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allochka39001 [22]

3 years ago

11

An office window has dimensions 3.1 m by 2.1 m. As a result of the passage of a storm, the outside air pressure drops to 0.954 a

tm, but inside the pressure is held at 1.0 atm. What net force pushes out on the window

1 answer:

Virty [35]3 years ago

3 0

Answer:

Net forces which pushes the window is 30342.78 N.

Explanation:

Given:

Dimension of the office window.

Length of the window = $3.1$ m

Width of the window = $2.1$ m

Area of the window = $(3.1\times 2.1) = 6.51\ m^2$

Difference in air pressure = Inside pressure - Outside pressure

= $(1.0-0.954)$ atm = $0.046$ atm

Conversion of the pressure in its SI unit.

⇒ $1$ atm = $101325$ Pa

⇒ $0.046$ atm = $0.046\times 101325 =4660.95$ Pa

We have to find the net force.

We know,

⇒ Pressure = Force/Area

⇒ $Pressure=\frac{Force }{Area}$

⇒ $Force =Pressure\times Area$

⇒ Plugging the values.

⇒ $Force =4660.95\times 6.51$

⇒ $Force=30342.78$ Newton (N)

So,

The net forces which pushes the window is 30342.78 N.

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2. Suppose that a parallel-plate capacitor has circular plates with radius R = 30 mm and a plate separation of d = 5.0 mm. Suppo

seropon [69]

Answer:

The maximum value of the induced magnetic field is $2.901\times10^{-13}\ T$ .

Explanation:

Given that,

Radius of plate = 30 mm

Separation = 5.0 mm

Frequency = 60 Hz

Suppose the maximum potential difference is 100 V and r= 130 mm.

We need to calculate the angular frequency

Using formula of angular frequency

$\omega=2\pi f$

Put the value into the formula

$\omega=2\times\pi\times60$

$\omega=376.9\ rad/s$

When r>R, the magnetic field is inversely proportional to the r.

We need to calculate the maximum value of the induced magnetic field that occurs at r = R

Using formula of magnetic filed

$B_{max}=\dfrac{\mu_{0}\epsilon_{0}R^2\timesV_{max}\times\omega}{2rd}$

Where, R = radius of plate

d = plate separation

V = voltage

Put the value into the formula

$B_{max}=\dfrac{4\pi\times10^{-7}\times8.85\times10^{-12}\times(30\times10^{-3})^2\times100\times376.9}{2\times130\times10^{-3}\times5.0\times10^{-3}}$

$B_{max}=2.901\times10^{-13}\ T$

Hence, The maximum value of the induced magnetic field is $2.901\times10^{-13}\ T$ .

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3 years ago

When determining the number of significant digits in a measurement,

Black_prince [1.1K]

B) All nonzero digits are significant.

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3 years ago

Read 2 more answers

A length of copper wire carries a current of 11 A, uniformly distributed through its cross section. Calculate the energy density

NISA [10]

Answer:

a) $0.983 \frac{J}{m^3}$

b) $u_E =7.329x10^-3 \frac{J}{m^3}$

Explanation:

The energy density is "the energy per unit volume, in the electric field. The energy stored between the plates of the capacitor equals the energy per unit volume stored in the electric field times the volume between the plates".

A magnetic field is a "vector field that describes the magnetic influence of electric charges in relative motion and magnetized materials".

Part a

For this case we can assume use the equation for the magnetic field in terms of the energy per unit of volume.

$B=\sqrt{2\mu_o u}$

Where μ0 represent the permeability constant, also known as the magnetic constant. If we solve for u we got:

$u=\frac{B^2}{2\mu_o}$

We also know that the magnetic field can be expressed in terms of the current and the radius of action R like this:

$B=\frac{\mu_o i}{2\pi R}$

Replacing this on the formula for u we have:

$u=\frac{1}{2\mu_o}(\frac{\mu_o i}{2\pi R})^2$

And simplyfing we got:

$u=\frac{\mu_o i^2}{8\pi^2 R^2}$

Replacing the values given we have:

$u=\frac{(4\pix10^{-7} \frac{H}{m} (11A)^2}{8\pi^2 (0.0014m)^2} =0.983 \frac{J}{m^3}$

Part b

The density current is given by this formula $J=i/A$ and the resistance by $R=\frac{\rho l}{A}$

If we use the equation for the energy density we have this:

$u_E =\frac{1}{2}\varepsilon_o E^2 =\frac{\varepsilon}{2}(\rho J)^2=\frac{\varepsilon}{2}(\frac{iR}{l})^2$

And replacing the values given we have:

$u_E =\frac{8.85x10^{-12}\frac{F}{m}}{2}(\frac{11A(3700\frac{\Omega}{m})}{l})^2 =7.329x10^-3 \frac{J}{m^3}$

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3 years ago

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Sidana [21]

Answer:

it depends most of the time but i don't really have a clue either

3 0

2 years ago

masha68 [24]

Answer:

18301.4Kg

Explanation:

Step 1:

Data obtained from the question. This include the following:

Mass 1 (M1) = 0.512Kg

Mass 2 (M2) =..?

Distance apart (r) = 0.0250m

Force (F) = 0.001N

Gravitational force constant (G) = 6.67x10^-11Nm²/Kg²

Step 2:

Determination of the mass, M2 needed to create a force of 0.001N.

This can be obtained as follow:

F = GM1M2/r²

0.001 = 6.67x10^-11 x 0.512 xM2/0.025²

Cross multiply

6.67x10^-11 x 0.512 xM2 = 0.001x0.025²

Divide both side by 6.67x10^-11 x 0.512

M2 = (0.001x0.025²)/(6.67x10^-11x0.512)

M2 = 18301.4Kg

Therefore, a mass of 18301.4Kg is needed

8 0

3 years ago