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faltersainse [42]
3 years ago
11

a car whose mass is 1000kg is traveling at a constant speed of 10m/s. Neglecting any friction how much force will the engine hav

e to supply to keep going the same speed ?
Physics
2 answers:
AURORKA [14]3 years ago
7 0
This next statement is a big deal.  It should be up on a board, surrounded
by flashing red and yellow lights, and hung on the wall of every Science
classroom.   Although we never see it in our daily lives, it's fundamental to
the workings of the universe, and it's also Newton's first law of motion:

<em>Without friction, it doesn't take <u>ANY</u> force to keep a moving object
moving.  </em>
<em>Force is only required to <u>change</u> the object's speed, or to
<u>change</u> the direction </em>
<em>in which it's moving.</em>

The answer to the question is:  On a level road, and neglecting any friction,
the engine doesn't have to supply ANY force to keep the car going at the
same speed.
serg [7]3 years ago
5 0
Since you already know:

Force  =  mass *  acceleration.

acceleration =  (Velocity Change) / time.

Since we were told that velocity is constant, therefore Velocity Change = ZERO.

Therefore acceleration = ZERO.

Force = 1000 kg * ZERO = ZERO.

Force = 0 N.

Cheers.
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Westkost [7]

Answer:

Required rate of return = 18.5 %

Explanation:

given,                            

rate of inflection = 4 %

risk free rate = 3 %                      

market risk premium = 5 %                    

firm has a beta  = 2.30                                  

rate of return has averaged 15.0% over the last 5 years

now,                                                                      

Nominal risk free rate = risk free rate  + inflation

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Required rate of return = Nominal risk free rate + β (RPM)

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5 0
3 years ago
A 5,257 kg rocket blasts off to the moon with an acceleration of 76 m/s ^2 what is the net force on the rocket
frutty [35]

Newton's subsequent law expresses that power is corresponding to what exactly is needed for an object of consistent mass to change its speed. This is equivalent to that item's mass increased by its speed increase.

We use Newtons, kilograms, and meters each second squared as our default units, albeit any proper units for mass (grams, ounces, and so forth) or speed (miles each hour out of every second, millimeters per second², and so on) could unquestionably be utilized also - the estimation is the equivalent notwithstanding.

Hence, the appropriate answer will be 399,532.

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hey

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