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VARVARA [1.3K]
3 years ago
8

Two asteroids are 50,000 m apart. one has a mass of 5 x 10^8 kg. If the force of gravity between them is 8.67 x 10 ^ -2 N, what

is the mass of the other asteroid?

Physics
2 answers:
Tema [17]3 years ago
6 0
6.5 *10^9 kg
is the mass of the second one

9966 [12]3 years ago
6 0

Answer:

The mass of the other asteroid is m_2=6.49\times 10^9\ kg.

Explanation:

Given that,

Mass of one asteroid, m=5\times 10^8\ kg

The separation between two asteroids, r = 50,000 m

The force of gravity between asteroids, F=8.67\times 10^{-2}\ N

We need to find the mass of the other asteroid. The gravitational force acting between two masses is given by :

F=G\dfrac{m_1m_2}{r^2}

m_2 is the mass of other asteroid

m_2=\dfrac{Fr^2}{Gm_1}

m_2=\dfrac{8.67\times 10^{-2}\times (50000)^2}{6.67\times 10^{-11}\times 5\times 10^8}

m_2=6.49\times 10^9\ kg

So, the mass of the other asteroid is m_2=6.49\times 10^9\ kg. Hence, this is the required solution.

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What are two ways you can increase power by climbing the stairs?
Nina [5.8K]

Answer:

change the

  • time
  • distance
  • force.

Explanation:

P = W/time

W = F*d

You have control over how fast you go up the stairs.

You also have control over how far up the stairs you go.

Therefore the answer is

  • time
  • distance

If you don't like distance as an answer, you can carry something up the stairs -- anything that increases F will do.

8 0
2 years ago
A shell is fired at an angle of 35° above the horizontal at a velocity of 40 m/s. (a) What is it's speed at the highest point of
Fudgin [204]

Answer:

Part a)

v_f = v_x = 32.77 m/s

Part b)

T = 4.68 s

Explanation:

Part a)

Shell is fired at speed of 40 m/s at angle of 35 degree

so here we have

v_x = 40 cos35 = 32.77 m/s

v_y = 40 sin35 = 22.94 m/s

since gravity act opposite to vertical speed of the shell so at the highest point of its trajectory the vertical component of the speed will become zero

so at the highest point the speed is given

v_f = 32.77 m/s

Part b)

After completing the motion we know that the displacement of the object will be zero in Y direction

so we have

\Delta y = 0

0 = v_y t - \frac{1}{2}gt^2

T = \frac{2v_y}{g}

T = \frac{2(22.94)}{9.81} = 4.68 s

7 0
3 years ago
While eating a peanut butter and jelly sandwich peanut butter drips on Dana shirt she tries to remove the peanut butter from her
MakcuM [25]

Answer:

The water does not remove the peanut butter because the peanut butter is thicker substance than the water is, therefore it would not remove immediately.. It could also be because the brown dye or color in the peanut butter color or the peanut butter itself has already gone through her clothing Dana would have to remove it as much as she could with a towel water and some soap then throw it to the washer afterward.

Hope this helps, good luck:)

6 0
3 years ago
The tongue weight of a trailer should be what percent of the gross trailer weight rating
mario62 [17]

Answer:

between 10 and 15 percent

Explanation:

How to put your load

- First load the heavy

The safe trailer starts loading correctly. Uneven weight can affect steering, brakes and swing control.

In general, 60% of the weight of the load should be in the front half of the trailer and 40% in the rear half (unless the manufacturer indicates something different). When you place the load, you want it to be balanced from side to side, keeping the center of gravity near the ground and on the axle of the trailer.

-  Hold your load

After balancing the load, you must hold it in place. An untapped load can move when the vehicle is moving and cause trailer instability.

- Trailer weight

To avoid overloading the trailer, look for the recommended weight rating. It is located on the VIN plate in the trailer chassis, usually on the tongue. Confirm the Gross Vehicle Weight Classification (GVWR) before towing.

GVWR: is the total weight that the trailer can support, including its weight. You can also find this number as the Gross Trailer Weight (GTW). The weight of the tongue should be 10-15% of the GTW.

7 0
2 years ago
Can someone please help me with this physics question? I'm desperate!
Lelu [443]

Answer:

a) 2·√10 seconds

b) Linda should be approximately 30.6 meters

c) Jenny's speed at the 100-m mark is approximately 6.325 m/s

Explanation:

The speed with which Linda is running = 8.6 m/s

The point Jenny starts = The 80-m mark

The acceleration of Jenny = 1.0 m/s²

a) The time it takes Jenny to run from the 80-m mark to the 100-m mark, <em>t</em>, is given as follows

Δs = u·t + (1/2)·a·t²

Δs = Distance = 100-m - 80-m = 20-m

u = The initial velocity of Jenny = 0

a = Jenny's acceleration = 1.0 m/s²

∴ 20 = 0×t + (1/2) × 1 × t² = t²/2

20 = t²/2

t = √(20 × 2) = 2·√10

The time it takes Jenny to run from the 80-m mark to the 100-m mark = 2·√10 seconds

b) The distance Linda runs in t = 2·√10 seconds, d = v × t

Given that Linda's velocity, v = 8.6 m/s, we have;

d = 8.0 × 2·√10 = 16·√10

The distance Linda runs in t = 2·√10 seconds = 16·√10 meters ≈ 50.6 meters

Therefore, Linda should be approximately (50.6 - 20) meters = 30.6 meters behind Jenny when Jenny starts running

c) Jenny's speed at the 100 m mark is given as follows;

v = u + a·t

t = 2·√10 seconds, a = 1.0 m/s², u = 0

∴ v = 0×t + 1.0×2·√10 = 2·√10 ≈ 6.325

Jenny's speed at the 100-m mark ≈ 6.325 m/s

3 0
2 years ago
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