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Lina20 [59]
3 years ago
11

. If a pendulum-driven clock gains 5.00 s/day, what fractional change in pendulum length must be made for it to keep perfect tim

e
Physics
1 answer:
inn [45]3 years ago
6 0

Answer:

The appropriate response will be "Length must be increased by 0.012%".

Explanation:

The given values is:

ΔT = 5 s/day

Now,

⇒ \frac{\Delta T}{T} =\frac{5}{24\times 60\times 60}

On multiplying both sides by  "100", we get

⇒ \frac{\Delta T}{T}\times 100 =\frac{500}{24\times 60\times 60}

⇒ \frac{\Delta T}{T}\times 100=0.005787 (%)

∵  T=2\pi\sqrt{\frac{l}{g} }

On substituting the values, we get

⇒ \frac{\Delta T}{T}% = \frac{1}{2}\times \frac{\Delta l}{l}%

On applying cross multiplication, we get

⇒ \frac{\Delta l}{l}% = 2\times \frac{\Delta T}{T}%

⇒        = 2\times 0.05787

⇒        = 0.011574

⇒        = 0.012%

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A waffle iron heated by coils

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3 years ago
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A plant box falls from the windowsill 25.0 m above the sidewalk and hits the cement 3.0 s later. What is the box's velocity when
mamaluj [8]

Answer:

22m/s

Explanation:

To find the velocity we employ the equation of free fall: v²=u²+2gh

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Substituting for the values in the question we get:

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8 0
2 years ago
A force is applied to a block sliding along a surface (Figure 2). The magnitude of the force is 15 N, and the horizontal compone
Lemur [1.5K]

Answer:

Fy = 14.3 [N]

Explanation:

To be able to solve this problem we must know that the force is a vector and has magnitude and direction, therefore it can be decomposed into the force in the X & y components:

When we have the components on the horizontal and vertical axes we must use the Pythagorean theorem.

F = \sqrt{F_{x}^{2} +F_{y}^{2}  }

where:

F = 15 [N]

Fx = horizontal component = 4.5 [N]

Fy = vertical component [N]

15=\sqrt{4.5^{2}+F_{y}^{2}}\\ 15^{2}= (\sqrt{4.5^{2}+F_{y}^{2}})^{2} \\225 = 4.5^{2}+F_{y} ^{2}\\  F_{y}^{2} =225 -4.5^{2}\\ F_{y}^{2}=204.75\\F_{y}=\sqrt{204.75}\\  F_{y}=14.3 [N]

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At what displacement of a sho is the energy half kinetic and half potential? what fraction of the total energy of a sho is kinet
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As we know that KE and PE is same at a given position

so we will have as a function of position given as

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also the PE is given as function of position as

PE = \frac{1}{2}m\omega^2x^2

now it is given that

KE = PE

now we will have

\frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2}m\omega^2x^2

A^2 - x^2 = x^2

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x = \frac{A}{\sqrt2}

so the position is 0.707 times of amplitude when KE and PE will be same

Part b)

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we can use the formula

KE = \frac{1}{2}m\omega^2(A^2 - x^2)

now to find the fraction of kinetic energy

f = \frac{KE}{TE} = \frac{A^2 - x^2}{A^2}

f = \frac{A^2 - (\frac{A}{3})^2}{A^2}

f_k = \frac{8}{9}

now since total energy is sum of KE and PE

so fraction of PE at the same position will be

f_{PE} = 1 - f_k

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Answer:

Models,Mathematics

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Explanation:

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