Complete play the table by writing the location orientation size and type of image formed by the lenses below.

A solenoid of length 0.250 m and radius 0.0250 m is comprised of 440 turns of wire. Determine the magnitude of the magnetic fiel

Bad White [126]

Answer: 0.02654

Explanation:

in the attachment

6 0

4 years ago

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Please can anybody tell me what are these lab equipments

olga nikolaevna [1]

Answer:

5 is the tripoid stand

Thanks have a bangtastic day

4 0

3 years ago

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Assume that in 2010 the United States will need 2.0×1012 watts of electric power produced by thousands of 1000 MW power plants.

Alex73 [517]

Answer:

1752.14 tonnes per year.

Explanation:

To solve this exercise it is necessary to apply the concepts related to power consumption and power production.

By conservation of energy we know that:

$\dot{P} = \bar{P}$

Where,

$\dot{P} =$ Production of Power

$\bar{P} =$ Consumption of power

Where the production of power would be,

$\dot{P} = m \dot{E}\eta$

Where,

m = Total mass required

$\dot{E} =$ Energy per Kilogram

$\eta =$ Efficiency

The problem gives us the aforementioned values under a production efficiency of 45%, that is,

$\dot{P} = \bar{P}$

$m \dot{E}\eta = \bar{P}$

Replacing the values we have,

$m(8*10^13)(0.45) = 2*10^{12}$

Solving for m,

$m = \frac{ 2*10^{12}}{(8*10^13)(0.45)}$

$m = 0.0556 \frac{kg}{s}$

We have the mass in kilograms and the time in seconds, we need to transform this to tons per year, then,

$m = 0.556\frac{kg}{s}*(\frac{3.1536*10^7s}{1year})(\frac{1ton}{1000kg})$

$m = 1752.14$ tonnes per year.

8 0

3 years ago

A rotating space station is said to create "artificial gravity" - a loosely-defined term used for an acceleration that would be

FrozenT [24]

Answer:

$\omega=0.31\frac{rad}{s}$

Explanation:

The artificial gravity generated by the rotating space station is the same centripetal acceleration due to the rotational motion of the station, which is given by:

$a_c=\frac{v^2}{r}(1)$

Here, r is the radius and v is the tangential speed, which is given by:

$v=\omega r(2)$

Here $\omega$ is the angular velocity, we replace (2) in (1):

$a_c=\frac{(\omega r)^2}{r}\\\\a_c=\omega^2r$

Recall that $r=\frac{d}{2}=\frac{200m}{2}=100m$ .

Solving for $\omega$ :

$\omega=\sqrt{\frac{a_c}{r}}\\\omega=\sqrt{\frac{9.8\frac{m}{s^2}}{100m}}\\\omega=0.31\frac{rad}{s}$

3 0

4 years ago

A child on a playground swings through a total of 32°. If the displacement is equal on each side of the equilibrium position, wh

sashaice [31]

Let L = length of the swing.

Let the lowest position of the swing be the reference point.
Therefore in the lowest position,
The height of the swing = 0,
The support for the swing is at height = L.

At 32° swing relative to the vertical,
The height of the swing is
A = L - Lcos(32°) = L[1 - cos(32°)] = 0.152L

Answer:
The amplitude is 0.152*L, where L = length of the swing.

6 0

4 years ago