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Semenov [28]
2 years ago
14

Two skaters, one with mass 52.5 kg and the other with mass 26.3 kg, stand on an ice rink holding a pole with a length of 5.00 m

and a negligible mass. Starting from the ends of the pole, the skaters pull themselves along the pole until they meet. How far will the 26.3 kg skater move?
Physics
1 answer:
Marizza181 [45]2 years ago
5 0

Answer: 10.02m

Explanation: Given

M1 = 26.3kg

M2 = 52.5kg

d = 5m

If the systems centre of mass does not change, the two skaters will definitely meet at the centre of mass.

If the 26.3kg skater is a m away from the centre of the mass, then the 52.5kg skater is (5-a) m away from the centre of the mass. Thus,

M1a + M2(5-a) = 0

26.3a + 52.5(5-a) = 0

26.3 a + 262.5 - 52.5a = 0

-26.2a + 262.5 = 0

26.2a = 262.5

a = 262.5/26.2

a = 10.02m

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A satellite is in a circular orbit 21000 km above the Earth’s surface; i.e., it moves on a circular path under the influence of
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Answer:

(orbital speed of the satellite) V₀ = 3.818 km

Time (t) = 4.5 × 10⁴s

Explanation:

Given that:

The radius of the Earth is 6.37 × 10⁶ m;    &

the acceleration of gravity at the satellite’s altitude is 0.532655 m/s

We can calculate the orbital speed of the satellite by using the formula:

Orbital Speed (V₀) = √(r × g)

radius of the orbit (r) = 21000 km + 6.37 × 10⁶ m

                                  = (2.1 × 10⁷ + 6.37 × 10⁶) m

                                  = 27370000

                                  = 2.737 × 10⁷m

Orbital Speed (V₀) = √(r × g)

Orbital Speed (V₀) = √(2.737 × 10⁷  × 0.532655 )

                              = 3818.215

                              = 3.818 × 10³

                             = 3.818 Km

To find the time it takes to complete one orbit around the Earth; we use the formula:

Time (t) = 2 π × \frac{r}{V_o}

            = 2 × 3.14 × \frac{2.737*10^7}{3.818*10^3}

            = 45019.28

            = 4.5 × 10 ⁴ s

6 0
3 years ago
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