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3 years ago

8

A transformer is to be used to provide power for a computer disk drive that needs 6.4 V (rms) instead of the 120 V (rms) from th

e wall outlet. The number of turns in the primary is 405, and it delivers 500 mA (the secondary current) at an output voltage of 6.4 V (rms). Find the current in the primary.

1 answer:

Amanda [17]3 years ago

8 0

Answer:

The current in the primary is 0.026 A

Explanation:

Using the formula

I1 = (V1/V2)*I2

we have

I1 = (6.4/120)*0.500

I1 = 0.026 A

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The gage pressure of an automobile tire is measured to be 210kPa before a trip and 220kPa after the trip at a location where the

Katarina [22]

Answer:

$T_{f} = 312.348\,K\,(39.198\,K)$

Explanation:

Let assume that gas inside the automobile tire behaves as an ideal gas. Due to the absence of leakages, the number of moles remains constant during the trip. Air temperature can be found by using the following relation:

$\frac{P_{o}}{T_{o}}=\frac{P_{f}}{T_{f}}$

Final temperature is cleared with the expression:

$T_{f} = \frac{P_{f}}{P_{o}}\cdot T_{o}$

$T_{f} = \frac{220\,kPa}{210\,kPa}\cdot (298.15\,K)$

$T_{f} = 312.348\,K\,(39.198\,K)$

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3 years ago

A double-slit arrangement produces interference fringes for sodium light (λ = 589 nm) that are 0.48° apart. What is the angular

larisa [96]

Answer:

0.38°

Explanation:

$\theta$ = Angle

m = Number

d = Distance

n = Refractive index of liquid = 1.25

a denotes air

l denotes liquid

In the case of double split interferance we have the relation

$m\lambda=dsin\theta$

For air

$m\lambda_a=dsin\theta_a$

For liquid

$m\lambda_l=dsin\theta_l$

Dividing the two equations

$\frac{m\lambda_a}{m\lambda_l}=\frac{dsin\theta_a}{dsin\theta_l}\\\Rightarrow \frac{\lambda_a}{\lambda_l}=\frac{sin\theta_a}{sin\theta_l}$

Wavelength ratio = $n$

$n=\frac{sin\theta_a}{sin\theta_l}\\\Rightarrow \frac{sin0.48}{1.25}=sin\theta_l\\\Rightarrow \theta_l=sin^{-1}\frac{sin0.48}{1.25}\\\Rightarrow \theta_l=0.38^{\circ}$

The angular separation is 0.38°

5 0

4 years ago

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A geneticist looks through a microscope to determine the phenotype of a fruit fly. The microscope is set to an overall magnifica

Ugo [173]

Answer:

$f_{e}$ = 1.9 cm

Explanation:

The magnification of a microscope is the product of the magnification of the eyepiece by the magnifier with the objective

M = M₀ $m_{e}$

Where M₀ is the magnification of the objective and $m_{e}$ is the magnification of the eyepiece.

The eyepiece is focused to the near vision point (d = 25 cm)

$m_{e}$ = 25 / $f_{e}$

The objective is focused on the distances of the tube (L)

M₀ = -L / f₀

Substituting

M = - L/f₀ 25/ $f_{e}$

1) Let's look for the focal length of the eyepiece (faith)

$f_{e}$ = - L 25 / f₀ M

M = 400X = -400

$f_{e}$ = - 12 25 /0.40 (-400)

$f_{e}$ = 1.875 cm

Let's approximate two significant figures

$f_{e}$ = 1.9 cm

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4 years ago

Please help ASAP WILL GIVE BRAINLIEST

Westkost [7]

Answer:

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Explanation: It makes the most sense. Plz mark brainliest

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3 years ago

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Suppose you pour 0.250 kg of 20.0°C water into a 0.600 kg aluminum pan off the stove with a temperature of 173°C. Assume that th

lapo4ka [179]

Answer:

$T_f=5.0116^{\circ}C$

Explanation:

Given:

mass of water, $m_w=0.25\ kg$

initial temperature of water, $T_i_w=20^{\circ}C$

initial temperature of pan, $T_i_p=173^{\circ}C$

mass of pan, $m_p=0.6\ kg$

mass of water evapourated, $m_v=0.03\ kg$

specific heat of water, $c_w=4186\ J.kg^{-1}.K^{-1}$

specific heat of aluminium pan, $c_a=900\ J.kg^{-1}.K^{-1}$

latent heat of vapourization, $L=2256000\ J.kg^{-1}$

<u>Using the equation of heat:</u>

<em>Here, initially certain mass of water is vapourised first and then the remaining mass of water comes in thermal equilibrium with the pan.</em>

$m_p.c_a.(T_{ip}-T_f)=m_v.L+(m_w-m_v).c_w.(T_f-T_{iw})$

$0.6\times 900\times (173-T_f)=0.03\times 2256000+(0.25-0.03)\times 4186\times (T_f-20)$

$T_f=5.0116^{\circ}C$

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3 years ago