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Semenov [28]
3 years ago
7

A huge cannon is assembled on an airless planet having insignificant axial spin. The planet has a radius of 5.00 × 106 m and a m

ass of 1.46 × 1023 kg. The cannon fires a projectile straight up at 2000 m/s. An observation satellite orbits the planet at a height of 1000 km. What is the projectile's speed as it passes the satellite? (G= 6.67 × 10-11 N · m2/kg2)
Physics
1 answer:
Nata [24]3 years ago
8 0

Answer:

1830 m /s

Explanation:

The potential energy of the projectile at  the surface of planet

- G x 1.46 x 10²³ m  / 5 x 10⁶

Kinetic energy at surface

1/2 m v²

Total energy = - G x 1.46 x 10²³ m  / 5 x 10⁶ + 1/2  m v²

- 6.67 X 10⁻¹¹ x 1.46 x 10²³ /5 x 10⁶  m + 1/2 m x 2000²

- 1.947 x 10⁶ m + 2m x 10⁶

=  .053  x 10⁶ m

Potential energy of projectile at 1000 km height

= - G x 1.46 x 10²³ m  / 6 x 10⁶

= -  1.6225 x 10⁶  m J  

Total energy

= - 1.6225 x 10⁶ + 1/2 m V ²

Applying conservation of energy in gravitational field at surface and height

- 1.6225 x 10⁶  m + 1/2 m V ² = .053  x 10⁶ m

1/2 V ²  = ( .053 + 1.6225 ) x 10⁶

1/2 V² = 1.6755  x 10⁶ m /s

V = 1.83 X 10³ m/s

1830 m /s

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A pump is used to lift 100 KG of water from a wel 60 m deep,in 20 S If force of gravity on 1 KG is 10 N,find
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Explanation:

Given,

  • m = 100 kg
  • g = 10 N/kg¹
  • h = 60 m
  • t = 20 s

To Find:

a) Work done by the pump

b) Potential energy stored in the water

c)Power spent by the pump

d)Power rating of the pump.

Solution:

  • a) Work done by the pump

We know that,

\rm \: Work  \: done = Force * Distance  \: moved

  • f = 100 kg * 10N/kg
  • d = 60 m

\rm \: Work\; Done =(100 \: kg \times  \cfrac{10N}{kg} ) \times 60 \: m

\rm \: Work\; Done =1000 \times 60 \: joule

\boxed{\rm \: Work\; Done =60000 \: joule}

[The unit'll be joule since N×M = J]

  • b) Potential energy stored in the water

\rm \: P.E = m \cdot g \cdot  h

  • m = 100 kg
  • g = 10N/kg
  • h = 60

\rm \: P.E =100 \:kg \:  \times  \cfrac{10 \: N}{kg}  \times 60

\boxed{\rm \: P.E =60000  \: joule}

  • same condition here as well, N×M = J
  • c) Power of the Pump

\rm \: P = W/T

  • where P = Power; W = Work done & T = Time taken
  • As we got the value of work done on question (a),& ATQ time taken is 20 S.

\rm \: P =  \cfrac{60000 \: joule}{20 \: seconds}  =\boxed{\rm { 3000 \: Watts }}\: or \: \boxed{\rm 3 \: kW}

  • d) Power rating of the pump = 3 kW

Assumption: The pump is 100% efficient & works well.

6 0
2 years ago
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