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elixir [45]
3 years ago
15

A deep space probe travels in a straight line at a constant speed of over 16,000 m/s. Assuming there is no friction in space, if

a loose part were released from the spacecraft without a push in any direction, what would happen to it? A. It would slowly coast to a stop. B. It would head off on its own trajectory. C. It would stop immediately as soon as it was no longer in contact with the probe. D. It would continue along with the spacecraft.
Physics
1 answer:
77julia77 [94]3 years ago
6 0
C because when the part gets out of the probe it would no longer stay contacted
You might be interested in
In Thomson’s experiment, why was the glowing beam repelled by a negatively charged plate?
Svetllana [295]

The glowing beam was repelled by a negatively charged plate because they were negatively charged

<h3>What are the nature of charges?</h3>

The nature of charges refers to the properties of charges.

There are two types of charges:

  • negative charges
  • positive charges

The law of electricity states that opposite charges attract whereas like charges repel.

Therefor, in Thomson’s experiment, the glowing beam was repelled by a negatively charged plate because they were negatively charged

In conclusion, like charges repel while opposite charges attract.

Learn more about charges at: brainly.com/question/12781208

#SPJ1

5 0
1 year ago
A horizontal pipe contains water at a pressure of 110 kPa flowing with a speed of 1.4 m/s. When the pipe narrows to one half its
Pavel [41]

Answer:

a

  v_2 =  5.6 \  m/s

b

   P_2 = 80600 \  Pa

Explanation:

From the question we are told that  

     The pressure of the water in the pipe is  P_1= 110 \  kPa  =  110 *10^{3 } \  Pa

      The speed of the water  is v_1 =  1.4 \  m/s

       The original area of the pipe is  A_1 =  \pi \frac{d^2 }{4}

       The  new area of the pipe is  A_2 = \pi *  \frac{[\frac{d}{2} ]^2}{4}  =  \pi *  \frac{\frac{d^2}{4} }{4} = \pi \frac{d^2}{16}

         

Generally the continuity equation is mathematically represented as

       A_1 *  v_1 =  A_2 * v_2

Here v_2 is the new velocity  

So

        \pi * \frac{d^2}{4}   *  1.4  = \pi * \frac{d^2}{16}   * v_2

=>     \frac{d^2}{4}   *  1.4  =  \frac{d^2}{16}   * v_2

=>    d^2    *  1.4  =  \frac{d^2}{4}   * v_2

=>    1.4  = 0.25    * v_2

=>     v_2 =  5.6 \  m/s

Generally given that the height of the original pipe and the narrower pipe are the same , then we will b making use of the  Bernoulli's equation for constant height to calculate the pressure

This is mathematically represented as

       

             P_1 + \frac{1}{2}  *  \rho *  v_1 ^2  =  P_2 + \frac{1}{2}  *  \rho *  v_2 ^2

Here \rho is the density of water with value  \rho =  1000  \  kg /m^3

             P_2 =  P_1 + \frac{1}{2} *  \rho [ v_1^2 - v_2^2 ]

=>          P_2 =  110 *10^{3} + \frac{1}{2} *  1000 *  [ 1.4 ^2 - 5.6 ^2 ]

=>          P_2 = 80600 \  Pa

4 0
2 years ago
A supersonic jet is at an altitude of 14 kilometers flying at 1,500 kilometers per hour toward the east. At this velocity, how f
Kaylis [27]

Answer:

The jet will fly 2400 km.

Explanation:

Given the velocity of the jet flying toward the east is 1,500 kmph toward the east.

We need to find the distance covered in 1.6 hours.

In our problem we are given speed and time, we can easily determine the distance using the following formula.

Distance=Speed\times Time

Distance=1500\times 1.6=2400\ km

So, the supersonic jet will travel 2400 km in 1.6 hours toward the east from its starting point.

3 0
3 years ago
GETS BRAINILIST!!!!Which of the following will increase the gravitational force exerted by one object on another? Increasing the
jeka94

Answer:

C. Decreasing the distance between the objects.

Explanation:

Gravitational force depends on mass and distance. The farther the objects are apart, the less gravitational force/pull they have.

5 0
3 years ago
The tundra is located _______.
Leona [35]

Answer:

d: both north and south

Explanation:

8 0
3 years ago
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