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3 years ago

What provides the force necessary to start a building or bridge oscillating?

Physics

1 answer:

Andrew [12]3 years ago

4 0

Usually, the forces that start the oscillation of buildings are the wind and microearthquakes.

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A polarized light that has an intensity I0 = 60.0 W/m² is incident on three polarizing disks whose planes are parallel and cente

nikitadnepr [17]

Answer:

The transmitted intensity through all polarizers is $I_3 =41.31 W/m^2$

Explanation:

According to Malu's law the intensity of a polarized light having an initial intensity $I_0$ is mathematically represented as

$I = I_0cos^2 \theta$

Now considering the polarizer(The polarizing disk) the equation above becomes

$I = I_0 (cos^2 \theta)^n$

Where n is the number of polarizers

Substituting $60.0W/m^2$ for the initial intensity 3 for the n and 20° for the angle of rotation

$I_3 = 60 (cos^220)^3$

$=41.31 W/m^2$

6 0

3 years ago

What could be used as another word for electrical potential

tekilochka [14]

Answer:

hey mate

answer is probably voltage as per me

Explanation:

Voltage, electric potential difference, electric pressure or electric tension is the difference in electric potential between two points, which is defined as the work needed per unit of charge to move a test charge between the two points

5 0

2 years ago

Read 2 more answers

Calculate the electric field at the center of a square

pantera1 [17]

Answer:

$E_y=1175510.2\ N.C^{-1}$

The Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

Explanation:

Given:

side of a square, $a=52.5\ cm$

charge on one corner of the square, $q_1=+45\times 10^{-6}\ C$

charge on the remaining 3 corners of the square, $q_2=q_3=q_4=-27\times 10^{-6}\ C$

Distance of the center from each corners $=\frac{1}{2} \times diagonals$

$diagonal=\sqrt{52.5^2+52.5^2}$

$diagonal=74.25\cm=0.7425\ m$

∴Distance of center from corners, $b=0.3712\ m$

Now, electric field due to charges is given as:

$E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}$

For charge $q_1$ we have the field lines emerging out of the charge since it is positively charged:

$E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}$

$E_1=2938775.5\ N.C^{-1}$

Force by each of the charges at the remaining corners:

$E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}$

$E_2=E_3=E_4=1763265.3\ N.C^{-1}$

Now, net electric field in the vertical direction:

$E_y=E_1-E_4$

$E_y=1175510.2\ N.C^{-1}$

Now, net electric field in the horizontal direction:

$E_y=E_2-E_3$

$E_y=0\ N.C^{-1}$

So the Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

8 0

3 years ago

QUICKKK I NEEEED HELP PLZ!!!!!!!!!!!!

avanturin [10]

There would be martial law (just elaborate on the definition) and the population would go awry(elaborate on subject)

4 0

3 years ago

Pls help and thank u need asap!

aleksandr82 [10.1K]

Answer:

ill help if u help me?its 4am

5 0

2 years ago