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Zolol [24]
3 years ago
5

You push a 1.30 kg physics book 2.80 m along a horizontal tabletop with a horizontal push of 1.55 N while the opposing force of

friction is 0.340 N. How much work does your 1.55 N push do on the book
Physics
1 answer:
Rzqust [24]3 years ago
4 0

Answer:

<h2>3.36J</h2>

Explanation:

Step one:

given data

mass m= 1.3kg

distance moved s= 2.8m

opposing frictional force= 0.34N

assume g= 9.81m/s^2

we know that work done= force *distance moved

1. work done to push the book= 1.55*2.8=4.34J

2. Work against friction = force of friction x distance

                                       = 0.34*2.8=0.952J

Step two:

the work done on the book is the net work, which is

Network done= work done to push the book- Work against friction

Network done= 4.32-0.952=3.36J

<u>Therefore the work of the 1.55N 3.36J</u>

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Answer:

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Explanation:

The charge on any capacity, q = CV, thus,

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The charge on the 280 pF capacitor is q = C*v

q = 280*10^-12 * 3.6*10^3

q = 1.008x10^-6 C

When they are connected as stated, the net total charge remaining will be 1.008*10^-6 - 2.52*10^-7 = 7.56*10^-7 C

Since the capacitors are in parallel, the equivalent capacitance will be 70 + 280 pF = 350 pF

Remember, q = CV, then V = q/C

V = 7.56*10^-7 C / 350*10^12 F

V = 2160 V

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The energy After is 1/2 Cv²

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E = 3.266*10^-4 J + 1.306*10^-3 J

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so the loss is 2.268 - 1.632 = 0.636 kJ

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Let the positive direction be directed in the direction of the block's acceleration, which is downward.

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