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Pani-rosa [81]
3 years ago
8

A rocket rises vertically, from rest, with an acceleration of 3.2 m/s2 until it runs out of fuel at an altitude of 1300 m. After

this point, its acceleration is that of gravity, downward. (a) What is the velocity of the rocket when it runs out of fuel? (Assume up is positive.)
(b) How long does it take to reach this point?
(c) What maximum altitude does the rocket reach?
(d) How much time (total) does it take to reach maximum altitude?
(e) With what velocity does the rocket strike the Earth? (Assume up is positive.)
(f) How long (total) is it in the air?

Physics
1 answer:
svetlana [45]3 years ago
6 0

Explanation:

Below is an attachment containing the solution.

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Suppose a gliding 2-kg cart bumps into, and sticks to, a stationary 5-kg cart. If the speed of the gliding cart before the colli
Thepotemich [5.8K]

Answer:

Therefore,

Final velocity of the coupled carts after the collision is

v_{f}=4\ m/s

Explanation:

Given:

Mass of  Glidding Cart = m₁ = 2 kg

Mass of Stationary Cart = m₂ = 5 kg

Initial velocity of Glidding Cart = u₁ = 14 m/s

Initial velocity of Stationary Cart = u₂ = 0 m/s

To Find:

Final velocity of the coupled carts after the collision = v_{f}=?

Solution:

Law of Conservation of Momentum:

For a collision occurring between two objects in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision.

It is denoted by "p" and given by

Momentum = p = mass × velocity

Hence by law of Conservation of Momentum we hame

Momentum before collision = Momentum after collision

Here after collision both are stuck together so both will have same final velocity,

m_{1}\times u_{1}+m_{2}\times u_{2}=(m_{1}+m_{2})\times v_{f}

Substituting the values we get

2\times 14 + 5\times 0 =(2+5)\times v_{f}

v_{f}=\dfrac{28}{7}=4\ m/s

Therefore,

Final velocity of the coupled carts after the collision is

v_{f}=4\ m/s

8 0
3 years ago
Suppose you wanted to hold up an electron against the force of gravity by the attraction of a fixed proton some distance above i
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Answer:

The value is  r =  5.077 \  m

Explanation:

From the question we are told that

   The  Coulomb constant is  k =  9.0 *10^{9} \  N\cdot  m^2  /C^2

   The  charge on the electron/proton  is  e =  1.6*10^{-19} \  C

    The  mass of proton m_{proton} =  1.67*10^{-27} \  kg

    The  mass of  electron is  m_{electron } =  9.11 *10^{-31} \ kg

Generally for the electron to be held up by the force gravity

   Then    

       Electric force on the electron  =  The  gravitational Force

i.e  

            m_{electron} *  g  = \frac{ k *  e^2  }{r^2 }

         \frac{9*10^9 *  (1.60 *10^{-19})^2  }{r^2 }  =     9.11 *10^{-31 }  *  9.81

         r =  \sqrt{25.78}

         r =  5.077  \  m

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What is helpful for motivating a person to achieve a fitness or health goal?
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Answer:

I think it's b

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A fisherman notices that wave crests pass the bow of his anchored boat every 5.0 s. He measures the distance between two crests
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Answer:          |

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