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eimsori [14]
3 years ago
7

Recently, the media, always trying to make things sound sensational, have started to call totally eclipsed moon "the blood moon.

" Why does the Moon look reddish to us when there is a total lunar eclipse?
Physics
1 answer:
Mariulka [41]3 years ago
6 0

Answer:

See explanation

Explanation:

- One meaning of a "blood moon" is based on its red glow. This blood moon occurs during a total lunar eclipse. During a total lunar eclipse, Earth lines up between the Moon and the Sun. This hides the Moon from the sunlight.

- When this happens, the only light that reaches the Moon's surface is from the edges of the Earth's atmosphere. The air molecules from Earth's atmosphere scatter out most of the blue light. The remaining light reflects onto the Moon's surface with a red glow, making the Moon appear red in the night sky.

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One has a frequency of 490 Hz and the other is 488 Hz. How many beats per second are heard?
tresset_1 [31]
When sounds at two different frequencies are combined,
two new sounds are created ... at the sum and difference
of the original frequencies.

Combining two sounds at 490 Hz and 488 Hz creates
beats at  978 Hz and 2 Hz.

The fluttering "wah wah" effect of the 2 Hz beat is much more
noticeable than the new sound at 978 Hz.
7 0
3 years ago
light traveling through air encounters a second medium which slows the light to 100,000 miles/second. What is the index of the s
IceJOKER [234]

Answer:

300000 hope it is help full

Explanation:

300000

6 0
1 year ago
Read 2 more answers
How much mass should be attached to a vertical ideal spring having a spring constant (force constant) of 39.5 n/m so that it wil
mrs_skeptik [129]
The frequency of a simple harmonic oscillator such as a spring-mass system is given by
f= \frac{1}{2 \pi}   \sqrt{ \frac{k}{m} }
where 
k is the spring constant
m is the mass attached to the spring.

Re-arranging the formula, we get:
m= \frac{k}{4 \pi^2 f^2}
and since we know the constant of the spring:
k=39.5 N/m
and the frequency of oscillation:
f=1.00 Hz
we can find the value of the mass attached to it:
m= \frac{39.5 Hz}{4 \pi^2 (1.00 Hz)^2} = 1.00 kg
7 0
3 years ago
A 3.0 kg object moving 8.0 m/s in the positive x direction has a one-dimensional elastic collision with an object (mass = M) ini
finlep [7]
<h2>Option 2 is the correct answer.</h2>

Explanation:

Elastic collision means kinetic energy and momentum are conserved.

Let the mass of object be m and M.

Initial velocity object 1 be u₁,  object 2 be u₂

Final velocity object 1 be v₁,  object 2 be v₂

Initial momentum = m x u₁ + M x u₂ = 3 x 8 + M x 0 = 24 kgm/s

Final momentum = m x v₁ + M x v₂ = 3 x v₁ + M x 6 = 3v₁ + 6M

Initial kinetic energy = 0.5 m x u₁² + 0.5 M x u₂² = 0.5 x 3 x 8² + 0.5 x M x 0² = 96 J

Final kinetic energy = 0.5 m x v₁² + 0.5 M x v₂² = 0.5 x 3 x v₁² + 0.5 x M x 6² = 1.5 v₁² + 18 M

We have

            Initial momentum = Final momentum

            24 = 3v₁ + 6M

            v₁ + 2M = 8

             v₁ = 8 - 2M

            Initial kinetic energy = Final kinetic energy

            96 = 1.5 v₁² + 18 M

            v₁² + 12 M = 64

Substituting  v₁ = 8 - 2M

           (8 - 2M)² + 12 M = 64    

           64 - 32M + 4M² + 12 M = 64    

            4M² = 20 M

               M = 5 kg

Option 2 is the correct answer.  

6 0
2 years ago
The particle, initially at rest, is acted upon only by the electric force and moves from point a to point b along the x axis, in
bezimeni [28]

1) Potential difference: 1 V

2) V_b-V_a = -1 V

Explanation:

1)

When a charge moves in an electric field, its electric potential energy is entirely converted into kinetic energy; this change in electric potential energy is given by

\Delta U=q\Delta V

where

q is the charge's magnitude

\Delta V is the potential difference between the initial and final position

In this problem, we have:

q=4.80\cdot 10^{-19}Cis the magnitude of the charge

\Delta U = 4.80\cdot 10^{-19}J is the change in kinetic energy of the particle

Therefore, the potential difference (in magnitude) is

\Delta V=\frac{\Delta U}{q}=\frac{4.80\cdot 10^{-19}}{4.80\cdot 10^{-19}}=1 V

2)

Here we have to evaluate the direction of motion of the particle.

We have the following informations:

- The electric potential increases in the +x direction

- The particle is positively charged and moves from point a to b

Since the particle is positively charged, it means that it is moving from higher potential to lower potential (because a positive charge follows the direction of the electric field, so it moves away from the source of the field)

This means that the final position b of the charge is at lower potential than the initial position a; therefore, the potential difference must be negative:

V_b-V_a = - 1V

8 0
2 years ago
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