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inna [77]
2 years ago
6

Is the box moving at a constant speed? Explain how you know. What does this tell you about the kinetic energy of the system

Physics
1 answer:
mars1129 [50]2 years ago
8 0

Answer:

The box is moved at constant speed, then the change in kinetic energy is zero.

Explanation:

Let suppose that box moves at constant speed, meaning that its kinetic energy (K), measured in joules, is expressed by the following equation:

K = \frac{1}{2}\cdot m\cdot v^{2} (1)

Where:

m - Mass, measured in kilograms.

v - Speed, measured in meters per second.

If the box moves at constant speed, then we notice that v_{f}^{2}-v_{o}^{2} = 0, therefore, \Delta K = 0. In a nutshell, if the box is moved at constant speed, then the change in kinetic energy is zero.

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This is for physics, i need help fast!!
jarptica [38.1K]
Speed = distance / time
S= 40 000m / 5400s
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7 0
3 years ago
If we put negative charge between two similar positive charges then what is it's equilibrium? And how?​
Gnesinka [82]

Your question has been heard loud and clear.

Well it depends on the magnitude of charges. Generally , when both positive charges have the same magnitude , their equilibrium point is towards the centre joining the two charges. But if magnitude of one positive charge is higher than the other , then the equilibrium point will be towards the charge having lesser magnitude.

Now , a negative charge is placed in between the two positive charges. So , if both positive charges have same magnitude , they both pull the negative charge towards each other with an equal force. Thus the equilibrium point will be where the negative charge is placed because , both forces are equal , and opposite , so they cancel out each other at the point where the negative charge is placed. However if they are of different magnitudes , then the equilibrium point will be shifted towards the positive charge having less magnitude.

Thank you

5 0
3 years ago
Read 2 more answers
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garik1379 [7]
Can you please stop pasting this question, just go to his profile and ask him.
7 0
3 years ago
Challenge! A marshmallow is dropped from a 5-meter high pedestrian bridge and 0.83 seconds later, it lands right on the head of
WINSTONCH [101]

a₀).  You know ...
         -- the object is dropped from 5 meters
             above the pavement;
         --  it falls for 0.83 second.

a₁).  Without being told, you assume ...
         -- there is no air anyplace where the marshmallow travels,
             so it free-falls, with no air resistance;
         -- the event is happening on Earth,
            where the acceleration of gravity is  9.81 m/s² .

b).  You need to find how much LESS than 5 meters
       the marshmallow falls in 0.83 second.
    
c).  You can use whatever equations you like.
       I'm going to use the equation for the distance an object falls in
       ' T ' seconds, in a place where the acceleration of gravity is ' G '.

d).  To see how this all goes together for the solution, keep reading:


The distance that an object falls in ' T ' seconds
when it's dropped from rest is

                                 (1/2 G) x (T²) .

On Earth, ' G ' is roughly  9.81 m/s², so in 0.83 seconds,
such an object would fall

                               (9.81 / 2) x (0.83)² = 3.38 meters .

It dropped from 5 meters above the pavement, but it
only fell 3.38 meters before something stopped it.
So it must have hit something that was

                         (5.00 - 3.38)  =  1.62 meters

above the pavement.  That's where the head of the unsuspecting
person was as he innocently walked by and got clobbered.

7 0
3 years ago
The electric field intensity between two large,
Ivahew [28]

Answer:

300 is the answer

Explanation:

Hope that this answer will help you

8 0
3 years ago
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