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3 years ago

Measuring the amount of deuterium in the universe allows us to set a limit on _________.

Physics

1 answer:

Mademuasel [1]3 years ago

4 0

Answer:

the density of ordinary (baryonic) matter in the universe

Explanation:

Deuterium detection is of interest because the amount of it may be related to the amount of dark matter in the universe, but precise measurements have been difficult to obtain. Due to the way in which deuterium was created in the Big Bang, an exact measurement of the amount of deuterium would allow scientists to set limits on the models of the great explosion.

Also, an exact measure of deuterium would be an indicator of the cosmic density of barions (ordinary matter), and that density of barions would indicate whether ordinary matter is dark and is found in regions such as black holes, gas clouds or brown dwarfs, or it is bright and can be found in the stars. This information will help scientists who try to understand the very beginning of our universe.

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two charges experience exert a force of 1n on each other when they are 1m apart. what force will these charges experience if the

Kisachek [45]

The force experienced by the charges when they are placed 2 m apart will be 0.25 N.

The force between two charges is given by Coulomb’s Law.

According to this law, the force between two charges varies inversely as the square of the distance between the charges and is directly proportional to the product of the magnitude of the two charges.

The above statement is represented by the following equation,

F= K Q1 x Q2/r^2

Here, F = Force between the two charges

K = Constant

Q1 and Q2 = Magnitude of the two charges

r = Separation between the two charges

According to this equation, the force is inversely proportional to the square of the distance between two charges.

Initially, when the two charges are 1 m apart, the force is 1 N.

When the distance between the two charges is doubled,

then according to Coulomb’s Law, the force should decrease by 4 times.

Hence, if the charges are placed 2 meters apart, the force becomes 1/4= 0.25 N.

To know more about "Coulomb's Law", refer to the following link:

brainly.com/question/9261306?referrer=searchResults

#SPJ4

8 0

1 year ago

A hydrogen fuel cell supplies power for a small motor. the fuel cell delivers a current of 0.5 a and a voltage of 0.43 v. what i

gogolik [260]

We want to know what is the power supplied by the power cell if the current I=0.5 A and the voltage V=0.43 V. The equation for power P is P= I*V, so:

P=I*V=0.5*0.43=0.215 W

So the correct answer is that the power cell is supplying the motor with P=0.215 W of power.

8 0

3 years ago

Find B when θ=35, E=10V, t=5, N=250, A=1.20m^2

givi [52]

Answer:

its most definitely c. trust me

Explanation:

3 0

2 years ago

A cannon is fired from the edge of a cliff, which is 60m above the sea. The cannonball's initial velocity is 88.3m/s and it is f

wel

Answer:

a. 11.29 s b. 94.72 m/s at -39.8° c. 821.57 m

Explanation:

a. Using y - y₀ = ut - 1/2gt² where u = vertical component of velocity = v₀sinθ where v₀ = 88.3 m/s and θ = 34.5°, y₀ = + 60 m and y = water surface = 0 m, g = 9.8 m/s² and t = time it takes the cannon to reach the water surface.

So y - y₀ = ut - 1/2gt²

y - y₀ = (v₀sinθ)t - 1/2gt²

substituting the values of the variables into the equation, we have

0 - 60 = (88.3 m/s × sin34.5°)t - 1/2 × 9.8 m/s²× t²

- 60 = 50t - 4.9t²

So, 4.9t² - 50t - 60 = 0

Using the quadratic formula to find t,

$t = \frac{-(-50) +/- \sqrt{(-50)^{2} - 4 X 4.9 X -60} }{2 X 4.9} \\t = \frac{50 +/- \sqrt{2500 + 1176} }{9.8} \\t = \frac{50 +/- \sqrt{3676} }{9.8} \\t = \frac{50 +/- 60.63 }{9.8} \\t = \frac{50 + 60.63 }{9.8} or t = \frac{50 - 60.63 }{9.8} \\t = \frac{110.63 }{9.8} or t = \frac{-10.63 }{9.8} \\t = 11.29 sor -1.085$

Since t cannot be negative, t = 11.29 s

b. We first need to find the impact vertical velocity component. Using

v = u - gt where u = initial vertical velocity component = v₀sinθ and t = 11.29 s and g = 9.8 m/s². So,

v = v₀sinθ - gt

= 88.3 m/s × sin34.5° - 9.8 m/s² × 11.29 s

= 50.01 m/s - 110.64 m/s

= -60.63 m/s

Since the horizontal velocity is constant u' = v₀cosθ = 88.3 m/s × cos34.5° = 72.77 m/s.

The impact velocity is thus the resultant of the horizontal velocity and final impact velocity. So, V = √(v² + u'²)

= √((-60.63 m/s)² + (72.77 m/s)²)

= √((3676 m²/s² + 5295.48 m²/s²)

= √(8971.48 m²/s²

= 94.72 m/s

The angle θ = tan⁻¹(v/u') = tan⁻¹(-60.63 m/s ÷ 72.77 m/s) = tan⁻¹(-0.8332) = -39.8°

So the impact velocity is 94.72 m/s at -39.8°

c. The horizontal distance out from the base of the cliff that the ball strikes the water is the range, R = u't = 72.77 m/s × 11.29 s = 821.57 m

5 0

3 years ago

HELP ME

Mazyrski [523]

Answer:

I THINK IT'S D....

HOPE SO

3 0

3 years ago