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Pani-rosa [81]
3 years ago
15

Give three examples of how gravity is at work in our solar system.

Physics
2 answers:
musickatia [10]3 years ago
6 0
Well.....
Gravity from the sun pulls the planets torward it while inertia pulls it outward....but I guess that would be why it orbits sorry if this doesn't help but uh
mariarad [96]3 years ago
3 0
Gravity is the force that keeps all of the planets, dwarf planets, moons, asteroids and comets orbiting the Sun.<span> The force of gravity is inversely proportional to the distance of any two objects. This means that objects that are close to the Sun experience a stronger gravitational pull, and therefore, orbit the Sun at a greater speed</span>
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A car is traveling at 75 m/s. 50 seconds later it is traveling at 25 m/s. What is the car’s acceleration?
scoray [572]

Answer:

a = -1 m/s^2

Explanation:

Vi = 75 m/s

Vf = 25 m/s

t = 50 s

Plug those values into the following equation:

Vf = Vi + at

25 = 75 + 50a

---> a = -1 m/s^2

5 0
3 years ago
Explain any two areas affected by migration​
mel-nik [20]

Answer:

<h2>Migration is affected by various factors like age, sex, marital status, education, occupation, employment etc. Age and sex are main demographic factors that affect the migration. Men, generally, migrate to other places quite often though there are more women who migrate to husbands' places after marriage.</h2>
4 0
3 years ago
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Most earthquakes happen:
umka2103 [35]

Along plate edges, at points where oceanic or continental plates meet ot at the edges of the plates

4 0
3 years ago
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A 1.60 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 1
bija089 [108]

1.

Answer:

Part a)

\rho = 1.35 \times 10^{-5}

Part b)

\alpha = 1.12 \times 10^{-3}

Explanation:

Part a)

Length of the rod is 1.60 m

diameter = 0.550 cm

now if the current in the ammeter is given as

i = 18.7 A

V = 17.0 volts

now we will have

V = I R

17.0 = 18.7 R

R = 0.91 ohm

now we know that

R = \rho \frac{L}{A}

0.91 = \rho \frac{1.60}{\pi(0.275\times 10^{-2})^2}

\rho = 1.35 \times 10^{-5}

Part b)

Now at higher temperature we have

V = I R

17.0 = 17.3 R

R = 0.98 ohm

now we know that

R = \rho \frac{L}{A}

0.98 = \rho' \frac{1.60}{\pi(0.275\times 10^{-2})^2}

\rho' = 1.46 \times 10^{-5}

so we will have

\rho' = \rho(1 + \alpha \Delta T)

1.46 \times 10^{-5} = 1.35 \times 10^{-5}(1 + \alpha (92 - 20))

\alpha = 1.12 \times 10^{-3}

2.

Answer:

Part a)

i = 1.55 A

Part b)

v_d = 1.4 \times 10^{-4} m/s

Explanation:

Part a)

As we know that current density is defined as

j = \frac{i}{A}

now we have

i = jA

Now we have

j = 1.90 \times 10^6 A/m^2

A = \pi(\frac{1.02 \times 10^{-3}}{2})^2

so we will have

i = 1.55 A

Part b)

now we have

j = nev_d

so we have

n = 8.5 \times 10^{28}

e = 1.6 \times 10^{-19} C

so we have

1.90 \times 10^6 = (8.5 \times 10^{28})(1.6 \times 10^{-19})v_d

v_d = 1.4 \times 10^{-4} m/s

8 0
4 years ago
A sample contains 36 g of a radioactive isotope. How much radioactive isotope remains in the sample after 3 half-lives?
Gre4nikov [31]

Answer:

<u>Option "C":</u> "4.5 g"

Explanation:

N0 = 36 g, Let half-life is T.

t = 3 T, n is number of half lives = t / T = 3

<u>By using the decay law of radioactivity</u>

N / N0 = (1 / 2)^n

where

"N0" be the "initial amount"

"N" be the "amount left"

"n" be the "number of half-lives"

N / 36 = (1/2)^3

N / 36 = 1 / 8

N = 36 / 8 = 4.5 g

3 0
3 years ago
Read 2 more answers
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