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3 years ago

What is cosmic background radiation?

2 answers:

n200080 [17]3 years ago

8 0

Cosmic background radiation is the energy left over from the big bang and it's everywhere we look so if we consider ourselves the center, no matter which way we look we see this cosmic background radiation which is left over energy from the big bang.

Kobotan [32]3 years ago

6 0

Cosmic background radiation is electromagnetic radiation from the sky with no discernible source. The origin of this radiation depends on the region of the spectrum that is observed.

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Your sister has been taking her friend's ADHD medication. She is struggling with trying to quit using the medication and asks yo

salantis [7]

The correct answer for this is A. <span>Accompany her to talk with your parents or another trusted adult to ask for help

when a friend is under the problems of addiction, it's always best to get help from a trusted adult.</span>

5 0

3 years ago

Read 2 more answers

The universe could be considered an isolated system because

Stolb23 [73]

Answer:

A(many people think that no energy or matter exists outside the universe)

Explanation:

4 0

3 years ago

Two particles, one with charge -5.45 × 10^-6 C and one with charge 4.39 × 10^-6 C, are 0.0209 meters apart. What is the magnitud

Levart [38]

Explanation:

Given that,

Charge 1, $q_1=-5.45\times 10^{-6}\ C$

Charge 2, $q_2=4.39\times 10^{-6}\ C$

Distance between charges, r = 0.0209 m

1. The electric force is given by :

$F=k\dfrac{q_1q_2}{r^2}$

$F=9\times 10^9\times \dfrac{-5.45\times 10^{-6}\times 4.39\times 10^{-6}}{(0.0209)^2}$

F = -492.95 N

2. Distance between two identical charges, $r=0.0209\ m$

Electric force is given by :

$F=\dfrac{kq_3^2}{r^2}$

$q_3=\sqrt{\dfrac{Fr^2}{k}}$

$q_3=\sqrt{\dfrac{492.95\times (0.0209)^2}{9\times 10^9}}$

$q_3=4.89\times 10^{-6}\ C$

Hence, this is the required solution.

3 0

2 years ago

Using Rayleigh's criterion, what is the smallest separation between two pointlike objects that a person could clearly resolve at

atroni [7]

Answer:

y = 52.44 10⁻⁶ m

Explanation:

It is Rayleigh's principle that two points are resolved if the maximum of the diffraction pattern of one matches the minimum the diffraction pattern of the other

Based on this principle we must find the angle of the first minimum of the diffraction expression

a sin θ= m λ

The first minimum occurs for m = 1

sin θ = λ / a

Now let's use trigonometry the object is a distance L = 0.205 m

tan θ = y / L

Since the angles are very small, let's approximate

tan θ = sin θ/cos θ = sin θ

sin θ = y / L

We substitute in the diffraction equation

y / L = λ / a

y = λ L / a

Let's calculate

y = 550 10⁻⁹ 0.205 / 2.15 10⁻³

y = 52.44 10⁻⁶ m

7 0

3 years ago

A woman is 160160cm tall. What is the minimum vertical length of a mirror in which she can see her entire body while standing up

kipiarov [429]

This question is incomplete, the complete question is;

A woman is 160cm tall. What is the minimum vertical length of a mirror in which she can see her entire body while standing upright.

Hint: Consider the ray diagram below, of the rays that enable her to see her feet and the top of her head.

Use what you know about the law of reflection, together with a bit geometry.

The missing Image is uploaded along this answer below.

Answer:

the minimum vertical length of a mirror in which she can see her entire body while standing upright is 80 cm

Explanation:

Given the data in the question and illustrated in the image below,

From image 2;

The distance from the woman's eyes to the top of her head is represented as b and a represent the distance from her eyes to her feet.

Therefore, since her height is 160 cm

a + b = 160 ------ let this be equation 1

i.e AD + DG = 160 cm

Now, from the same image 2, we will notice that triangle ABC and tringle CBD are similar, so

∠ABC = ∠CBD

AC = CD

since AD = a and AC + CD = A

AC = CD = a/2

Also, triangle DEF and FEG are si,ilar

∠DEF = ∠FEG

DF = FG

since DG = b and DF + FG = b

DF = b/2

so the minimum vertical length of a mirror in which the woman can see her entire body while standing upright will be;

⇒ a/2 + b/2

⇒ a + b / 2

from equation 1, a + b = 160

⇒ a + b / 2 = 160 / 2 = 80

Therefore, the minimum vertical length of a mirror in which she can see her entire body while standing upright is 80 cm

4 0

3 years ago