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3 years ago

This is junior year english

Physics

2 answers:

pychu [463]3 years ago

6 0

D. is the correct answer.

Alla [95]3 years ago

6 0

I think the answer is D

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A transformer is used to light a lamp rated 40w, 240v from a 400v A.C supply. Calculate:

zubka84 [21]

Answer:1.81

(a) Explanation:the turn ratio= input voltage÷output voltage.

400÷220=1.81.

Don't know how to solve b part...

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2 years ago

When caring for wounds, doctors need to make sure wounds are not infected by bacteria. Which electromagnetic wave would doctors

musickatia [10]

Ultravoilent rays is an type of an electromegnetic wave with wavelength 10nm .Ultraviolet wave is used by doctors in order to keep the wounds away from bacteria.

<h3>What is ultravoilent wave ?</h3>

Ultravoilent rays is an type of an electromegnetic wave with wavelength 10nm. Ultravoilent wave is directly comes frim the sun. It is classified into the UV(A) , UV(B),UV(B).

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5 0

1 year ago

Read 2 more answers

A cable is 100-m long and has a cross-sectional area of 1.0 mm2. A 1000-N force is applied to stretch the cable. Young's modulus

Blizzard [7]

Answer:

1 m

Explanation:

L = 100 m

A = 1 mm^2 = 1 x 10^-6 m^2

Y = 1 x 10^11 N/m^2

F = 1000 N

Let the cable stretch be ΔL.

By the formula of Young's modulus

$Y=\frac{F\times L}{A\times\Delta L}$

$\Delta L=\frac{F\times L}{A\times\Y}$

$\Delta L=\frac{1000\times 100}{10^{-6}\times10^{11}}$

ΔL = 1 m

Thus, the cable stretches by 1 m.

4 0

2 years ago

A capacitor is charged until it holds 5.0 j of energy. it is then connected across a 10-kω resistor. in 13.6 ms , the resistor d

prohojiy [21]

Answer:

The capacite is C=5.32 uF using the equations of voltage and energy in capacitance

Explanation:

The energy holds is 5 J and the resistor dissipates 2J so the energy total is 3J

Using:

$V_{t}= V_{o}e^{\frac{-t}{R*C} }$

Voltage in this case is the energy dissipated so

$E_{t}= E_{o}e^{\frac{-t}{R*C} }$

$\frac{\sqrt{E_t} }{\sqrt{E_o} } = e^{\frac{-t}{R*C} }$

$\frac{\sqrt{3 J} }{\sqrt{5J} } = e^{\frac{-13.6ms}{10kw*C} }$

Using the equation to find capacitance

$ln 0.775= e^{\frac{-13.6 x10^{3} }{10x10^{3}*C }} \\ln(0.775)= ln * e^{\frac{-13.6 x10^{3s} }{10x10^{3}*C }} \\\\ln(0.775)= {\frac{-13.6 x10^{3} }{10x10^{3}*C }} \\C= \frac{-13.6 x10^{-3} }{10x10x^{3}*ln(0.775) }$