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4 years ago

Which of the following best describes the dependant variable?

1 answer:

5 0

Hello!

The independent variable is the variable deliberately changed.

The dependent variable is the variable that responds to change. So the answer is A.

Hope this helps. Any questions please just ask! Thank you!

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A) A 5.75 mL sample of mercury has a measured mass of 77.05 g. The density is ___________.

Kitty [74]

Answer:

Density of the sample will be 13.4 kg/L

Explanation:

We have given volume of the sample $V=5.75mL=5.75\times 10^{-3}L$

Mass of the sample $M=77.05gram =77.05\times 10^{-3}kg$

We have to find the density of the sample

Density of the sample is given by

$Density=\frac{mass}{volume}=\frac{77.05\times 10^{-3}}{5.75\times 10^{-3}}=13.4kg/L$

So density of the sample will be 13.4 kg/L

4 0

4 years ago

A particle moves in a straight line so that time t seconds its displacement is x metres from a fixed point is given by x=3t^2 +2

Andru [333]

Answer:

-5 meters

Explanation:

x = 3t² + 2t − 5

The initial displacement is when t = 0.

x = 3(0)² + 2(0) − 5

x = -5

6 0

4 years ago

Neutrons incident on a heavy nucleus with spin J 0 show a resonance at an incident energy ER = 250 eV in the total cross-section

ivolga24 [154]

Answer:

elastic partial width is 2.49 eV

Explanation:

given data

ER E = 250 eV

spin J = 0

cross-section magnitude σ = 1300 barns

peak P = 20ev

to find out

elastic partial width W

solution

we know here that

σ = λ²× W / ( E × π × P ) ...................1

put here all value

σ = (0.286)² × W × $10^{-16}$ / ( 250 × π × 20 )

1300 × $10^{-24}$ = (0.286)² × W × $10^{-16}$ / ( 250 × π × 20 )

solve it and we get W

W = 249.56 × $10^{-2}$

so elastic partial width is 2.49 eV

8 0

4 years ago

There has long been an interest in using the vast quantities of thermal energy in the oceans to run heat engines. A heat engine

adell [148]

Answer:

Explanation:

The maximum efficient power plant will be the plant based on carnot cycle whose efficiency is given by the following formula

Efficiency = (T₁ - T₂) / T₁

T₁ is temperature of hot reservoir and T₂ is temperature of cold reservoir.

Putting the given values

efficiency of power plant = (35 - 5) / (273 + 35 )

= 30 / 308

= .097

= 9.7 %

3 0

4 years ago

In the equation F = Kq1 q2/r2 solve for q2. Solve for r.

Tpy6a [65]

Answer:

r = √(k q₁ q₂ / F)

Explanation:

F = k q₁ q₂ / r²

Multiply both sides by r²:

F r² = k q₁ q₂

Divide both sides by F:

r² = k q₁ q₂ / F

Take the square root of both sides:

r = √(k q₁ q₂ / F)

3 0

3 years ago