The speed of the object increases
Explanation:
We can answer this question by applying the work-energy theorem, which states that the work done on an object is equal to the change in kinetic energy of the object. Mathematically:

where
W is the work done on the object
are the final and initial kinetic energy of the object, respectively
m is the mass of the object
v is its final speed
u is its initial speed
In this case, the force does a positive amount of work on the object, so

This also implies that

And so

And therefore

which means that the speed of the object increases.
Learn more about work:
brainly.com/question/6763771
brainly.com/question/6443626
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Answer:
<u>because of the doppler effect</u>
Explanation:
<em>Remember</em>, the doppler effect refers to the changes in sound (frequency of sound) observed by a person who is in a position relative to the wave source.
In this example, we notice as the train comes closer to the boy, the sound becomes louder also increasing the pitch slightly, the doppler effect sets in when the train passes the boy because the boy notices a decrease in the pitch of the moving train.
We learn from the change in the observed sound of the train that the frequency of the sound is determined by the distance of the observer from the wave source.
In other words, the closer the source of the sound to the observer; the faster it travels to the observer, however, the farther it is; the lesser it is; the greater the sound heard.
Aaron's car is moving at speed of 30 m/s
His reaction time is given as 0.7 s
but when he is tired the reaction time is doubled
Now we need to find the distance covered by his car when he is tired during the time when he react to apply brakes
So here since during this time speed is given as constant so we can say that distance covered can be product of speed and time
So here we can use



So the car will move to 42 m during the time when he apply brakes
Please state the options and I will answer to the best of my abilities XD
Answer:

the force you applied to your car =1350N