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3 years ago

RC time constant circuit if R 50 KOC-21 a TOSS c. 1.05 s . what is the expected RC value b. 10.55 d. 0.105 s

Physics

1 answer:

Afina-wow [57]3 years ago

7 0

Answer:

Time constant of RC circuit is 0.105 seconds.

Explanation:

It is given that,

Resistance, $R=50\ K\Omega=5\times 10^4\ \Omega$

Capacitance, $C=2.1\ \mu F=2.1\times 10^{-6}\ F$

We need to find the expected time constant for this RC circuit. It can be calculated as :

$\tau=R\times C$

$\tau=5\times 10^4\times 2.1\times 10^{-6}$

$\tau=0.105\ s$

So, the time constant for this RC circuit is 0.105 seconds. Hence, this is the required solution.

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A hockey player skates with an average speed of 8.25 m/s. The distance the player will travel in 5.60 s is __________ .

OverLord2011 [107]

<h2>Greetings!</h2>

To find this value, you need to remember the speed formula:

3 = 6 / 2

Speed = distance ÷ time

Rearrange to make distance the subject:

Distance = speed * time

Simply plug these values into this:

5.6 * 8.25 = 46.2

<h3>So the player will travel 46.2 metres!</h3>
<h2>Hope this helps!</h2>

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3 years ago

A circular radar antenna on a Coast Guard ship has a diameter of 2.10 m and radiates at a frequency of 16.0 GHz. Two small boats

Anna35 [415]

Answer:

d = 76.5 m

Explanation:

To find the distance at which the boats will be detected as two objects, we need to use the following equation:

$\theta = \frac{1.22 \lambda}{D} = \frac{d}{L}$

<u>Where:</u>

θ: is the angle of resolution of a circular aperture

λ: is the wavelength

D: is the diameter of the antenna = 2.10 m

d: is the separation of the two boats = ?

L: is the distance of the two boats from the ship = 7.00 km = 7000 m

To find λ we can use the following equation:

$\lambda = \frac{c}{f}$

<u>Where:</u>

c: is the speed of light = 3.00x10⁸ m/s

f: is the frequency = 16.0 GHz = 16.0x10⁹ Hz

$\lambda = \frac{c}{f} = \frac{3.00 \cdot 10^{8} m/s}{16.0 \cdot 10^{9} s^{-1}} = 0.0188 m$

Hence, the distance is:

$d = \frac{1.22 \lambda L}{D} = \frac{1.22*0.0188 m*7000 m}{2.10 m} = 76.5 m$

Therefore, the boats could be at 76.5 m close together to be detected as two objects.

I hope it helps you!

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3 years ago

Which explanation is least likely to be evidence for geological processes changing Earth's surface?

horsena [70]

Pollution isn’t a geological process.

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2 years ago

Read 2 more answers

A narrow beam of light from a laser travels through air (n = 1.00) and strikes the surface of the water (n = 1.33) in a lake at

Natalka [10]

Answer:

A) d = 11.8m

B) d = 4.293 m

Explanation:

A) We are told that the angle of incidence;θ_i = 70°.

Now, if refraction doesn't occur, the angle of the light continues to be 70° in the water relative to the normal. Thus;

tan 70° = d/4.3m

Where d is the distance from point B at which the laser beam would strike the lakebottom.

So,d = 4.3*tan70

d = 11.8m

B) Since the light is moving from air (n1=1.00) to water (n2=1.33), we can use Snell's law to find the angle of refraction(θ_r)

So,

n1*sinθ_i = n2*sinθ_r

Thus; sinθ_r = (n1*sinθ_i)/n2

sinθ_r = (1 * sin70)/1.33

sinθ_r = 0.7065

θ_r = sin^(-1)0.7065

θ_r = 44.95°

Thus; xonsidering refraction, distance from point B at which the laser beam strikes the lake-bottom is calculated from;

d = 4.3 tan44.95

d = 4.293 m

4 0

3 years ago

A thin film of soap with n = 1.34 hanging in the air reflects dominantly red light with λ = 642 nm. What is the minimum thicknes

wariber [46]

Answer:

The minimum thickness of the film and the wavelength of the light in air are $1.197\times10^{-7}\ m$ and 371 nm.

Explanation:

Given that,

Refractive index of soap= 1.34

Refractive index of glass= 1.55

Wavelength = 642 nm

(I). We need to calculate the minimum thickness

Using formula of thickness

$t=\dfrac{(2m+1)\lambda}{4n}$

Where, m = 0 for constrictive

Put the value into the formula

$t=\dfrac{(642\times10^{-9})}{4\times1.34}$

$t=1.197\times10^{-7}\ m$

(II). We need to calculate the wavelength

Using formula of wavelength

$\lambda=\dfrac{2nt}{m}$

Where, m = 1

Put the value into the formula

$\lambda=\dfrac{2\times1.55\times1.197\times10^{-7}}{1}$

$\lambda=371\ nm$

Hence, The minimum thickness of the film and the wavelength of the light in air are $1.197\times10^{-7}\ m$ and 371 nm.

6 0

3 years ago