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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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Compared to a plane gravitational pull on the ground to 7 miles in sky is what
Citrus2011 [14]
When you're in an airplane that's 7 miles up off the ground, the strength of gravity plunges to only 99.6 percent of its strength all the way down on the ground. A big heavy person, who weighs 200 pounds down at the airport, weighs only 199 pounds 4.7 ounces in a plane at the altitude of 7 miles.
4 0
3 years ago
A rocket moves upward from rest with an acceleration of 40 m/s2 for 5 seconds. It then runs out of fuel and continues to move up
Snezhnost [94]

Answer:

Maximum height of rocket  = 2538.74 m

Explanation:

We have equation of motion s = ut + 0.5 at²

For first 5 seconds

          s = 0 x 5 + 0.5 x 40 x 5² = 500 m

Now let us find out time after 5 seconds rocket move upward.

We have the equation of motion v = u + at

After 5 seconds velocity of rocket

         v = 0 + 40 x 5 = 200 m/s

After 5 seconds the velocity reduces 9.8m/s per second due to gravity.

Time of flying after 5 seconds

          t=\frac{200}{9.81}=20.38s

Distance traveled in this 20.38 s

          s = 200 x 20.38 - 0.5 x 9.81 x 20.38² = 2038.74 m

Maximum height of rocket = 500 +2038.74 = 2538.74 m

6 0
3 years ago
Dolphin echolocation is similar to ultrasound. Reflected sound waves
s344n2d4d5 [400]

Answer:

Waves with high frequencies have shorter wavelengths that work better  than low frequency waves for successful echolocation.

Explanation:

To understand why high-frequency waves work better  than low frequency waves for successful echolocation, first we have to understand the relation between frequency and wavelength.

The relation between frequency and wavelength is given by

λ = c/f

Where λ is wavelength, c is the speed of light and f is the frequency.

Since the speed of light is constant, the wavelength and frequency are inversely related.

So that means high frequency waves have shorter wavelengths, which is the very reason for the successful echolocation because waves having shorter wavelength are more likely to reach and hit the target and then reflect back to the dolphin to form an image of the object.

Thus, waves with high frequencies have shorter wavelengths that work better  than low frequency waves for successful echolocation.

3 0
3 years ago
In a television set, electrons are accelerated from rest through a potential difference of 20 kV. The electrons then pass throug
Svetradugi [14.3K]

Answer: Fmax = 5.54*10^-12 N

Explanation: From the question, we have the potential difference (V) =20kv = 20,000v and strength of magnetic field (B) =0.41 T.

The maximum force experienced by a charge of magnitude (q) is given as

Fmax = qvB

Where v = velocity of electron.

The velocity of the electron can be gotten by using the work energy theorem.

The kinetic energy of the electron (mv²/2) equals the work done needed to accelerate it.

mv²/2 = qV.

Where m = mass of an electronic charge = 9.11×10^-31 kg, q = magnitude of an electronic charge = 1.609×10^-19 c, v = velocity of electron, V = potential difference = 20,000v.

By substituting the parameters, we have that

(9.11×10^-31 × v²)/2 = 1.609×10^-19 × 20000

(9.11×10^-31 × v²) = 1.609×10^-19 × 20000 ×2

v² = (1.609×10^-19 × 20000 ×2)/9.11×10^-31

v² = 64.36*10^(-16)/9.11×10^-31

v² = 7.0647×10^15

v = √7.0647×10^15

v = 8.40×10^7 m/s

Fmax = 1.609×10^-19 × 8.40×10^7 × 0.41

Fmax = 5.54*10^-12 N

7 0
3 years ago
One gallon of gasoline in an automobiles engine produces on average 9.50 kg of carbon dioxide, which is a greenhouse gas; that i
Musya8 [376]

The annual production of carbon dioxide is 124121.49×10^{6}[/tex] kg.

First we calculate the fuel consumed by each car in a year

Fuel consumed=6990/21.4=326.63 gallon

Now we calculate the amount of fuel consumed by 40 million cars in a year

Fuel consumed=326.63*40*10^6=13065.42 million gallon,

Now we can calculate the annual production of carbon dioxide in the USA

CO2 production rate=9.50*13065.42=124121.49*10^6 kg

Therefore the annual production of carbon dioxide in USA is 124121.49×10^{6}[/tex] kg

4 0
3 years ago
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