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m_a_m_a [10]
3 years ago
9

A person is lying on a diving board 3.50 m above the surface of the water in a swimming pool. She looks at a penny that is on th

e bottom of the pool directly below her. To her, the penny appears to be a distance of 5.00 m from her.
Physics
1 answer:
sergij07 [2.7K]3 years ago
4 0

Answer:

1.995 m

Explanation:

Distance of penny as seen by the person = 5 m

Height of person from water surface = 3.50 m

Apparent depth of penny = 5 - 3.50 = 1.5 m

refractive index of water, n = 1.33

real depth / apparent depth = n

real depth = 1.33 x 1.5 = 1.995 m

Thus, the actual depth of water at that point is 1.995 m.

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B. Resistivity

Explanation:

Resistance offered by a substance of unit area per unit length.

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Explanation:distance-time =speed

a,V =s/t

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b,v=s/t

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What is the apparent magnitude of the brightest star in the Big Dipper?
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Consider the motion of a 4.00-kg particle that moves with potential energy given by U(x) = + a) Suppose the particle is moving w
gtnhenbr [62]

Correct question:

Consider the motion of a 4.00-kg particle that moves with potential energy given by

U(x) = \frac{(2.0 Jm)}{x}+ \frac{(4.0 Jm^2)}{x^2}

a) Suppose the particle is moving with a speed of 3.00 m/s when it is located at x = 1.00 m. What is the speed of the object when it is located at x = 5.00 m?

b) What is the magnitude of the force on the 4.00-kg particle when it is located at x = 5.00 m?

Answer:

a) 3.33 m/s

b) 0.016 N

Explanation:

a) given:

V = 3.00 m/s

x1 = 1.00 m

x = 5.00

u(x) = \frac{-2}{x} + \frac{4}{x^2}

At x = 1.00 m

u(1) = \frac{-2}{1} + \frac{4}{1^2}

= 4J

Kinetic energy = (1/2)mv²

= \frac{1}{2} * 4(3)^2

= 18J

Total energy will be =

4J + 18J = 22J

At x = 5

u(5) = \frac{-2}{5} + \frac{4}{5^2}

= \frac{4-10}{25} = \frac{-6}{25} J

= -0.24J

Kinetic energy =

\frac{1}{2} * 4Vf^2

= 2Vf²

Total energy =

2Vf² - 0.024

Using conservation of energy,

Initial total energy = final total energy

22 = 2Vf² - 0.24

Vf² = (22+0.24) / 2

Vf = \sqrt{frac{22.4}{2}

= 3.33 m/s

b) magnitude of force when x = 5.0m

u(x) = \frac{-2}{x} + \frac{4}{x^2}

\frac{-du(x)}{dx} = \frac{-d}{dx} [\frac{-2}{x}+ \frac{4}{x^2}

= \frac{2}{x^2} - \frac{8}{x^3}

At x = 5.0 m

\frac{2}{5^2} - \frac{8}{5^3}

F = \frac{2}{25} - \frac{8}{125}

= 0.016N

8 0
3 years ago
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