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vodomira [7]
3 years ago
10

What is the name of the small early planets, which formed through gravitational attraction reaching sizes of a few miles to even

tually the size of our moon?
Physics
1 answer:
Svetradugi [14.3K]3 years ago
6 0

Answer:

Planetesimals  (Ex, Mercury, Venus , Mars & Earth)

Explanation:

They are known as Planetesimals , They are small in size and have rocky surface. The examples of Planetesimals  are Mercury, Venus Mars and Earth.

They are small early planets , which also has gravitational attraction.

Thanks

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A 0.6 kg basketball moving 7.2 m/s to the right collides with a 0.04 kg tennis
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B

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A block with a weight of 9.0 N is at rest on a horizontal surface. A 1.2 N upward force is applied to the block by means of an a
lesya [120]

Answer:

a)   9 - 1.2 = 7.8 N

b) Since the force exerted by the box is it's weight, it acts in a downward direction. So the box will exert a force downward (perpendicular to the horizontal surface).

Explanation:

A 9N block would exert 9N of normal force on the horizontal plane under normal conditions. But in this case, we have a spring taking away some of the force by applying an upward force on the box.

So the force exerted by the box on the surface would now be:

a)   9 - 1.2 = 7.8 N

b) Since the force exerted by the box is it's weight, it acts in a downward direction. So the box will exert a force downward (perpendicular to the horizontal surface).

8 0
3 years ago
An archer standing on a 15 degree slope shoots an arrow at an angle of 26 degrees above the horizontal. How far below its origin
aivan3 [116]

Answer:

The arrow will hit 112.07 m from the point of release.

Explanation:

The equation for the position of an object in a parabolic movement is as follows:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

Where:

x0 = initial horizontal position

v0 = initial velocity

α = launching angle

y0 = initial vertical position

t = time

g = acceleration due to gravity

We know that at the final time the y-component of the vector "r" (see figure") is -1.88 m. The x-component of that vector will be the horizontal distance traveled by the arrow. Using the equation of the y-component of "r", we can obtain the final time and with that time we can calculate the value of the x-component (horizontal distance).

Then:

y = y0 + v0 · t · sin α + 1/2 · g · t²

Since the origin of the frame of reference is located at the point where the arrow is released, y0 = 0. Notice that the angle α = 26° + 15° = 41° ( see figure)

-1.88 m = 33 m/s · sin 41° · t - 1/2 · 9.8 m/s² · t²    (g is downward)

0 = -4.9 m/s² · t² + 33 m/s · sin 41° · t + 1.88 m

Solving the quadratic equation:

t = 4.5 s   ( the negative value is discarded)

Now, with this time we can calculate the horizontal distance:

x = x0 + v0 · t · cos α    (x0 = 0, the same as y0)

x = 33 m/s · 4.5 s · cos 41° = 112.07 m

6 0
3 years ago
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