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3 years ago

9. How does water pollution contribute to water shortages?

Physics

1 answer:

Grace [21]3 years ago

8 0

Answer:

<h2>if the chemicals seep into the groundwater or in underground aquifers.</h2><h2 />

Explanation:

<h3>One of the largest causes of water scarcity across the world is pollution. ... Pesticides and other fertilizer use by farmers can also lead to water pollution if the chemicals seep into the groundwater or in underground aquifers. The misuse of water resources is another massive issue that leads to water scarcity.</h3>

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What should we do to be a computer engineer ‍ ‍?

ycow [4]

Answer:

to be an eginere u would have to go to college and study hard

Explanation:

6 0

3 years ago

Read 2 more answers

Consider an object with s=12cm that produces an image with s′=15cm. Note that whenever you are working with a physical object, t

Leni [432]

A. 6.67 cm

The focal length of the lens can be found by using the lens equation:

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}$

where we have

f = focal length

s = 12 cm is the distance of the object from the lens

s' = 15 cm is the distance of the image from the lens

Solving the equation for f, we find

$\frac{1}{f}=\frac{1}{12 cm}+\frac{1}{15 cm}=0.15 cm^{-1}\\f=\frac{1}{0.15 cm^{-1}}=6.67 cm$

B. Converging

According to sign convention for lenses, we have:

- Converging (convex) lenses have focal length with positive sign

- Diverging (concave) lenses have focal length with negative sign

In this case, the focal length of the lens is positive, so the lens is a converging lens.

C. -1.25

The magnification of the lens is given by

$M=-\frac{s'}{s}$

where

s' = 15 cm is the distance of the image from the lens

s = 12 cm is the distance of the object from the lens

Substituting into the equation, we find

$M=-\frac{15 cm}{12 cm}=-1.25$

D. Real and inverted

The magnification equation can be also rewritten as

$M=\frac{y'}{y}$

where

y' is the size of the image

y is the size of the object

Re-arranging it, we have

$y'=My$

Since in this case M is negative, it means that y' has opposite sign compared to y: this means that the image is inverted.

Also, the sign of s' tells us if the image is real of virtual. In fact:

- s' is positive: image is real

- s' is negative: image is virtual

In this case, s' is positive, so the image is real.

E. Virtual

In this case, the magnification is 5/9, so we have

$M=\frac{5}{9}=-\frac{s'}{s}$

which can be rewritten as

$s'=-M s = -\frac{5}{9}s$

which means that s' has opposite sign than s: therefore, the image is virtual.

F. 12.0 cm

From the magnification equation, we can write

$s'=-Ms$

and then we can substitute it into the lens equation:

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}\\\frac{1}{f}=\frac{1}{s}+\frac{1}{-Ms}$

and we can solve for s:

$\frac{1}{f}=\frac{M-1}{Ms}\\f=\frac{Ms}{M-1}\\s=\frac{f(M-1)}{M}=\frac{(-15 cm)(\frac{5}{9}-1}{\frac{5}{9}}=12.0 cm$

G. -6.67 cm

Now the image distance can be directly found by using again the magnification equation:

$s'=-Ms=-\frac{5}{9}(12.0 cm)=-6.67 cm$

And the sign of s' (negative) also tells us that the image is virtual.

H. -24.0 cm

In this case, the image is twice as tall as the object, so the magnification is

M = 2

and the distance of the image from the lens is

s' = -24 cm

The problem is asking us for the image distance: however, this is already given by the problem,

s' = -24 cm

so, this is the answer. And the fact that its sign is negative tells us that the image is virtual.

3 0

4 years ago

In a simple circuit a 6-volt dry cell pushes charge through a single lamp which has a resistance of 3 Ω. According to Ohms law t

RSB [31]

Answer:

The answers and workings is in the Explanation section

Explanation:

<em>In a simple circuit a 6-volt dry cell pushes charge through a single lamp which has a resistance of 3 Ω. According to Ohms law the current through the circuit is _____________ Amps. </em>

According to Ohm’s law V =I*R

Where V = Voltage, I = Current and R = Resistance

I = V/R =6/3 =2 Amps of current

Answer = 2 Amps

<em> If a second identical lamp is connected in series, the 6-volt battery must push a charge through a total resistance of ________ Ω</em>. <em>The current in the circuit is then __________ Amps. </em>

Since the second lamp is connected in series with the first, the total resistance will R₂ = 3 + 3 = 6Ω

2 resistance in series R₂ = 6Ω

The current in the circuit with the two lamps connected in series is I₂ =V/R₂ =6/6 = 1 Amps

The current is 1 Amps

Answer = 6Ω and 1 Amps

<em>If a third identical lamp is connected in series, the total resistance is now _________Ω. The current through all three lamps in series is now _________ Amps. </em>

Since the third lamp is connected in series with the first and second, the total resistance will R₃ = 3 + 3 + 3 = 9Ω

total resistance of the 3 lamps R₃ = 9Ω

The current in the circuit with the three lamps connected in series is

I =V/R₃ =6/9 = 0.67 Amps

The current through the 3 lamps I₃ = 0.67 Amps

Answer = 9Ω and 0.67 Amps

<em>The current through each individual lamp is __________ Amps. </em>

Since all 3 lamps are connected in series, the same current will flow through each of the 3 lamps, and that current is I₃

The current through each individual lamp is 0.67 Amp

Answer = 0.67 Amp

<em>What is the power when a voltage of 120 volts drives a current of 3 amps through a device? </em>

The formula for power P = I*V =120*3 = 360 Watts

power P = 360 Watts

Answer = 360 Watts

<em>What is the current when a 90-W light bulb is connected to 120 V? </em>

From P =I*V, make I the subject of the formula, I = P/V =90/120 = 0.75

Current= 0.75 Amps

Answer = 0.75 Amps

<em>How much current does a 75-W light bulb draw when connected to 120 V?</em>

Current I =P/V = 75/120 = 0.625 Amps.

Answer = 0.625 Amps

<em>If part of an electric circuit dissipates energy at 6 W when it draws a current of 3 A, what voltage is impressed across it? </em>

Voltage V =P/I =6/3 =2 Volts

Answer = 2 Volts

<em> If a 60 W light bulb at 120 V is left on in your house to prevent burglary, and the power company charges 10 cents per kilowatt-hour, how much will it cost to leave the bulb on for 30 days? Show your work. </em>

24 hours make 1 day, so the number of hours the bulb was left on = 24 *30 = 720 hours

The power rating of the bulb is 60W = 60/1000 = 0.06 KiloWatt

Total power consumed in kilowatt-hour = 0.06 * 720 = 43.2 kilowatt-hour

Cost for 30 days = 0.1*43.2 = $4.32 ( note that 10 Cents = $0.1)

Answer = $4.32

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3 years ago

Consider 2 cars traveling directly towards each other. Car A has twice the mass of Car B. In

galben [10]

Answer:

If Car A has half the mass of Car B, then it must have twice the velocity to come to a stop when colliding.

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3 years ago

A bucket of water experiencing a gravitational force of 525 N is pulled up from a water well. The net force in the y-direction i

lukranit [14]

Answer:

6n!!!!!!!!!!!!!!!!!!

Explanation:

nnnn

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3 years ago