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1 year ago

What role does gravity have in the motion of planets around the sun?

Physics

1 answer:

mina [271]1 year ago

8 0

Answer:

The gravity helps the planets stay together near the Sun, without it Earth would be floating away with the planet eventually becoming frozen up, thus the role gravity have in the motion of planets around the sun is by keeping them together.

Explanation:

Hope this helps!

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A point charge with a charge q1 = 4.00 μC is held stationary at the origin. A second point charge with a charge q2 = -4.10 μC mo

GrogVix [38]

Answer:

W = -0.480 J

Explanation:

given,

q₁ = 4 μC

q₂ = -4.10 μC

$W = kq_1q_2(\dfrac{1}{a}+\dfrac{1}{b})$

$b = \sqrt{(0.27-0)^2+(0.27-0)^2}$

b = 0.381

k = 8.99 × 10⁹ Nm²/C²

$W = 8.99\times 10^9\times 4\times 10^{-6}\times (-4.1 \times 10^{-6})(\dfrac{1}{0.17}+\dfrac{1}{0.381})$

$W = [-147.436\times (5.88-2.62)\times 10^{-3}]J$

W = -0.480 J

Work done by the electric force W = -0.480 J

4 0

2 years ago

Darren filled ocean water, fresh water, bottled water, and tap water into four different containers. He then dropped identical g

elena-14-01-66 [18.8K]

I would say container 1

6 0

2 years ago

Read 2 more answers

These questions !plz !! i need help!!!

Nataly_w [17]

(6) Wagon B is at rest so it has no momentum at the start. If v is the velocity of the wagons locked together, then

(140 kg) (15 m/s) = (140 kg + 200 kg) v

==> v ≈ 6.2 m/s

(7) False. If you double the time it takes to perform the same amount of work, then you halve the power output:

E / (2t ) = 1/2 × E/t = 1/2 P

3 0

2 years ago

Two bicycle tires are set rolling with the same initial speed of 4.0 m/s along a long, straight road, and the distance each trav

umka2103 [35]

Answer:

The coefficient of rolling friction will be "0.011".

Explanation:

The given values are:

Initial speed,

$v_i = 4.0 \ m/s$

then,

$v_f=\frac{4.0}{2}$

$=2.0 \ m/s$

Distance,

s = 18.2 m

The acceleration of a bicycle will be:

⇒ $a=\frac{v_f^2-v_i^2}{2s}$

On substituting the given values, we get

⇒ $=\frac{(2.0)^2-(4.0)^2}{2\times 18.2}$

⇒ $=\frac{4-8}{37}$

⇒ $=\frac{-4}{37}$

⇒ $=0.108 \ m/s^2$

As we know,

⇒ $f=ma$

and,

⇒ $\mu_rmg=ma$

⇒ $\mu_r=\frac{a}{g}$

On substituting the values, we get

⇒ $=\frac{0.108}{9.8}$

⇒ $=0.011$

7 0

2 years ago

A round object of mass 10 kg and radius 0.5 m rolls without slipping down a hill from a height of 4.5 m. If its speed at the bot

Mice21 [21]

Answer:

moment of inertia is 2.72 kg m²

Explanation:

given data

mass m = 10kg

height h = 4.5 m

radius r = 0.5 m

speed v = 6.5 m/s

to find out

moment of inertia

solution

we apply here conservation of energy

that is

mgh = 1/2 ×mv² + 1/2 × Iω²

here I is moment of inertia we find and

we know ω = Velocity / radius = 6.5 / 0.5 = 13

and g = 9.8

so put here all these value

10 (9.8) 4.5 = 1/2 ×(10)(6.5)² + 1/2 × I(13)²

441 = 211.25 + 1/2 × I( 169 )

I = 2.72

so moment of inertia is 2.72 kg m²

7 0

3 years ago