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emmainna [20.7K]
3 years ago
6

Before the big bang, the universe was much ____ and _____ than it is now.

Physics
2 answers:
kherson [118]3 years ago
7 0
My answer is that Before the big bang, the universe was much cooler and denser. There are three circumstances before the big bang theory, and it shows that the flatness problem and horizon problem affect the universe to be cooler and denser. In flatness problem, it states there that the universe seems to have enough energy density in the form of matter and in horizon problem it affects the coolness of the universe. 
Dahasolnce [82]3 years ago
5 0
The answer to the question is that before the big bang, the universe was much hotter and more dense than it is now. Letter B.
It is because after the big bag occurred, the universe became cooler and less dense.
a. - does not correspond in the answer because the universe became less dense after the big bang.
c - the universe became cool and less dense after the big bang so being cool and less dense does not correspond to the question.
d - cooler does not answer the question because it only became cooler after the big bang.
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As a bat flies toward a wall at a speed of 6.0 m/s, the bat emits an ultrasonic sound wave with frequency 30.0 kHz. What frequen
Leto [7]

Answer:

f_b = 29.98 Hz

Explanation:

speed of bat = 6 m/s

sound wave frequency emitted by bat = 30.0 kHz

as we know,

speed of sound (c)= 343 m/s

f_w = f(\dfrac{c+v_r}{c+v_s})

f_w = 30(\dfrac{343+0}{343+6})

f_w = 29.48 Hz

now frequency received by bat is equal to  

f_b = f(\dfrac{c+v_r}{c+v_s})

f_b = 29.48(\dfrac{343+6}{343+0})

f_b = 29.98 Hz

hence the frequency hear by bat will be 29.98 Hz

7 0
3 years ago
What is the slope of the line plotted below?
lubasha [3.4K]

Answer:

Slope is in fact -2

Explanation:

Rise over Run

6 0
3 years ago
If a high jumper needs to make his center of gravity rise 1.50 m, how fast must he be able to sprint? Assume all of his kinetic
sergiy2304 [10]

Answer:

v = 5.42 m/s

Explanation:

given,

height of the jumper = 1.5 m

velocity of sprinter = ?

kinetic energy can be transformed into potential energy

m g h = \dfrac{1}{2}mv^2

g h = \dfrac{1}{2}v^2

v =\sqrt{2gh}

v =\sqrt{2\times 9.8 \times 1.5}

v = 5.42 m/s

Speed of the sprinter is equal to v = 5.42 m/s

7 0
3 years ago
Greg throws a 2.8-kg pumpkin horizontally off the top of the school roof in order to hit Mr. H's car. The car has parked a dista
Igoryamba

Answer:

The horizontal velocity is v = 9.2 m/s

Explanation:

From the question we are told that

     The mass of the pumpkin is  m = 2.8 \ kg

      The distance of the the car from the building's base is  d = 13.4 \ m

       The height of the roof is h = 10.4 \ m

       

The height is mathematically represented as

         h = \frac{1}{2} gt^2

Where g is the acceleration due to gravity which has a value of g =9.8 \ m/s^2

substituting values

          10.4= 0.5 * 9.8 * t

making the time taken the subject of the formula

         t = \frac{10.4}{0.5 * 9.8 }

          t = 1.457 \ s

The speed at which the pumpkin move horizontally can be represented mathematically  as

                         v = \frac{d}{t}

substituting values

                     v =\frac{13.4}{1.457}

                     v = 9.2 m/s

7 0
3 years ago
A 7 kg sled is initially at rest on a horizontal road. the sled is pulled a distance of 2.9 m by a force of 42 n applied to the
Artemon [7]
I have no idea on this question.
6 0
3 years ago
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