Name 3 renewable energy resources

Rectangle A has side lengths of 6\text{ cm}6 cm6, space, c, m and 3.5\text{ cm}3.5 cm3, point, 5, space, c, m. The side lengths

VikaD [51]

Answer:

Explanation:

Gotta work on writing the problem, kida hard to read with the coding text. Anyway, a rectangle with one side of 6 cm and another at 3.5 cm, then rectangle B is proportional to that. Which basically means rectangle B will be multiplied or divided by some number.

To make the concept a little simpler imagine it was a square with sides of length 2. a proportional square twice as big would have lengths of 4 and half as big would have lengths of 1.

In this case we need measurements where one is 6 multiplied or divided by a number x, and then 3.5 has the same action done on it with the same number. so again, twice as big would be 12 and 7 while half as big would be 3 and 1.75 anyway, let's look at the options.

A) 3 and 1.75

B) 5 and 2.5

C) 7 and 7

D) 12 and 5

E) 5.25 and 9

Just to have ti written, the original rectangle is 6 by 3.5

So A is obvious since I used it as an example, this is a proportional rectangle half the size of A.

The rest are harder to check except C. 7 and 7 make a square, but we know the side lengths have to be different, so C is easy to tell it's not right.

B we have to check. Since it's not as obvious as A, we have to check with math more carefully. What do we have to multiply 6 by to get 5? Also, you always want to make the largest side of each rectangle proportional to each other. Anyway, 6 times 5/6 is 5, so for B to be right, 3.5 times 5/6 needs to be equal to 2.5. Does it? Nope, its 2.9166 repeating. So B is not right.

I'm not going to go through so much explanation for the others, if you don't get something let me know.

D we could also know quickly isn't right since I told you the measurements if the rectangle is changed to one with a side length of 12. We would need the other to be 7, so 12 and 5 isn't right.

E is our last possibility, since we need to pick two, but let's check anyway. we need to multiply 6 by 7/8 to get 5.25. 3.5 times 7/8 is 3.0625, so not 9. Good thing we checked here, because only one answer looks to be right. Are you sure you put the right numbers? Or did I misinterpret some?

8 0

4 years ago

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A group of environmental scientists wants to study the impact of population growth on the environment. Which discipline will con

Julli [10]

Answer:

A. Social Science

Explanation:

3 0

3 years ago

PHYSICS CIRCUIT QUESTION PLEASE HELP!! 20 Points!

dimulka [17.4K]

This really calls for a blackboard and a hunk of chalk, but
I'm going to try and do without.

If you want to understand what's going on, then PLEASE
keep drawing visible as you go through this answer, either
on the paper or else on a separate screen.

The energy dissipated by the circuit is the energy delivered by
the battery. We'd know what that is if we knew I₁ . Everything that
flows in this circuit has to go through R₁ , so let's find I₁ first.

-- R₃ and R₄ in series make 6Ω.
-- That 6Ω in parallel with R₂ makes 3Ω.
-- That 3Ω in series with R₁ makes 10Ω across the battery.
-- I₁ is 10volts/10Ω = 1 Ampere.

-- R1: 1 ampere through 7Ω ... V₁ = I₁ · R₁ = 7 volts .

-- The battery is 10 volts.
    7 of the 10 appear across R₁ .
   So the other 3 volts appear across all the business at the bottom.

-- R₂: 3 volts across it = V₂.
   Current through it is I₂ = V₂/R₂ = 3volts/6Ω = 1/2 Amp.

-- R3 + R4: 6Ω in the series combination
   3 volts across it
   Current through it is I = V₂/R = 3volts/6Ω = 1/2 Ampere

-- Remember that the current is the same at every point in
a series circuit. I₃ and I₄ must be the same 1/2 Ampere,
because there's no place in the branch where electrons can
be temporarily stored, no place for them to leak out, and no
supply of additional electrons.

-- R₃: 1/2 Ampere through it = I₃ .
   1/2 Ampere through 2Ω ... V₃ = I₃ · R₃ = 1 volt

-- R₄: 1/2 Ampere through it = I₄
   1/2 Ampere through 4Ω ... V₄ = I₄ · R₄ = 2 volts

Notice that I₂ is 1/2 Amp, and (I₃ , I₄) is also 1/2 Amp.
So the sum of currents through the two horizontal branches is 1 Amp,
which exactly matches I₁ coming down the side, just as it should.
That means that at the left side, at the point where R₁, R₂, and R₃ all
meet, the amount of current flowing into that point is the same as the
amount flowing out ... electrons are not piling up there.

Concerning energy, we could go through and calculate the energy
dissipated by each resistor and then addum up. But why bother ?
The energy dissipated by the resistors has to come from the battery,
so we only need to calculate how much the battery is supplying, and
we'll have it.

The power supplied by the battery = (voltage) · (current)

   = (10 volts) · (1 Amp) = 10 watts .

"Watt" means "joule per second".
The resistors are dissipating 10 joules per second,
and the joules are coming from the battery.

   (30 minutes) · (60 sec/minute) = 1,800 seconds

   (10 joules/second) · (1,800 seconds) = 18,000 joules in 30 min

The power (joules per second) dissipated by each individual resistor is

   P = V² / R
   or
   P = I² · R ,

whichever one you prefer. They're both true.

If you go through the 4 resistors, calculate each one, and addum up, you'll
come out with the same 10 watts / 18,000 joules total.

They're not asking for that. But if you did it and you actually got the same
numbers as the battery is supplying, that would be a really nice confirmation
that all of your voltages and currents are correct.

7 0

3 years ago

Standing still, Bruce, the quarterback, gets tackled by Biff, the 90.0-kg tackle, who is traveling at 7.0 m/s. Upon collision, B

Doss [256]

$\\ \sf\longmapsto \Delta P=P$

$\\ \sf\longmapsto m1v1=m2v2$

$\\ \sf\longmapsto 90(7)=m2(10)$

$\\ \sf\longmapsto 10m2=630$

$\\ \sf\longmapsto m2=\dfrac{630}{10}$

$\\ \sf\longmapsto m2=63kg$

Bruces mass is 63kg

5 0

3 years ago

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Jessie draws a ray diagram to show how an image of a candle is produced by a concave mirror when the candle is placed in front o

irina1246 [14]

There are two rules of reflection for the concave mirror:
1)<span>Any incident ray traveling parallel to the principal axis on the way to the mirror will pass through the focal point upon reflection.
2)</span><span>Any incident ray passing through the focal point on the way to the mirror will travel parallel to the principal axis upon reflection.
Also, keep in that for concave mirror center of curvature is 2f.
Using these two rules we can construct the image.
You should always use these two "special" rays, I will mark then 1 and 2 on the picture.
From the picture, we can see that if you place the object between the focus and vertex you get the virtual image.
The answer is: object should be between the focal point and the vertex</span>

8 0

3 years ago

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