All categories

3 years ago

When we look at a star

1 answer:

4 0

I'm not really sure what specific answer they're looking for, but if it's an open-ended question, then let's think about it this way...

A light year, is the distance it takes for light to travel in a year. If an object is 50,000 light years away, then by the time the light travels to us, 50,000 years has passed. We are looking at a 50,000 year old image of that object. (ignoring gravity and spatial expansion fun stuffs)

You might be interested in

Enter the expression 2cos2(θ)−1, where θ is the lowercase Greek letter theta.

stiv31 [10]

2cos2(θ)−1 = 0
2cos2(θ)=1
cos2(θ)=1/2
2θ=arc cos1/2
2θ=60 degrees
θ=30 degrees

4 0

2 years ago

Why does the output of a microphone increase as the frequency of the sound waves which it receives increases

Sloan [31]

Answer:

See Explanation

Explanation:

The frequency of sound waves received by the microphone influences the output or pitch of the sound obtained from the microphone.

The higher the frequency of the sound received by the microphone, the higher the output of the microphone and vice versa. This is because, the higher the frequency of sound, the higher the oscillations produced and the greater the output of the microphone.

The rise and fall in the pitch of sound waves as the frequency of sound waves varies is called inflection.

7 0

2 years ago

___meaning two' sugar (or saccharide) units in a chain.

ludmilkaskok [199]

The answer is c

Explanation:

3 0

2 years ago

I do believe that i need help

iris [78.8K]

WHAT IS THAT omg Helpppp

3 0

2 years ago

I attach a 4.1 kg block to a spring that obeys Hooke's law and supply 3.8 J of energy to stretch the spring. I release the block

borishaifa [10]

Answer:

The amplitude of the oscillation is 2.82 cm

Explanation:

Given;

mass of attached block, m = 4.1 kg

energy of the stretched spring, E = 3.8 J

period of oscillation, T = 0.13 s

First, determine the spring constant, k;

$T = 2\pi \sqrt{\frac{m}{k} }$

where;

T is the period oscillation

m is mass of the spring

k is the spring constant

$T = 2\pi \sqrt{\frac{m}{k} } \\\\k = \frac{m*4\pi ^2}{T^2} \\\\k = \frac{4.1*4*(3.142^2)}{(0.13^2)} \\\\k = 9580.088 \ N/m\\\\$

Now, determine the amplitude of oscillation, A;

$E = \frac{1}{2} kA^2$

where;

E is the energy of the spring

k is the spring constant

A is the amplitude of the oscillation

$E = \frac{1}{2} kA^2\\\\2E = kA^2\\\\A^2 = \frac{2E}{k} \\\\A = \sqrt{\frac{2E}{k} } \\\\A = \sqrt{\frac{2*3.8}{9580.088} }\\\\A = 0.0282 \ m\\\\A = 2.82 \ cm$

Therefore, the amplitude of the oscillation is 2.82 cm

8 0

2 years ago