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Aliun [14]
3 years ago
6

HELP ASAP

Physics
1 answer:
kipiarov [429]3 years ago
4 0

Answer:

This can be part of your paragraph.

Explanation:

From the cornea, the light passes through the pupil. The iris, or the colored part of your eye, controls the amount of light passing through. From there, it then hits the lens. This is the clear structure inside the eye that focuses light rays onto the retina.

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Two speakers face each other, and they each emit a sound of wavelength λ. One speaker is 180∘ out of phase with respect to the o
kolbaska11 [484]
The question ask to calculate and choose among the following is the distance of the microphone from the left - most speaker  in order to pick up the loudest  sound and the best answer would be letter C. I hope you are satisfied with my answer and feel free to ask for more if you have questions and further clarifications 
4 0
3 years ago
Where can you find most of earths freshwater
Snezhnost [94]

Answer:

HEY PLS LISTEN THERE IS A GUY GOING AROUND TELLING GIRLS TO JOIN HIS ZOOM ON BRAINLY AND MAKING ACCS AND STUFF TO MAKE PEOPLE JOIN HIS ZOOM.THE CODE IS 387 187 6280 HES A PEDXO AND TELLING GIRLS TO SHOW THERE BODIES TO HIM. SO PLEASE BE AWARE! COPY AND PASTE THIS TO SPREAD THE WORD!

Explanation:

HEY PLS LISTEN THERE IS A GUY GOING AROUND TELLING GIRLS TO JOIN HIS ZOOM ON BRAINLY AND MAKING ACCS AND STUFF TO MAKE PEOPLE JOIN HIS ZOOM.THE CODE IS 387 187 6280 HES A PEDXO AND TELLING GIRLS TO SHOW THERE BODIES TO HIM. SO PLEASE BE AWARE! COPY AND PASTE THIS TO SPREAD THE WORD!

3 0
3 years ago
Calculate the rate of heat generation in kW due to the burning of the fuel when you drive a car rated at 34 MPG (miles per gallo
alexira [117]

Answer:

Explanation:

Let us calculate gallon used in one hour .

It travels 70 miles in one hour

in 70 miles it uses 70 / 34 gallons of fuel

70 / 34 gallons = 70 / 34 x 3.7854 kg

= 7.8 kg

heat generated = 7.8 x 44 x 10⁶ J

= 343.2 x 10⁶ J

This is heat generated in one hour

heat generated in one second = 343.2 x 10⁶ / 60 x 60 J/s

= 95.33 x 10³ J /s

= 95.33 kW.

4 0
3 years ago
Two identical charges, each -8.00 E-5 C, are separated by a distance of 20.0 cm (100 cm = 1 m). What is the force of repulsion?
Rom4ik [11]

F = 1440 N. The repulsion force between two identical charges, each -8.00x10⁻⁵C separated by a distance of 20.0 cm is 1440 N.

The easiest way to solve this problem is using Coulomb's Law given by the equation F=k\frac{|q_{1}*q_{2}|}{r^{2} }, where k is the constant of proportionality or Coulomb's constant, q₁ and q₂ are the charges magnitude, and r is the distance between them.

We have to identical charges of -8.00x10⁻⁵C, are separated by a distance of 20.0 cm, and we need to know the force of repulsion between the charges.

First, we have to convert 20.0 cm to meters.

(20.0 cm x 1m)/100cm = 0.20 m

Using the Coulomb's Law equation:

F = 9.00x10^{9}\frac{Nm^{2}}{C^{2}} \frac{|-8.00x10^{-5}C*-8.00x10^{-5}C|}{(0.20m)^{2} }

F = 9.00x10^{9}\frac{Nm^{2}}{C^{2}}(1.6x10^-7\frac{C^{2} }{m^{2} } })\\F = 1440N

5 0
3 years ago
If a sound is 30 dB and its absolute pressure was 66 x 10-9 Pa, what must have been the reference pressure?
Ksenya-84 [330]

Given:

I = 30dB

P = 66 × 10^{-9} Pa

Solution:

Formula used:

I = 20\log_{10}(\frac{P}{P_{o}})           (1)

where,

I = intensity of sound

P = absolute pressure

P_{o} = reference pressure

Using Eqn (1), we get:

30 = 20\log _{10}\frac{66\times 10^{-9}}{P^{o}}

P_{o} = \frac{66\times 10^{-9}}{10^{1.5}}

P_{o} = 2.08 × 10^{-9} Pa

4 0
4 years ago
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