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Vesna [10]
3 years ago
15

This is one of the four planets Mercury,Venus,Earth,and Mars, whose orbits lie nearest the Sun

Physics
2 answers:
galben [10]3 years ago
6 0

Answer:

Mercury

Explanation:

Mercury is the closest planet to the sun followed be Venus. Though Mercury is the closest Venus is actually the hotest planet

qwelly [4]3 years ago
4 0
Mercy is the closest to orbit the sun.

Source: The internet. (Google, safari)
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Answer:

it's true

Explanation:

A strong body allows you to perform movements and activities that require power without getting tired.

3 0
2 years ago
What role did gravity played in the formation of the solar system​
hodyreva [135]

Gravity is the cause of . . .

-- star formation,

-- planet formation,

-- orbits.

So without gravity, there pretty much wouldn't be any solar or any system.

5 0
3 years ago
Please help brainliest, rattings, thanks, ect. its answer choice
Svetllana [295]

D: a force equal to the force

3 0
3 years ago
What’s the answer to question 12 please
matrenka [14]

Answer:

C) amplitude

Explanation:

"The amplitude is a measure of the strength or intensity of the wave. For example, when looking at a sound wave, the amplitude will measure the loudness of the sound. The energy of the wave also varies in direct proportion to the amplitude of the wave."-Ducksters

7 0
3 years ago
In a game of angry birds you launch a bird with an angle of 53 degrees to horizontal. Unfortunatly, its not a good shot and the
Alisiya [41]

Answer:

The maximum height covered is 3.25 m.

The horizontal distance covered is 9.81 m.

The total time in the air is 1.63 seconds.

Explanation:

The launch speed, u_0= 10 m/s.

Angle of launch with the horizontal, \theta = 53 ^{\circ}

So, the vertical component of the initial velocity,

u_0\sin\theta=10 \sin 53 ^{\circ}\cdots(i).

The horizontal component of the initial velocity,

u_0\cos\theta=10 \cos 53 ^{\circ}

Let, t be the time of flight, to the horizontal distance covered

D=10 \cos (53 ^{\circ})t\cdots(ii).

Not, applying the equation of motion in the vertical direction.

s= ut +\frac 1 2 at^2

Where s is the displacement in time t, u is the initial velocity and a is the acceleration.

In this case, u =10 \sin 53 ^{\circ} (from equation (i), s=0 (as the final height is same as the launch height) and a = -9.81 m/s^2 (negative sign is due to the downward direction).

\Rightarrow 0 = 10 (\sin 53 ^{\circ})t-\frac 1 2 (9.81)t^2

\Rightarrow t= \frac {2\times 10 (\sin 53 ^{\circ})}{9.81}=1.63 seconds.

So, the total time in the air is 1.63 seconds.

From equation (i),

Total horizontal distance covered is

D=10 \cos (53 ^{\circ})\times 1.63 = 9.81 m.

Now, for the maximum height, H, applying the equation of motion as

v^2=u^2+2as

Here, v is the final velocity and v=0 (at the maximum height), and h=H.

So, 0^2=(10 \sin 53 ^{\circ})^2-2(9.81)H

\Rightarrow H = \frac {(10 \sin 53 ^{\circ})^2}{2\times 9.81}

\Rightarrow H = 3.25 m.

Hence, the maximum height covered is 3.25 m.

8 0
3 years ago
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