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PIT_PIT [208]
3 years ago
9

Suppose that a 1000 kg car is traveling at 25 m/s. Its brakes can apply a force of 5000N. What is the minimum distance required

for the car to stop?
Physics
1 answer:
stealth61 [152]3 years ago
6 0
The car's kinetic energy is

                             (1/2) (mass) (speed)²

                       =    (1/2) (1,000 kg) (25 m/s)²

                       =      (500 kg)  (625 m²/s²)

                       =        312,500 joules .

THAT's the work the brakes have to do in order to stop the car.
They have to absorb that kinetic energy and send it somewhere.

Work done by the brakes = (force) x (distance)

             312,500 joules  =  (5,000 N) x (distance)

                     Distance  =  (312,500 joules) / (5,000 N)

                                   =     62.5 meters .

The brakes soak up the car's kinetic energy, turn it to heat,
and let it blow away in the wind.

Too bad you paid good money to buy that energy in gasoline,
and it ended up blowing away in the wind.  You could have 
stayed home, and just opened some windows and let some
money blow away while you ate chips and watched TV.
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ikadub [295]
The severity of the seasons on Earth is given not by the distance Earth-Sun but by the tilt of the Earth axis. This happens because that the sun rays are oblique in winter and perpendicular in summer (thus the same quantity of sun rays heats a bigger surface in winter - oblique rays). 
The present tild of the Earth axis is  23.5 degrees (from the vertical). If the axis were tilt at 157 degree this would be equivalent  to 180-157 =23 degree. Thus the severity of the seasons would be approximately the same but the seasons would be reversed (for example instead of winter we would have summer, instead of summer we would have winter). 
7 0
3 years ago
" A bowl of soup at 200Á F. is placed in a room of constant temperature of 60Á F. The
Dahasolnce [82]
<span>T(t)=60+140<span>e<span>−0.075t</span></span></span> <span>T(12)=60+140<span>e<span>−0.075∗12</span></span></span> <span>T(12)=60+140<span>e<span>−0.9</span></span></span> <span><span>T(12)=60+140(0.4065696597)
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7 0
3 years ago
804 n of force are applied to a 51.7 kg. What is the acceleration that the object experiences?
Andreyy89

We can use Newton II here  (where F=m*a), that F is the net (or resultant) force on the object, m is the mass of the object and a is the acceleration the object experiences.

This means, in this case there would be no friction and absolutely no other force which gives a component in the plane of motion, only then can you assume that F=804N.

Now using F= m*a

804 = 51.7*a

Therefore a = 804/51.7 = 15.55 m/s²


7 0
3 years ago
What was the average velocity for the entire trip?
just olya [345]

Answer:

<u><em>3.721 m/s</em></u>

This is the explanation of the ans

6 0
2 years ago
You are bungee jumping from a bridge. Initially, while you are falling the slack bungee cord isn’t exerting any forces or torque
harina [27]

Answer:

he fall movement we see that both the force is different from zero, and the torque is different from zero.

When analyzing the statements the d is true

Explanation:

Let's pose the solution of this problem, to be able to analyze the firm affirmations.

When the person is falling, the weight acts on them all the time, initially the rope has no force, but at the moment it begins to lash it exerts a force towards the top that is proportional to the lengthening of the rope.

The equation for this part is

                 Fe - W = m a  

                 k x - mg = m a

As the axis of rotation is located at the top where they jump, there is a torque.

What is it

                Fe y - W y = I α

angular and linear acceleration are related

       a = α r

       Fe y - W y = I a / r

In the fall movement we see that both the force is different from zero, and the torque is different from zero.

When analyzing the statements the d is true

4 0
3 years ago
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