0

Answer:

The correct answer is the option B) It is a lunar eclipse, in which the Earth casts a shadow on the Moon.

Explanation:

A lunar eclipse is an astronomical phenomenon that occurs when the Earth passes directly between the Moon and the Sun, causing the Earth's shadow produced by sunlight to be cast over the Moon completely covering it. For this to happen, the three celestial bodies must be formed in a straight line.

Every year there are between 2 and 7 lunar eclipses. According to the position of the Moon with respect to the Earth's shadow, 3 types of lunar eclipses occur.

The difference between a lunar eclipse and a solar eclipse is that, in the first case, the Earth stands between the Moon and the Sun casting a shadow on the Moon, while in a solar eclipse, the Moon stands between the Sun and Earth, casting its shadow to a small part of the latter.

Finally, the correct answer is the option B) It is a lunar eclipse, in which the Earth casts a shadow on the Moon.

iogann1982 [59]4 years ago

6 0

The moon is IN the Earth's shadow.
Anybody on the night side of Earth saw the moon disappear during the past few hours.
This is a lunar eclipse. (B)

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Two long, parallel transmission lines, 40.0cm apart, carry 25.0-A and 73.0-A currents.A). Find all locations where the net magne

In-s [12.5K]

Answer:

a) If the currents are in the same direction, the magnetic field is zero at x = 0.298 m = 29.8 cm

That is, in between the wires, 29.8 cm from the 73.0 A wire and 10.2 cm from the 25.0 A wire.

b) If the currents are in opposite directions, the magnetic field is zero at x = 0.608 m = 60.8 cm

That is, along the positive x-axis, 60.8 cm from the 73.0 A wire and 20.8 cm from the 25.0 A wire.

Explanation:

The origin is at the 73.0 A wire and the 25.0 A wire is at x = 0.40 m

The magnetic field in a current carrying wire at a distance r from the wire is given by

B = (μ₀I/2πr)

μ₀ = magnetic constant = (4π × 10⁻⁷) H/m

a) If the currents are in the same direction, at what positions is the magnetic field equal to 0.

According to laws describing the direction.of magnetic fields, this position will be at some point between the two wires.

The magnetic field due to the 73.0 A wire points out of the book, at points along the positive x-axis while the magnetic field due to the 25.0 A wire points into the plane of the book, moving in the negative x-direction.

Hence,

For the 73.0 A wire, I₁ = 73.0 A, r₁ = x

For the 25.0 A wire, I₂ = 25.0 A, r₂ = (0.4 - x)

B = B₁ - B₂ = 0

(μ₀/2π) [(I₁/r₁) - (I₂/r₂)] = 0

(I₁/r₁) = (I₂/r₂)

(I₁/x) = [I₂/(0.4-x)]

(73/x) = [25/(0.4-x)]

73(0.4-x) = 25x

29.2 - 73x = 25x

73x + 25x = 29.2

98x = 29.2

x = (29.2/98) = 0.298 m

b) If the currents are in the opposite directions, at what positions is the magnetic field equal to 0?

According to laws describing the direction.of magnetic fields, this position will be at some point beyond the second wire (since we're initially concerned about the positive x-direction).

The magnetic field due to the 73.0 A wire points out of the book, at points along the positive x-axis while the magnetic field due to the 25.0 A wire (whose direction is now in the opposite direction to the current in the first wire) is also along the positive x-direction.

Hence,

For the 73.0 A wire, I₁ = 73.0 A, r₁ = x

For the 25.0 A wire, I₂ = 25.0 A, r₂ = (x - 0.4)

B = B₁ - B₂ = 0

(μ₀/2π) [(I₁/r₁) - (I₂/r₂)] = 0

(I₁/r₁) = (I₂/r₂)

(I₁/x) = [I₂/(x-0.4)]

(73/x) = [25/(x-0.4)]

73(x-0.4) = 25x

73x - 29.2 = 25x

73x - 25x = 29.2

48x = 29.2

x = (29.2/48) = 0.608 m

Hope this Helps!!!

5 0

4 years ago

Suppose that you'd like to find out if a distant star is moving relative to the earth. The star is much too far away to detect a

rjkz [21]

Answer:

The speed will be "18km/s". A further explanation is given below.

Explanation:

According to the question, the values are:

Wavelength,

$\lambda = 656.46 \ nm$

$\Delta \lambda = 0.04$

$c=3\times 10^8$

As we know,

⇒ $\frac{\Delta \lambda}{\lambda} =\frac{v}{c}$

On substituting the values, we get

⇒ $\frac{656.46}{0.04} =\frac{v}{3\times 10^8}$

⇒ $v=\frac{656.46}{0.04} (3\times 10^8)$

⇒ $=16411.5\times 3\times 10^8$

⇒ $=18280 \ m/s$

or,

⇒ $=18 \ km/s$

8 0

3 years ago

A crane used 250,000 Joules of work to move a beam to the top of a building in 20 seconds. How much power did the crane use?

maria [59]

Answer:

12500W

Explanation:

Given parameters:

Work done = 250000J

Time taken = 20s

Unknown:

Power of the crane = ?

Solution:

Power is the defined as the rate at which work is being done;

Mathematically;

Power = $\frac{work done}{time }$

insert the parameters and solve;

Power = $\frac{250000}{20}$ = 12500W

7 0

3 years ago

As you approach your vehicle and perform checks it is not necessary to always

Greeley [361]

1.) Have your keys in hand before approaching or entering your car.

2. Be alert to other pedestrians and drivers.

3.) Search for signs of movement between, beneath and around objects to both sides of your vehicle

4.) check the spare tire for proper inflation

4 0

3 years ago

Two protons are 1 × 10−10 m apart (about one atomic radius). Which interaction between two protons is stronger, the gravitationa

anastassius [24]

Answer: The electric repulsion between the two protons is stronger than the gravitational attraction.

Explanation: Please see the attachments below

4 0

4 years ago