Ask question

Ask question

All categories

2 years ago

11

In this experiment, the ______ was intentionally manipulated. This was the independent variable.

1 answer:

Marizza181 [45]2 years ago

3 0

Answer:

Fan speed

Explanation:

because is the dependent variable

You might be interested in

What do you mean by shearing stress??

lord [1]

Answer:

Shear stress, force tending to cause deformation of a material by slippage along a plane or planes parallel to the imposed stress. The resultant shear is of great importance in nature, being intimately related to the downslope movement of earth materials and to earthquakes.

8 0

3 years ago

Read 2 more answers

Is this statement true or false? Electric currents flow easily through materials that are conductors and through materials that

Alex777 [14]

False because currents do not flow easily through insulators. If it only said conductors, then it would be true.

3 0

3 years ago

Read 2 more answers

A gas at a pressure of 2.10 atm undergoes a quasi static isobaric expansion from 3.70 to 5.40 L. How much work is done by the ga

BartSMP [9]

Answer:

Total work done in expansion will be $3.60\times 10^5J$

Explanation:

We have given pressure P = 2.10 atm

We know that 1 atm $=1.01\times 10^5Pa$

So 2.10 atm $=2.10\times 1.01\times 10^5=2.121\times 10^5Pa$

Volume is increases from 3370 liter to 5.40 liter

So initial volume $V_1=3.70liter$

And final volume $V_2=5.40liter$

So change in volume $dV=5.40-3.70=1.70liter$

For isobaric process work done is equal to $W=PdV=2.121\times 10^5\times 1.70=3.60\times 10^5J$

So total work done in expansion will be $3.60\times 10^5J$

5 0

3 years ago

A long thin rod of mass M and length L is situated along the y axis with one end at the origin. A small spherical mass m1 is pla

Yuri [45]

Answer:

$= - 5.65\times 10^{-2} J$

Explanation:

Given data:

L =2.00 *10^4 m

d = 18*10^4 m

M = 18 *10^6 kg

m_1 = 8*10^6 kg

Gravitational energy is given as

$U =- G \frac{m_1 m_2}{r}$

mass per unit length is given as

$\sigma = \frac{M}{L} = \frac{18 \times 10^6}{2\times 10^4 m} = 900 kg/m$

calculating potential energy

$dU ==-G\int_{16\times 10^4}^{18\times 10^4} \frac{m_1 *dm}{r}$

$=-G\int_{16\times 10^4}^{18\times 10^4} \frac{m_1 *\sigma dr}{r}$

$=-G*m_1*\sigma\int_{16\times 10^4}^{18\times 10^4} \frac{dr}{r}$

$=-G*m_1*\sigma \left | ln r \right |_{16\times 10^4}^{18\times 10^4}$

$=-G*m_1*\sigma ln(1.125)$

$=-6.673 \times 10^{-11}*8*10^6*900*ln(1.125)$

$= - 5.65\times 10^{-2} J$

5 0

3 years ago

A researcher finds that blood alcohol levels cause Progressive damage the liver all of mice and his study were fed the same amou

Juli2301 [7.4K]

Answer:hmm

Explanation:

4 0

2 years ago