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3 years ago

High levels of pollution will MOST likely

Physics

2 answers:

galben [10]3 years ago

4 0

Answer:

Explanation:

A) increase the rate of chemical weathering

Nataly [62]3 years ago

4 0

Answer:

Explanation:

increase the rate of chemical weathering is the right choice

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) Water falls from a height of 60m at the rate of 15kg/s to operate a turbine. The losses due to frictional force are 10% of ene

Angelina_Jolie [31]

Answer:

8100W

Explanation:

Let g = 10m/s2

As water is falling from 60m high, its potential energy from 60m high would convert to power. So the rate of change in potential energy is

$P = \dot{E} = \dot{m}gh = 15*10*60 = 9000 J/s$ or 9000W

Since 10% of this is lost to friction, we take the remaining 90 %

P = 9000*90% = 8100 W

3 0

3 years ago

What is the wavelenght and the frequencen of walkie-talkie?

GuDViN [60]

Nowadays, most of the walkie talkies ... and all R/C models ... operate in the "Family Radio Service" (FRS) band. That's 46-49 MHz. The wavelength is 6.12-6.52 meters.

8 0

3 years ago

Consider a perfectly reflecting mirror oriented so that solar radiation of intensity I is incident upon, and perpendicular to, t

VLD [36.1K]

Answer:

Frad = 2IA/C

Explanation:

see attached file

4 0

3 years ago

A) A real object 4.0 cm high stands 30.0 cm in front of a converging lens of focal length 23 cm. Find the image distance, the im

vfiekz [6]

Explanation:

Given that,

Height of object = 4.0 cm

Distance of the object u= -30.0 cm

Focal length = 23 cm

We need to calculate the image distance

Using lens's formula

$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

Where, u = object distance

v = image distance

f = focal length

Put the value into the formula

$\dfrac{1}{v}-\dfrac{1}{-30}=\dfrac{1}{23}$

$\dfrac{1}{v}=\dfrac{1}{23}-\dfrac{1}{30}$

$\dfrac{1}{v}=\dfrac{7}{690}$

$v=98.57\ cm$

We need to calculate the height of the image

Using formula of height

$\dfrac{h'}{h}=\dfrac{-v}{u}$

Where, h' = height of image

h = height of object

$\dfrac{h'}{4}=\dfrac{-98.57}{-30}$

$h'=\dfrac{98.57\times4}{30}$

$h'=13.14\ cm$

The image is real and inverted.

(b). Now, object distance u = 13.0 cm

We need to calculate the image distance

Using lens's formula

$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

Put the value into the formula

$\dfrac{1}{v}=\dfrac{1}{23}-\dfrac{1}{13.0}$

$\dfrac{1}{v}=-\dfrac{10}{299}$

$v=-29.9\ cm$

We need to calculate the height of the image

$\dfrac{h'}{4}=\dfrac{-(-29.9)}{-13.0}$

$h=-\dfrac{29.9\times4}{13.0}$

$h=-9.2\ cm$

The image is virtual and erect.

Hence, This is required solution.

6 0

3 years ago

6. When a positive charge is released and moves along an electric field line, it moves to a position of A) lower potential and l

Marysya12 [62]

Answer:

Since you would have to do work on the charge to bring it back to its original position, the charge moves to a position of lower potential and lower potential energy.

4 0

3 years ago