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Nana76 [90]
3 years ago
5

2. Stars normally convert hydrogen into helium through nuclear fusion. That requires incredibly hot temperatures and high pressu

re. At the moment, the temperature of empty space is nowhere near warm enough to fuse any elements together. What type of temperature and pressure conditions needed to be present in the early universe in order to create the first hydrogen atoms
Physics
1 answer:
jolli1 [7]3 years ago
6 0

Answer:

About 4,000 K and 10⁻¹⁷ atm

Explanation:

The Big Bang theory states that the Big Bang which is the origin of the universe was about 13.75 billion years ago, and the temperature a few seconds later was 10³²K

The first element began forming at about 3 minutes after the Big Bang with a temperature of 10⁹ K, the nuclei of simple elements

The nuclei of hydrogen and helium began combine with electrons at a temperature of 3,000 K to 4,000 K to form the first neutral atoms. The pressure of the universe at that stage was 10⁻¹⁷ atmospheres

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Setler79 [48]

Answer:

C because it make sens

C the light wave traveled through ice and then through a Dimond.

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2 years ago
When air is blown across the top of an open
Tanzania [10]
When air is blown across the top of an open <span>water bottle, air molecules in the bottle vibrate at a particular frequency and sound is produced in a process called "refraction". 
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8 0
3 years ago
A metal ball rolls from rest at Point A down the track to Point E as shown below.
Tju [1.3M]

Answer:

Explanation:

Velocity is at its greatest when kinetic energy is at its max which is when all the ball's energy has been transformed from potential energy to kinetic energy which is at the lowest point in its travels (assuming the ball is rolling down a ramp). You have no picture here so this answer is a general one, not a specific one.

8 0
3 years ago
A uniform solid disk of mass 5.00 kg and diameter 47.0 cm starts from rest and rolls without slipping down a 40.0 ∘ incline that
saveliy_v [14]

Answer:

The speed it reaches the bottom is

v=6.51m/s

Explanation:

Given: m=5.0kg, r=47cm\frac{1m}{100cm}=0.47m

Using the conservation of energy theorem

U_i=K_E+K_{ER}

m*g*h=\frac{1}{2}*m*v^2+\frac{1}{2}*I*w^2

v=r*w, I=\frac{1}{2}*m*r^2

m*g*h=\frac{1}{2}*m*(r*w)^2 +\frac{1}{2}*[\frac{1}{2} *m*r^2]*w^2

m*g*h=\frac{3}{4}*m*r^2*w^2

g*h=\frac{3}{4}*r^2*w^2

Solve to w'

w^2=\frac{4*g*h}{3*r^2}

h=x*sin(30)=6.5m*sin(30)=3.25m

w=\sqrt{\frac{4*9.8m/s^2*3.25m}{3*(0.235m)^2}}

w=27.74rad/s

v=27.74rad/s*0.235m=6.51m/s

7 0
3 years ago
A 2500 kg car traveling to the north is slowed down uniformly from an initial velocity of 20 m/s by a 5620 N braking force actin
Dennis_Churaev [7]

Answer:

the distance traveled by the car is 42.98 m.

Explanation:

Given;

mass of the car, m = 2500 kg

initial velocity of the car, u = 20 m/s

the braking force applied to the car, f = 5620 N

time of motion of the car, t = 2.5 s

The decelaration of the car is calculated as follows;

-F = ma

a = -F/m

a = -5620 / 2500

a = -2.248 m/s²

The distance traveled by the car is calculated as follows;

s = ut + ¹/₂at²

s = (20 x 2.5) + 0.5(-2.248)(2.5²)

s = 50 - 7.025

s = 42.98 m

Therefore, the distance traveled by the car is 42.98 m.

5 0
3 years ago
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