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Serhud [2]
3 years ago
14

The planets in the solar system are believed to have formed as a result of a process known as

Physics
1 answer:
Nonamiya [84]3 years ago
8 0
The answer is the Big Bang 
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Which statement does not describe the relationship between genes and
jek_recluse [69]

Answer: C. Inherited traits carry the instructions for individual genes.

Explanation: bejewels I know stuff. ☆ - ~ hope this helps

4 0
3 years ago
Read 2 more answers
4.) An apartment building is on fire and a guy is trapped on the fire escape ladder. There is a
Tems11 [23]

Answer:

5.3 m/s

Explanation:

First, find the time it takes for him to fall 7m.

y = y₀ + v₀ t + ½ at²

0 = 7 + (0) t + ½ (-9.8) t²

0 = 7 − 4.9 t²

t ≈ 1.20 s

Now find the velocity he needs to travel 6.3m in that time.

x = x₀ + v₀ t + ½ at²

6.3 = 0 + v₀ (1.20) + ½ (0) (1.20)²

v₀ ≈ 5.27 m/s

Rounded to two significant figures, the man must run with a speed of 5.3 m/s.

3 0
3 years ago
What is the impulse of a 3 kg object that starts from rest and moves to 20 m/s?
IrinaK [193]

Answer:

The impulse on the object is 60Ns.

Explanation:

Impulse is defined as the product of the force applied on an object and the time at which it acts. It is also the change in the momentum of a body.

F = m a

F = m(\frac{v_{2}  - v_{1} }{t})

⇒     Ft = m(v_{2} - v_{1})

where: F is the dorce on the object, t is the time at which it acts, m is the mass of the object, v_{1} is its initialvelocity and v_{2} is the final velocity of the object.

Therefore,

impulse = Ft = m(v_{2} - v_{1})

From the question, m = 3kg, v_{1} = 0m/s and v_{2} = 20m/s.

So that,

Impulse = 3 (20 - 0)

             = 3(20)

             = 60Ns

The impulse on the object is 60Ns.

8 0
3 years ago
Suppose that light from a laser with wavelength 633 nm is incident on a thin slit of width 0.500 mm. If the diffracted light pro
coldgirl [10]

Explanation:

It is given that,

Wavelength, \lambda=633\ nm=633\times 10^{-9}\ m

Slit width, a=0.5\ mm=0.0005\ m

Order, m = 2

If the diffracted light projects onto a screen at distance 1.50 m, L = 1.5 m

For the diffraction of light,

y=\dfrac{m\lambda L}{a}

y=\dfrac{2\times 633\times 10^{-9}\times 1.5}{0.0005}

y = 0.0037 m

So, the distance from the center of the diffraction pattern to the dark band is 0.0037 meters. Hence, this is the required solution.

3 0
3 years ago
A flag is waved 3.2 m above the surface of a flat pool of water. When viewed from under the water, what is the magnification of
jek_recluse [69]

Answer:

1.33

Explanation:

For an optical instrument, the magnification ratio of the apparent diameter of the image to that of the object.

Mathematically, from the given information;

Magnification= \dfrac{n_{water}}{n_{air}}

where;

n_{water} =1.33\\ \\ n_{air} = 1.00

= \dfrac{1.33}{1.00} \\ \\=  \mathbf{1.33}

5 0
2 years ago
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