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1 year ago

What cloud extending to the gravitational limits of the solar system would comets come from?.

Physics

1 answer:

ser-zykov [4K]1 year ago

4 0

The Oort cloud extends to the gravitational limits of the solar system would comets come from.

<h3>What is oort cloud?</h3>

It is a hypothetical idea of a cloud of mostly frozen planetesimals that would orbit the Sun at distances between 2,000 and 200,000 AU.

The Oort cloud reaches the solar system's outer gravitational boundaries, where comets originate.

Hence cloud extending to the gravitational limits of the solar system will be oorto cloud.

To learn more about the oort cloud refer;

brainly.com/question/23368033

#SPJ1

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a block of wood has a length of 4 cm a width of 5 cm and a height of 10 cm what is the volume of the wood

geniusboy [140]

Volume= Length X width X height.

Plug in the values for each and solve for the volume.

V= (L)(W)(H)

V=(4cm)(5cm)(10cm).

8 0

3 years ago

How much equal charge should be placed on the earth and the moon so that the electrical repulsion balances the gravitational for

kumpel [21]

As we know that electrostatic force between two charges is given as

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

here we know that electrostatic repulsion force is balanced by the gravitational force between them

so here force of attraction due to gravitation is given as

$F_g = 1.98 \times 10^{20} N$

here we can assume that both will have equal charge of magnitude "q"

now we have

$1.98 \times 10^{20} = \frac{kq^2}{r^2}$

$1.98 \times 10^{20} = \frac{(9\times 10^9)(q^2)}{(3.84 \times 10^8)^2}$

$1.98 \times 10^{20} = (6.10 \times 10^{-8}) q^2$

now we have

$q = 5.7 \times 10^{13} C$

6 0

3 years ago

A body is projected from the ground at an angle of 30° with the horizontal at an initial speed of 128 ft/s. Ignoring air frictio

Elanso [62]

Answer

given,

v = 128 ft/s

angle made with horizontal = 30°

now,

horizontal component of velocity

vx = v cos θ = 128 x cos 30° = 110.85 ft/s

vertical component of velocity

vy = v sin θ = 128 x sin 30° = 64 m/s

time taken to strike the ground

using equation of motion

v = u + at

0 =-64 -32 x t

t = 2 s

total time of flight is equal to

T = 2 t = 2 x 2 = 4 s

b) maximum height

using equation of motion

v² = u² + 2 a h

0 = 64² - 2 x 32 x h

64 h = 64²

h = 64 ft

c) range

R = v_x × time of flight

R = 110.85 × 4

R = 443.4 ft

4 0

3 years ago

A car travelling at 5.0 m/s starts to speed up. After 3.0 s its velocity has increased to 11 m/s. a) What is its acceleration? (

lions [1.4K]

initially, the car is traveling at 5.0 m/s.

so, we know acceleration for changing velocity is :

a = (v-v $_{o}$ )/t ..........(i)

where v is the final velocity

v $_{o}$ is the initial velocity

t is the time taken to change velocity

Now, as per the question :

initial velocity, v $_{o}$ =5.0 m/s

final velocity, v =11 m/s

time taken, t = 3 s

putting the values in equation (i),

a = ( 11-5 )/3

a = 2 m/s²

Therefore, a, after 3 s, is <em>2 m/s².</em>

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9 months ago

Solve 3* +5-220t = 0

slega [8]

Answer:

t = 27.5

Explanation:

$3 + 5 -220t = 0$

Well to solve for t we need to combine like terms and seperate t.

So 3+5= 8

8 - 220t = 0

We do +220 to both sides

8 = 220t

And now we divide 220 by 8 which is 27.5

Hence, t = 27.5

4 0

2 years ago