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Alex
2 years ago
5

A lens of focal length 15.0 cm is held 10.0 cm from a page (the object ). Find the magnification .

Physics
1 answer:
nevsk [136]2 years ago
3 0

Answer:

Magnification, m = 3

Explanation:

It is given that,

Focal length of the lens, f = 15 cm

Object distance, u = -10 cm

Lens formula :

\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}

v is image distance

\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{15}+\dfrac{1}{(-10)}\\\\v=-30\ cm

Magnification,

m=\dfrac{v}{u}\\\\m=\dfrac{-30}{10}\\\\m=3

So, the magnification of the lens is 3.

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What is the centripetal acceleration of the tip of the fan blade?

6.0 m/s2

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Answer is 96

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2 years ago
A disk of mass m and radius r is initially held at rest just above a larger disk of mass M and radius R that is rotating at angu
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Answer:

Explanation:

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2 years ago
Armand is monitoring a large sealed tank by way of a sensor that records the liquid level over time on a graph. He looks at the
timofeeve [1]

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i need ppoints

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4 0
2 years ago
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The air in a 6.00 L tank has a pressure of 2.00 atm. What is the final pressure, in atmospheres, when the air is placed in tanks
ser-zykov [4K]

Explanation:

Given that,

Initial volume of tank, V = 6 L

Initial pressure, P = 2 atm

We need to find the final pressure when the air is placed in tanks that have the following volumes if there is no change in temperature and amount of gas:

(a) V' = 1 L

It is a case of Boyle's law. It says that volume is inversely proportional to the pressure at constant temperature. So,

PV=P'V'\\\\P'=\dfrac{PV}{V'}\\\\P'=\dfrac{6\times 2}{1}\\\\P'=12\ atm

(b) V' = 2500 mL

New pressure becomes :

PV=P'V'\\\\P'=\dfrac{PV}{V'}\\\\P'=\dfrac{6\times 2}{2500\times 10^{-3}}\\\\P'=4.8\ atm

(c) V' = 750 mL

New pressure becomes :

PV=P'V'\\\\P'=\dfrac{PV}{V'}\\\\P'=\dfrac{6\times 2}{750\times 10^{-3}}\\\\P'=16\ atm

(d) V' = 8 L

New pressure becomes :

PV=P'V'\\\\P'=\dfrac{PV}{V'}\\\\P'=\dfrac{6\times 2}{8}\\\\P'=1.5\ atm

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3 0
2 years ago
Consider a bird that flies at an average speed of 10.7 m/sm/s and releases energy from its body fat reserves at an average rate
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Answer:

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Time required to burn the fat

t=\dfrac{E}{P}\\\Rightarrow t=\dfrac{157393.6}{3.7}\\\Rightarrow t=42538.811\ s

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s=vt\\\Rightarrow s=10.7\times 42538.811\\\Rightarrow s=455165.2777\ m

The bird will fly 455165.278 m

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2 years ago
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