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hodyreva [135]
3 years ago
12

A Martian rover is descending a hill sloped at 30° with the horizontal. It travels with a constant velocity of 2 meters/second.

Calculate its horizontal velocity.
Physics
2 answers:
kodGreya [7K]3 years ago
8 0
Vx = 2*cos30 = 1.73m's
Other words:
D = Vo*t = 1.4 * 21 = 31.5m. @ 30 Deg.
Dy = 31.5*sin30 = 15.75 m.
uysha [10]3 years ago
7 0

<u>Answer:</u> The horizontal velocity of rover is 1.73 m/s

<u>Explanation:</u>

Horizontal velocity is defined as the velocity of body proceeding in the horizontal direction. This velocity proceeds in the 'x' direction.

Mathematically,

V_x=V_o\cos \theta

where,

V_o = constant velocity of rover = 2 m/s

\cos \theta = cosine function of angle 30°

Putting values in above equation, we get:

V_x=2\times \cos (30^o)\\\\V_x=1.73m/s

Hence, the horizontal velocity of rover is 1.73 m/s

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At the same instant that a 0.50-kg ball is dropped from 25 m above Earth, a second ball, with a mass of 0.25 kg, is thrown strai
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Answer:

The center of mass of the two-ball system is 7.05 m above ground.

Explanation:

<u>Motion of 0.50 kg ball:</u>

Initial speed, u = 0 m/s

Time = 2 s

Acceleration = 9.81 m/s²

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Substituting in equation s = ut + 0.5 at²

                 s = 0 x 2 + 0.5 x 9.81 x 2² = 19.62 m

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<u>Motion of 0.25 kg ball:</u>

Initial speed, u = 15 m/s

Time = 2 s

Acceleration = -9.81 m/s²

Substituting in equation s = ut + 0.5 at²

                 s = 15 x 2 - 0.5 x 9.81 x 2² = 10.38 m

Height above ground = 10.38 m

We have equation for center of gravity

          \bar{x}=\frac{m_1x_1+m_2x_2}{m_1+m_2}

          m₁ = 0.50 kg

          x₁ = 5.38 m        

          m₂ = 0.25 kg

          x₂ = 10.38 m    

Substituting

         \bar{x}=\frac{0.50\times 5.38+0.25\times 10.38}{0.50+0.25}=7.05m

The center of mass of the two-ball system is 7.05 m above ground.  

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