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Tatiana [17]
3 years ago
9

Hello please help me out

Physics
2 answers:
agasfer [191]3 years ago
8 0
The answer is c.........
AveGali [126]3 years ago
3 0
The Answer Is C:Cytosine Nucleotides
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Why does diamond sparkles stars twinkles?​
Helga [31]

Answer:

Because they want attention

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143°C = _____<br><br> 416 K<br> -130 K<br> 0 K<br> 143 K
dezoksy [38]
The answer is 0k because 143c equals nothing
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1 year ago
Read 2 more answers
A person who weighs 509,45 N empties her lungs as much as
Masja [62]

Answer:

The weight of the girl = 1045.86 kg/m³

Explanation:

Density: This can be defined as the ratio of the mass of a body to the volume of that body. The S.I unit of density is kg/m³.

From Archimedes principle,

R.d = Density of the person/Density of water = Weight of the person in air/Upthrust.

⇒ D₁/D₂ = W/U............................... Equation 1.

Where D₁ = Density of the person, D₂ = Density of water, W = Weight of the person in air, U = Upthrust in water.

Making D₁ the subject of the equation,

D₁ = D₂(W/U)................................... Equation 2

<em>Given: D₂ = 1000 kg/m³ , W = 509.45 N, U = lost in weight = weight in air - weight in water = 509.45 - 22.34 = 487.11 N</em>

<em>Substituting these values into equation 2</em>

D₁ = 1000(509.45/487.11)

D₁ = 1045.86 kg/m³

Thus the weight of the girl = 1045.86 kg/m³

<em></em>

7 0
3 years ago
Pls heplp 70 points!!!!!
Rasek [7]

Answer

the answer is d for sure

5 0
3 years ago
8.) If a car moving at 50km/h skids 15m with locked brakes, how far does the same car moving at 100km/h
pantera1 [17]

(8) A car starting with a speed <em>v</em> skids to a stop over a distance <em>d</em>, which means the brakes apply an acceleration <em>a</em> such that

0² - <em>v</em>² = 2 <em>a</em> <em>d</em> → <em>a</em> = - <em>v</em>² / (2<em>d</em>)

Then the car comes to rest over a distance of

<em>d</em> = - <em>v</em>² / (2<em>a</em>)

Doubling the starting speed gives

- (2<em>v</em>)² / (2<em>a</em>) = - 4<em>v</em>² / (2<em>a</em>) = 4<em>d</em>

so the distance traveled is quadrupled, and it would move a distance of 4 • 15 m = 60 m.

Alternatively, you can explicitly solve for the acceleration, then for the distance:

A car starting at 50 km/h ≈ 13.9 m/s skids to a stop in 15 m, so locked brakes apply an acceleration <em>a</em> such that

0² - (13.9 m/s)² = 2 <em>a</em> (15 m) → <em>a</em> ≈ -6.43 m/s²

So the same car starting at 100 km/h ≈ 27.8 m/s skids to stop over a distance <em>d</em> such that

0² - (27.8 m/s)² = 2 (-6.43 m/s²) <em>d</em> → <em>d</em> ≈ 60 m

(9) Pushing the lever down 1.2 m with a force of 50 N amounts to doing (1.2 m) (50 N) = 60 J of work. So the load on the other end receives 60 J of potential energy. If the acceleration due to gravity is taken to be approximately 10 m/s², then the load has a mass <em>m</em> such that

60 J = <em>m g h</em>

where <em>g</em> = 10 m/s² and <em>h</em> is the height it is lifted, 1.2 m. Solving for <em>m</em> gives

<em>m</em> = (60 J) / ((10 m/s²) (1.2 m)) = 5 kg

(10) Is this also multiple choice? I'm not completely sure, but something about the weight of the tractor seems excessive. It would help to see what the options might be.

4 0
3 years ago
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