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3 years ago

11

Where did the elements in our solar system come from? they condensed out of hot gas clouds surrounding the early sun. they were

accreted from other solar systems. they were transformed from helium and hydrogen by the solar furnace?

1 answer:

Afina-wow [57]3 years ago

6 0

I think the correct answer would be that <u>they condensed out of hot gas clouds surrounding the early sun.</u><span />

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What is the relationship between an object’s mass and its gravitational potential energy?.

antiseptic1488 [7]

Answer:

If we’re talking about objects on the Earth, the gravitational potential energy is given by:

Explanation:

PEg=mgh

so the energy is proportional to the mass ( m ), but also to the strength of the gravitational field ( g ), and the height ( h ) to which the mass is lifted.

5 0

2 years ago

Please help! I’ll give brainliest! :)

jeka94

Answer:

C.) Gravity

Explanation:

The projectile is an object upon which the only force is gravity. Gravity acts to influence the vertical motion of the projectile.

6 0

2 years ago

Two bodies of masses 1000kg and 2000kg are separated 1km which is the gravitational force between them

denpristay [2]

Answer:

1.33×10⁻¹⁰ N

Explanation:

F = GMm / r²

where G is the gravitational constant,

M and m are the masses of the objects,

and r is the distance between them.

F = (6.67×10⁻¹¹ N/m²/kg²) (1000 kg) (2000 kg) / (1000 m)²

F = 1.33×10⁻¹⁰ N

3 0

3 years ago

What happens to the force between two charges when each charge is doubled and the distance between them is 1/4 its original

DIA [1.3K]

Answer:

F' = 64 F

Explanation:

The electric force between charges is given by :

$F=\dfrac{kq_1q_2}{r^2}$

Where

q₁ and q₂ are charges

r is the distance between charges

When each charge is doubled and the distance between them is 1/4 its original magnitude such that,

q₁' = 2q₁, q₂' = 2q₂ and r' = (r/4)

New force,

$F'=\dfrac{kq_1'q_2'}{r'^2}$

Apply new values,

$F'=\dfrac{k\times 2q_1\times 2q_2}{(\dfrac{r}{4})^2}\\\\=\dfrac{k\times 4q_1q_2}{\dfrac{r^2}{16}}\\\\=64\times \dfrac{kq_1q_2}{r^2}\\\\=64F$

So, the new force becomes 64 times the initial force.

7 0

2 years ago

Each plate of a parallel‑plate capacitor is a square of side 3.63 cm, and the plates are separated by 0.473 mm. The capacitor is

NARA [144]

Answer:

E= 55.53 x 10³ V/m

Explanation:

Given that

a= 3.63 cm

Area ,A= a²

distance ,d= 0.473 mm

Stored energy ,U = 8.49 nJ

Value of capacitor given as

$C=\dfrac{\varepsilon _oA}{d}$

By putting the values

$C=\dfrac{8.85\times 10^{-12}\times 3.63^2\times 10^{-4}}{0.473\times 10^{-3}}$

C=2.46 x 10⁻¹¹ F

$U=\dfrac{1}{2}CV^2$

V=Voltage difference

$V=\sqrt{\dfrac{2U}{C}}$

$V=\sqrt{\dfrac{2\times 8.49\times 10^{-9}}{2.46\times 10^{-11}}}$

V=26.27 V

V= E d

E=Electric filed

26.27 = E x 0.473 x 10⁻³

E= 55.53 x 10³ V/m

7 0

3 years ago