6x10^8 kg<br> —————<br> 2x10^4 m^3

What's your normal body temperature? it may not be 98.6°f, the oft-quoted average that was determined in the nineteenth century.

harkovskaia [24]

(98.6 - 98.2)ºF = .4ºF

.4ºF* 5ºC/9ºF = 0.222 ºC

6 0

4 years ago

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. A laser beam shines along the surface of a block of transparent material (see Fig. E33.8). Half of the beam goes straight to a

Neporo4naja [7]

Answer:

n = 1,875

Explanation:

The speed of light in vacuum is constant (c) and in a material medium it is

v = d / t

The refractive index of a material is defined by

n = c / v

Let's look for the speed of light in the material, in general the length that light travels is known, this value is high, x = 1, when we place a block on the road, a small amount is lengthened by the length of the block, which in general is despised

These measurements are made on a digital oscilloscope that allows to stop the signals and measure their differences, that is, the zero is taken when the first ray arrives and the time for the second ray is measured,

v = d / t

v = 1 / 6.25 10⁻⁹

v = 1.6 10⁸ m / s

we calculate the refractive index

n = 3 10⁸ / 1.6 10⁸

n = 1,875

6 0

3 years ago

A high-pass filter consists of a 1.66 μF capacitor in series with a 80.0 Ω resistor. The circuit is driven by an AC source with

Julli [10]

Explanation:

Given that,

Capacitor $C=1.66\ \mu F$

Resistor $R=80.0\ \Omega$

Peak voltage = 5.10 V

(A). We need to calculate the crossover frequency

Using formula of frequency

$f_{c}=\dfrac{1}{2\pi R C}$

Where, R = resistor

C = capacitor

Put the value into the formula

$f_{c}=\dfrac{1}{2\pi\times80.0\times1.66\times10^{-6}}$

$f_{c}=1198.45\ Hz$

(B). We need to calculate the $V_{R}$ when $f = \dfrac{1}{2f_{c}}$

Using formula of $V_{R}$

$V_{R}=V_{0}(\dfrac{R}{\sqrt{R^2+(\dfrac{1}{2\pi fC})^2}})$

Put the value into the formula

$V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times\dfrac{1}{2}\times1198.45\times1.66\times10^{-6}})^2}})$

$V_{R}=2.280\ Volt$

(C). We need to calculate the $V_{R}$ when $f = f_{c}$

Using formula of $V_{R}$

$V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times1198.45\times1.66\times10^{-6}})^2}})$

$V_{R}=3.606\ Volt$

(D). We need to calculate the $V_{R}$ when $f = 2f_{c}$

Using formula of $V_{R}$

$V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times2\times1198.45\times1.66\times10^{-6}})^2}})$

$V_{R}=4.561\ Volt$

Hence, This is the required solution.

8 0

4 years ago

Estimate the monthly cost of using a 700 W refrigerator that runs for 10 hours a day if the cost per kilowatt hours is cents $.2

Mrac [35]

Answer:

$4.2

Explanation:

Given data

Power= 700W

time= 10 hours

Cost per kilowatt hours is cents $0.20

Let us find the number of hours in a month

=24*30

=720 hours

Energ= power*time

Energy= 700/1000*30

Energy= 7*3

Enery= 21 kwh

1 kwh= 0.2

21kwh= x

cross multiply

x=21*0.2

x= $4.2

4 0

3 years ago

The shortstop plays between first and second base.<br> true or false

adell [148]

SINGLE TO RIGHT OR CENTER FIELD-On a single to center or right field the first baseman will be the cutoff to home. The shortstop will cover second base.

4 0

3 years ago

6x10^8 kg ————— 2x10^4 m^3