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3 years ago

True or false:The potential energy of an object is affected by its speed

1 answer:

7 0

True because the speed of a mass can change it's gravitational potential energy in the gravitational field of another massive object.

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Choose the correct word from the drop-down menu to complete each sentence.

ra1l [238]

Answer:

- Peat soil

- Loamy soil

- Clay soil

Explanation:

Peat soil is dark, highly decomposed organic matter found in soil.

Loamy soil is a mixture of materials that holds moisture and drains well and also contains coarse grains and allows water to drain quickly.

Clay soil contains medium grains and retains water. Also it contains fine grains and has little space for water.

4 0

3 years ago

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The weight of an apple at 4r from earth is of that on earth

iris [78.8K]

W = m.g = weight
g = Gme/Re**2 where G= universal gravitational constant , Re= radius of the earth
me= mass of the earth
therefore it weighs 16 times less

4 0

3 years ago

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could anyone determinate the period knowing that it performs 4000 vibrations in 0.5 minutes, Sorry my english is bad

dexar [7]

Answer:

f = 4000 / 30 sec = 133.3 vibrations/sec

P = 1 / f = .0075 sec period of 1 vibration

4 0

2 years ago

I need help

musickatia [10]

Answer:

Explanation:

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3 years ago

A rocket in deep space has an empty mass of 150 kg and exhausts the hot gases of burned fuel at 2500 m/s. It is loaded with 600

3241004551 [841]

Answer:

$v(10\,s) \approx 775.387\,\frac{m}{s}$

$v(20\,s)\approx 1905.350\,\frac{m}{s}$

$v(30\,s) \approx 4023.595\,\frac{m}{s}$

Explanation:

The speed of the rocket is given the Tsiolkovsky's differential equation, whose solution is:

$v (t) = v_{o} - v_{ex}\cdot \ln \frac{m}{m_{o}}$

Where:

$v_{o}$ - Initial speed of the rocket, in m/s.

$v_{ex}$ - Exhaust gas speed, in m/s.

$m_{o}$ - Initial total mass of the rocket, in kg.

$m$ - Current total mass of the rocket, in kg.

Let assume that fuel is burned linearly. So that,

$m(t) = m_{o} + r\cdot t$

The initial total mass of the rocket is:

$m_{o} = 750\,kg$

The fuel consumption rate is:

$r = -\frac{600\,kg}{30\,s}$

$r = -20\,\frac{kg}{s}$

The function for the current total mass of the rocket is:

$m(t) = 750\,kg - (20\,\frac{kg}{s} )\cdot t$

The speed function of the rocket is:

$v(t) = - 2500\,\frac{m}{s}\cdot \ln \frac{750\,kg -(20\,\frac{kg}{s} )\cdot t}{750\,kg}$

The speed of the rocket at given instants are:

$v(10\,s) \approx 775.387\,\frac{m}{s}$

$v(20\,s)\approx 1905.350\,\frac{m}{s}$

$v(30\,s) \approx 4023.595\,\frac{m}{s}$

7 0

3 years ago