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Marrrta [24]
3 years ago
11

The potential energy for a certain mass moving in one dimension is given by U(x)=(2.0J/m3)x3−(15J/m2)x2+(36J/m)x−23JU(x)=(2.0J/m

3)x3−(15J/m2)x2+(36J/m)x−23JU(x) = (2.0 {\rm J/m}^{3})x^{3}- (15 {\rm J/m}^{2})x^{2}+ (36 {\rm J/m})x - 23 {\rm J}. Find the location(s) where the force on the mass is zero.
Physics
1 answer:
Angelina_Jolie [31]3 years ago
8 0

Answer:x=2 and x=3

Explanation:

Given

Potential Energy for a certain mass is

U(x)=2x^3-15x^2+36x-23

and we know force is given by

F=-\frac{\mathrm{d} U}{\mathrm{d} x}

F=-(2\times 3x^2-15\times 2x+36)

For Force to be zero F=0

\Rightarrow 6x^2-30x+36=0

\Rightarrow x^2-5x+6=0

\Rightarrow x^2-2x-3x+6=0

\Rightarrow (x-2)(x-3)=0

Therefore at x=2 and x=3 Force on particle is zero.

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3 years ago
For exercise, an athlete lifts a barbell that weighs 400 N from the ground to a height of 2.0 m in a time of 1.6 s. Assume the e
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Answer:

Explanation:

(ΔK + ΔUg + ΔUs + ΔEch + ΔEth = W)

ΔK is increase in kinetic energy . As the athelete is lifting the barbell at constant speed change in kinetic energy is zero .

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ΔUg  is change in potential energy . It will be positive as weight is being lifted so its potential energy is increasing .

ΔUg = positive

ΔUs is change in the potential energy of sportsperson . It is zero since there is no change in the height of athlete .

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b )

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= 400 x 2 = 800 J

energy output = 800 J

c )

It is 25% of metabolic energy output of his body

so metalic energy output of body

= 4x 800 J .

3200 J

power = energy output / time

= 3200 / 1.6

= 2000 W .

d )

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2 ) Since he is doing the same exercise in less time so his power is increased . Hence in the second day his power is more .

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4 years ago
Two flat 4.0 cm × 4.0 cm electrodes carrying equal but opposite charges are spaced 2.0 mm apart with their midpoints opposite ea
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σ = ε0 x E

σ = 8.85 x 10^-12 x 2.5 x 10^6 = 22.125 x 10^-6 C/m^2

The surface charge density of each plate is ± σ / 2

So, the surface charge density on each = ± 22.125 x 10^-6 / 2

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Charge on each plate = Surface charge density on each plate x area of each plate

Charge on each plate = ± 11.0625 x 10^-6  x 16 x 10^-4 = ± 1.77 x 10^-8 C

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