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3 years ago

The larger the area is the smaller the pressure true or folse

Physics

2 answers:

Ilya [14]3 years ago

8 0

Answer:

False

Explanation:

The depending on what type of pressure is enclosed in a object, the smaller the object the more pressure because that pressure has limited room to escape.

bekas [8.4K]3 years ago

5 0

The larger the area the smaller the pressure is true since a greater amount of space with a certain amount of pressure exerted upon is lesser, due to the pressure being divided over the larger space.

Meanwhile, a smaller area but that has the same amount of pressure exerted upon it, will seem greater since the pressure is not very widespread and more of the pressure is concentrated in that smaller area.

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A 200-loop coil of cross sectional area 8.5 cm2 lies in the plane of the page. An external magnetic field of 0.060 T is directed

yuradex [85]

(a) 0.0068 Wb

Since the plane of the coil is perpendicular to the magnetic field, the magnetic flux through the coil is given by

$\Phi = NBA$

where

N = 200 is the number of loops in the coil

B is the magnetic field intensity

$A=8.5 cm^2 = 8.5\cdot 10^{-4} m^2$ is the area of the coil

At the beginning, we have

$B_i = 0.060 T$

so the initial magnetic flux is

$B_i = (200)(0.060 T)(8.5\cdot 10^{-4} m^2)=0.0102 Wb$

at the end, we have

$B_f=0.020 T$

so the final magnetic flux is

$B_f = (200)(0.020 T)(8.5\cdot 10^{-4} m^2)=0.0034 Wb$

So the magnitude of the change in the external magnetic flux through the coil is

$\Delta \Phi = |\Phi_f - \Phi_i|=|0.0034 Wb-0.0102 Wb|=0.0068 Wb$

(b) 0.567 V

The magnitude of the average voltage (emf) induced in the coil is given by Faraday-Newmann law

$\epsilon= \frac{\Delta \Phi}{\Delta t}$

where

$\Delta \Phi = 0.0068 Wb$ is the variation of magnetic flux

$\Delta t = 12 ms = 0.012 s$ is the time interval

Substituting into the formula, we find

$\epsilon=\frac{0.0068 Wb}{0.012 s}=0.567 V$

(c) 0.142 A

The average current in the coil can be found by using Ohm's law:

$I=\frac{V}{R}$

where

I is the current

V is the voltage

R is the resistance

Here we have:

V = 0.567 V (induced voltage)

$R=4.0 \Omega$ (resistance of the coil)

Solving for I, we find

$I=\frac{0.567 V}{4.0 \Omega}=0.142 A$

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2 years ago

Which of the following experiments or phenomena provided evidence for the particle nature of light?

Marysya12 [62]

<h3>Answer;</h3>

Expulsion of electrons with varying frequencies of light observed in the photoelectron effect.

<h3>Explanation;</h3>

The photoelectric effect supports a particle theory of light in that it behaves like an elastic collision between two particles, the photon of light and the electron of the metal.
Albert Einstein observed the photoelectric effect in which ultraviolet light forces a surface to release electrons when the light hits. He explained the reaction by defining light as a stream of photons, or energy packets.

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3 years ago

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If love is the answer, then what is the question?

kipiarov [429]

Answer:

what is happiness

Explanation:

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2 years ago

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On Earth, a spring stretches by 5.0 cm when a mass of 3.0 kg is suspended from one end.

Neko [114]

Answer:

Mass = 18.0 kg

Explanation:

From Hooke's law,

F = ke

where: F is the force, k is the spring constant and e is the extension.

But, F = mg

So that,

mg = ke

On the Earth, let the gravitational force be 10 m/ $s^{2}$ .

3.0 x 10 = k x 5.0

30 = 5k

⇒ k = $\frac{30}{5}$ ................ 1

On the Moon, the gravitational force is $\frac{1}{6}$ of that on the Earth.

m x $\frac{10}{6}$ = k x 5.0

$\frac{10m}{6}$ = 5k

⇒ k = $\frac{10m}{30}$ ............. 2

Equating 1 and 2, we have;

$\frac{30}{5}$ = $\frac{10m}{30}$

m = $\frac{900}{50}$

= 18.0

m = 18.0 kg

The mass required to produce the same extension on the Moon is 18 kg.

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2 years ago

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A 2kg book is held against a vertical wall. The coefficient of friction is 0.45. What is the minimum force that must be applied

Vika [28.1K]

We have that for the Question "A 2kg book is held against a vertical wall. The coefficient of friction is 0.45. What is the minimum force that must be applied on the book, perpendicular to the wall, to prevent the book from slipping down the wal" it can be said that the minimum force that must be applied on the book is

F=44N

From the question we are told

A 2kg book is held against a vertical wall. The coefficient of friction is 0.45. What is the minimum force that must be applied on the book, perpendicular to the wall, to prevent the book from slipping down the wal

Generally the equation for the Force is mathematically given as

$F=\frac{mg}{\mu}\\\\F=\frac{2*9.8}{0.45}\\\\$

F=44N

Therefore

the minimum force that must be applied on the book is

F=44N

For more information on this visit

brainly.com/question/23379286

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2 years ago