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ExtremeBDS [4]
3 years ago
10

What did scientists predict the big bang should have left ??

Physics
2 answers:
ELEN [110]3 years ago
4 0
<span>radiation, hydrogen, and helium </span>
Alecsey [184]3 years ago
3 0

Answer:

Cosmic Microwave Background Radiations (CMBR)

Explanation:

The big bang was that point from which the universe started to exist, releasing a huge amount of <u>Cosmic Microwave Background Radiations (CMBR) into the space</u>.

<u>CMBR is an electromagnetic radiation that are the the oldest remains of the big bang explosion</u>. With the continuous expansion of the universe after the explosion, this radiations ere constantly spreading in all the direction for the last 13.5 billion years approximately.

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Which symbol represents a type of electromagnetic radiation released during radioactive decay?
Tanya [424]
The answer for <span>electromagnetic radiation released during radioactive decay i</span>s C. He
8 0
3 years ago
In science, a broad idea that has been repeatedly verified so as to give scientists great confidence that it represents reality
Colt1911 [192]

In science, a broad idea that has been repeatedly verified so as to give scientists great confidence that it represents reality is called "a theory".

<u>Explanation:</u>

In science, the interpretation of a feature of the organic world that can be tested in repeat manner and analysed by applying agreed tests validation methods, calculation and observation in according to the scientific method, such process is called as a theory in science.

The difference lie between a theory and a hypothesis. Because hypothesis is an "educated guess". Overall it is either a proposed interpretation of an observed phenomenon, or a logical inference of a possible causal association between several phenomena.

6 0
3 years ago
I need help with these Physics problems​
adoni [48]

Answer:

1. 3 m

2. 27 s

Explanation:

1. "A car traveling at +33 m/s sees a red light and has to stop.  If the driver can accelerate at -5.5 m/s², how far does it travel?"

Given:

v₀ = 33 m/s

v = 0 m/s

a = -5.5 m/s²

Unknown: Δx

To determine the equation you need, look for which variable you don't have and aren't solving for.  In this case, we aren't given time and aren't solving for time.  So look for an equation that doesn't have t in it.

Equation: v² = v₀² + 2aΔx

Substitute and solve:

(0 m/s)² = (33 m/s)² + 2(-5.5 m/s²) Δx

Δx = 3 m

2. "A plane starting from rest at one end of a runway accelerates at 4.8 m/s² for 1800 m.  How long did it take to accelerate?"

Given:

v₀ = 0 m/s

a = 4.8 m/s²

Δx = 1800 m

Unknown: t

Equation: Δx = v₀ t + ½ a t²

Substitute and solve:

1800 m = (0 m/s) t + ½ (4.8 m/s²) t²

t ≈ 27 s

4 0
3 years ago
a machine with a mechanical advantage of 2.5 requires an input force of 120 newtons. What output force is applied by this machin
creativ13 [48]
If your machine has a mechanical advantage of 2.5, then WHATEVER force you apply to the input, the force at the output will be 2.5 times as great.

If you apply 1 newton to the machine's input, the output force is

                 (2.5 x 1 newton)  =  2.5 newtons.

If you apply 120 newtons to the machine's input, the output force is

                 (2.5 x 120 newtons)  =  300 newtons.

4 0
3 years ago
A real object is 10.0 cm to the left of a thin, diverging lens having a focal length of magnitude 16.0 cm. What is the location
amm1812

Answer:

A)6.15 cm to the left of the lens

Explanation:

We can solve the problem by using the lens equation:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}

where

q is the distance of the image from the lens

f is the focal length

p is the distance of the object from the lens

In this problem, we have

f=-16.0 cm (the focal length is negative for a diverging lens)

p=10.0 cm is the distance of the object from the lens

Solvign the equation for q, we find

\frac{1}{q}=\frac{1}{-16.0 cm}-\frac{1}{10.0 cm}=-0.163 cm^{-1}

q=\frac{1}{-0.163 cm^{-1}}=-6.15 cm

And the sign (negative) means the image is on the left of the lens, because it is a virtual image, so the correct answer is

A)6.15 cm to the left of the lens

6 0
3 years ago
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