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In a physics laboratory experiment, a coil with 170 turns enclosing an area of 10.9 cm2 is rotated during the time interval 3.50×10−2 s from a position in which its plane is perpendicular to Earth's magnetic field to one in which its plane is parallel to the field. The magnitude of Earth's magnetic field at the lab location is 5.60×10−5 T. What is the total magnitude of the magnetic flux (initial) through the coil before it is rotated?

Answers

Answer 1

Answer:

$N= 170 turns\nA=10.9cm^2\nt=3.5*10^(-2)s \n\B = 5.6*10^(-5)T$

The Magnetic flow $\Phi_(initial)$ is given by the formula,

$\Phi_(Initial)=BAsin\theta$

Replacing the values

$\Phi_(Initial) =(5.6*10^(-5))(10.9)((10^(-4)m^2)/(1cm^2)) sin90\°$

$\Phi_(Initial) =6.104*10^(-7) Wb$

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Explain how ultrasound devices use the Doppler effect to create images of body parts. HELP ME ASAP!! Will give brainliest answer!!

Complete the calculations for total magnification produced by various combinations of the eyepiece and objective lenses. You may assume that the magnification for the eyepiece is 10X for each question. 1. When the scanning (4X) objective is used the total magnification will be:________
2. When the low power (10X) objective is used the total magnification will be:________
3. When the high power (40X) objective is used the total magnification will be:________
4. When the oil immersion (100X) objective is used the total magnification will be:_________

Answers

Answer:

a) m_ttoal = 40x, b) m_total = 100X, c) m_total = 400X,

d) m_total = 1000 X

Explanation:

La magnificación o aumentos es el incremento de del tamaño de la imagen con respecto al tamaño original del objeto, en la mayoria del os sistema optico la magnificacion total es el producoto de la magnificación del objetivo por la magnificación del ocular

m_total = m_ objetivo * m=ocular

apliquemos esto a nuestro caso

1) m_total = 4 x * 10 x

m_ttoal = 40x

2) m_total = 10X * 10X

m_total = 100X

3)mtotal = 40X * 10X

m_total = 400X

4) m _totla = 100x * 10 X

m_total = 1000 X

en este ultimo caso para magnificación grandes es decalcificar el objeto

The total magnification produced by different combinations of eyepiece and objective lenses in a microscope.

1. When the scanning (4X) objective is used, the total magnification will be 40X because the eyepiece magnification is 10X and the objective magnification is 4X.

2. When the low power (10X) objective is used, the total magnification will be 100X because the eyepiece magnification is 10X and the objective magnification is 10X.

3. When the high power (40X) objective is used, the total magnification will be 400X because the eyepiece magnification is 10X and the objective magnification is 40X.

4. When the oil immersion (100X) objective is used, the total magnification will be 1000X because the eyepiece magnification is 10X and the objective magnification is 100X.

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Two resistors are to be combined in parallelto form an equivalent resistance of 400Ω. The resistors are takenfrom available stock on hand as acquired over the years. Readily available are two common resistorsrated at 500±50 Ωand two common resistors rated at 2000 Ω±5%. What isthe uncertainty in an equivalent 400 Ωresistance?(Hint: the equivalent resistance connected in parallel can be obtained by 1212TRRRRR=+)

Answers

Answer:

Δ $R_(e)$ = 84 Ω, $R_(e)$ = (40 ± 8) 10¹ Ω

Explanation:

The formula for parallel equivalent resistance is

1 / $R_(e)$ = ∑ 1 / Ri

In our case we use a resistance of each

R₁ = 500 ± 50 Ω

R₂ = 2000 ± 5%

This percentage equals

0.05 = ΔR₂ / R₂

ΔR₂ = 0.05 R₂

ΔR₂ = 0.05 2000 = 100 Ω

We write the resistance

R₂ = 2000 ± 100 Ω

We apply the initial formula

1 / $R_(e)$ = 1 / R₁ + 1 / R₂

1 / $R_(e)$ = 1/500 + 1/2000 = 0.0025

$R_(e)$ = 400 Ω

Let's look for the error (uncertainly) of Re

$R_(e)$ = R₁R₂ / (R₁ + R₂)

R’= R₁ + R₂

$R_(e)$ = R₁R₂ / R’

