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Kisachek [45]
3 years ago
14

ou have just moved into a new apartment and are trying to arrange your bedroom. You would like to move your dresser of weight 3,

500 NN across the carpet to a spot 5 mm away on the opposite wall. Hoping to just slide your dresser easily across the floor, you do not empty your clothes out of the drawers before trying to move it. You push with all your might but cannot move the dresser before becoming completely exhausted. How much work do you do on the dress
Physics
1 answer:
Dimas [21]3 years ago
5 0

So far, since you moved into the apartment until the end of this much of the story, you haven't done ANY work on the dresser yet.

I'll admit that you pushed, groaned and grunted, sweated and strained plenty.  You're physically and mentally exhausted, you're not interested in the dresser at the moment, and right now you just want to snappa cappa brew, crash on the couch, and watch cartoons on TV.  But if you've done your Physics homework, you know you haven't technically done any <u><em>work</em></u> yet.

In Physics, "Work" is the product of Force times Distance.

Since the dresser hasn't budged yet, the Distahce is zero.  So no matter how great the Force may be, it's multiplied by zero, so the <em>Work is zero</em>.

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System A has masses m and m separated by a distance r; system B has masses m and 2m separated by a distance 2r; system C has mas
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Answer:

System D --> System C --> System A --> System B

Explanation:

The gravitational force between two masses m1, m2 separated by a distance r is given by:

F=G \frac{m_1 m_2}{r^2}

where G is the gravitational constant. Let's apply this formula to each case now to calculate the relative force for each system:

System A has masses m and m separated by a distance r:

F=G\frac{m \cdot m}{r^2}=G \frac{m^2}{r^2}

system B has masses m and 2m separated by a distance 2r:

F=G\frac{m \cdot 2m}{(2r)^2}=G \frac{2m^2}{4r^2}=\frac{1}{2} G \frac{m^2}{r^2}

system C has masses 2m and 3m separated by a distance 2r:

F=G\frac{2m \cdot 3m}{(2r)^2}=G \frac{6m^2}{4r^2}=\frac{3}{2} G \frac{m^2}{r^2}

system D has masses 4m and 5m separated by a distance 3r:

F=G\frac{4m \cdot 5m}{(3r)^2}=G \frac{20m^2}{9r^2}=\frac{20}{9} G \frac{m^2}{r^2}

Now, by looking at the 4 different forces, we can rank them from the greatest to the smallest force, and we find:

System D --> System C --> System A --> System B

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A driver in a 2000 kg Porsche wishes to pass to pass a slow-moving school bus on a four-lane road. What is the average power in
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The average power is 3.0\cdot 10^6 W

Explanation:

First of all, we calculate the work done to accelerate the car; according to the work-energy theorem, the work done is equal to the change in kinetic energy of the car:

W=K_f -K_i= \frac{1}{2}mv^2-\frac{1}{2}mu^2

where :

K_f = \frac{1}{2}mv^2 is the final kinetic energy of the car, with

m = 2000 kg is the mass of the car

v = 60 m/s is the final speed of the car

K_i = \frac{1}{2}mu^2 is the initial kinetic energy of the car, with

u = 30 m/s is initial speed of the car

Soolving:

W=\frac{1}{2}(2000)(60)^2 - \frac{1}{2}(2000)(30)^2=2.7\cdot 10^6 J

Now we can find the power required for the acceleration, which is given by

P=\frac{W}{t}

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Solving:

P=\frac{2.7\cdot 10^6}{9}=3.0\cdot 10^6 W

Learn more about power:

brainly.com/question/7956557

#LearnwithBrainly

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