All categories

3 years ago

Why does the Moon appear red during a lunar eclipse?

Physics

1 answer:

dem82 [27]3 years ago

6 0

Answer:

Image result for Why does the Moon appear red during a lunar eclipse?

A lunar eclipse takes place when the sun, Earth and moon line up in space. The moon passes through Earth's shadow. ... Bottom line: The moon can look red during a total lunar eclipse because of sunlight that's filtered and refracted by Earth's atmosphere

You might be interested in

What is R2 in the circuit?<br> WILL GIVE BRAINLIEST !!!!

Nataly_w [17]

Answer:

1. Rₑq = 4 Ω

2. R₂ = 6 Ω

3. Vₜ = 12 V, V₁ = 12 V, V₂ = 12 V

4. Iₜ = 3 A, I₁ = 1 A, I₂ = 2 A

Explanation:

1. Determination of the equivalent resistance

Voltage (V) = 12 V

Current (I) = 3 A

Resistance (Rₑq) =?

V= IRₑq

12 = 3 × Rₑq

Divide both side by 3

Rₑq = 12 / 3

Rₑq = 4 Ω

Thus, the equivalent resistance (Rₑq) = 4 Ω

2. Determination of R₂.

Equivalent resistance (Rₑq) = 4 Ω

Resistance 1 (R₁) = 12 Ω

Resistance 2 (R₂)

Since the resistor are in parallel arrangement, the value of R₂ can be obtained as follow:

Rₑq = R₁ × R₂ / R₁ + R₂

4 = 12 × R₂ / 12 + R₂

Cross multiply

4(12 + R₂) = 12R₂

48 + 4R₂ = 12R₂

Collect like terms

48 = 12R₂ – 4R₂

48 = 8R₂

Divide both side by 8

R₂ = 48 / 8

R₂ = 6 Ω

3. Determination of the total voltage (Vₜ), V₁ and V₂.

From the question given above, the total voltage is 12 V

Since the resistors are arranged in parallel connection, the same voltage will go through them.

Thus,

Vₜ = V₁ = V₂ = 12 V

4. Determination of the total current (Iₜ), I₁ and I₂

From the question given above, the total current (Iₜ) is 3 A

Next, we shall determine I₁. Since the resistors are arranged in parallel connection, different current will pass through each resistor respective.

Vₜ = V₁ = 12 V

R₁ = 12 Ω

I₁ =?

V₁ = I₁R₁

12 = I₁ ×12

Divide both side by 12

I₁ = 12 / 12

I₁ = 1 A

Next, we shall determine I₂. This can be obtained as follow:

Iₜ = 3 A

I₁ = 1 A

I₂ =?

Iₜ = I₁ + I₂

3 = 1 + I₂

Collect like terms

I₂ = 3 – 1

I₂ = 2 A

5 0

3 years ago

Which of the following instrument measure time most accurately

valkas [14]

C. Quarts watches is the correct answer

8 0

3 years ago

A current of 2 A flows through a resistor. The voltage across the resistor is 18 V.

stepan [7]

Answer:

$R=9\ \Omega$

Explanation:

Given that,

Current, I = 2 A

Voltage across the resistor, V = 18 V

We need to find the value of resistance of the resistor. Let the resistance be R. We can find it using Ohm's law i.e.

V = IR

Where

R is the resistance of the resistor

$R=\dfrac{V}{I}\\\\R=\dfrac{18}{2}\\\\R=9\ \Omega$

So, the resistance of the resistor is equal to $9\ \Omega$ .

3 0

3 years ago

A stationary bomb explodes in space breaking into a number of small fragments. At the location of the explosion, the net force d

Kaylis [27]

Answer:

.D)The Vector sum of the linear momenta of the fragments must be zero.

Explanation:

.D)The Vector sum of the linear momenta of the fragments must be zero.

This statement is true. This is so because no external force is acting on the masses. The motion is created by internal force so momentum of fragments will be conserved.

A) this statement is false because kinetic energy was zero in the beginning ( the bomb was stationary in the beginning )

B ) This statement is false because it violates the law of conservation of momentum .( it does not violates only when all the fragments have equal mass )

C ) This statement is zero because kinetic energy is not a vector quantity so two kinetic energy when added can not sum up to zero.

6 0

3 years ago

A solid conducting sphere of radius 2.00 cm has a charge of 6.88 μC. A conducting spherical shell of inner radius 4.00 cm and ou

zepelin [54]

Explanation:

Given that,

Radius R= 2.00

Charge = 6.88 μC

Inner radius = 4.00 cm

Outer radius = 5.00 cm

Charge = -2.96 μC

We need to calculate the electric field

Using formula of electric field

$E=\dfrac{kq}{r^2}$

(a). For, r = 1.00 cm

Here, r<R

So, E = 0

The electric field does not exist inside the sphere.

(b). For, r = 3.00 cm

Here, r >R

The electric field is

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times6.88\times10^{-6}}{(3.00\times10^{-2})^2}$

$E=6.88\times10^{7}\ N/C$

The electric field outside the solid conducting sphere and the direction is towards sphere.

(c). For, r = 4.50 cm

Here, r lies between R₁ and R₂.

So, E = 0

The electric field does not exist inside the conducting material

(d). For, r = 7.00 cm

The electric field is

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times(-2.96\times10^{-6})}{(7.00\times10^{-2})^2}$

$E=5.43\times10^{6}\ N/C$

The electric field outside the solid conducting sphere and direction is away of solid sphere.

Hence, This is the required solution.

6 0

3 years ago