Let's look for the uncertainty of this equation

Δ $R_(e)$ / $R_(e)$ = ΔR₁ / R₁ + ΔR₂ / R₂ + ΔR’/ R’

The uncertainty of a sum is

ΔR’= ΔR₁ + ΔR₂

We substitute the values

Δ $R_(e)$ / 400 = 50/500 + 100/2000 + (50 +100) / (500 + 2000)

Δ $R_(e)$ / 400 = 0.1 + 0.05 + 0.06

Δ $R_(e)$ = 0.21 400

Δ $R_(e)$ = 84 Ω

Let's write the resistance value with the correct significant figures

$R_(e)$ = (40 ± 8) 10¹ Ω

A spectroscope:measures light from distant objects
makes object far away look closer
receives radio signals from objects in space

Answers

Answer:

Option A

Measures light from distant objects

Explanation:

A spectroscope is used to measure the use of light from a distant object to work out the object is made of.

It could be the single-most powerful tool astronomers use.

Professor Fred Watson from the Australian Astronomical Observatory says that "It lets you see the chemicals being absorbed or emitted by the light source"

One of the primary visible emissions from a distant planet occurs at 425 nm. Calculate the energy of a mole of photons of this emission.]

Answers

Answer:

Explanation:

Given

Wavelength of incoming light $\lambda =425\ nm$

We know

$speed\ of\ wave=frequency* wavelength$

$frequency=(speed)/(wavelength)$

$\mu =(3* 10^8)/(425* 10^(-9))$

$\mu =7.058* 10^(14)\ Hz$

Energy associated with this frequency

$E=h\mu$

where h=Planck's constant

$E=6.626* 10^(-34)* 7.058* 10^(14)$

$E=46.76* 10^(-20)\ Hz$

Energy of one mole of Photon $=N_a* E$

$=6.022* 10^(23)* 46.76* 10^(-20)$

$=281.58* 10^(3)$

$=281.58\ kJ$

Final answer:

To calculate the energy of a mole of photons of the emission at 425 nm, use the equation E = hc/λ, where E is the energy, h is Planck's constant, c is the speed of light, and λ is the wavelength. Convert the wavelength to meters, substitute the values into the equation, and calculate to find the energy of a single photon. Multiply this by Avogadro's number to find the energy of a mole of photons.

Explanation:

To calculate the energy of a mole of photons of the emission at 425 nm, we can use the equation E = hc/λ, where E is the energy, h is Planck's constant (6.63 x 10^-34 J·s), c is the speed of light (3.00 x 10^8 m/s), and λ is the wavelength (in meters).

Converting the wavelength to meters, we have 425 nm = 425 x 10^-9 m.

Substituting the values into the equation, we get E = (6.63 x 10^-34 J·s)(3.00 x 10^8 m/s) / (425 x 10^-9 m). Calculating this gives us the energy of a single photon of this emission. To find the energy of a mole of photons, we can multiply this value by Avogadro's number (6.02 x 10^23 photons/mol).

Therefore, the energy of a mole of photons of this emission is (6.63 x 10^-34 J·s)(3.00 x 10^8 m/s) / (425 x 10^-9 m) x (6.02 x 10^23 photons/mol).

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Analysis of the relationship between the fuel economy (mpg) and engine size (liters) for 35 models of cars produces the regression model ModifyingAbove mpg with caret equals 36.44 minus 3.829 times Engine size. If a car has a 5 liter engine, what does this model suggest the gas mileage would be?

Answers

Answer:

Gas mileage is 17.29

Explanation:

Given data:

The total number of the model is 35

The total size of the engine is 5 ltr

The regression model is given as

$36.44 - 3.829* engine\ size$

From the information given in question we have

Regression equation is : model- mpg $= 36.44 - 3.829* engine\ size$

Therefore for engine capacity of 5 liters;

Gas mileage $= 36.44 - 3.829* 5 = 17.29$

Gas mileage is 17.29

Answer:16.

Explanation:

A celestial body moving in an ellipical orbit around a star

Answers

A celestial body moving in an elliptical orbit around a star is a planet.

Depending on its size, composition, and the eccentricity of its orbit, that scanty description could apply to a planet, an asteroid, a comet, a meteoroid, or another star.

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