Water can form large dewdrops in nature how would droplets made of vegetable oil instead of water be different

Physics

2 answers:

Jlenok [28]1 year ago

6 0

Answer:

Explanation:

Natasha2012 [34]1 year ago

3 0

Answer:d

Explanation: oil would form droplets but only tiny ones because it’s surface tension is lower than that of water

You might be interested in

A 2 column table with 2 rows. The first column is labeled material with entries gold, iron. The second column is labeled specifi

topjm [15]

Iron atoms are heavier than gold atoms

Explanation:

Chart:

Materials Specific heat

Gold 0.129

Iron 0.45

The specific heat of a substance is the amount of heat required to raise the temperature of 1g of the substance by 1K.

We can see that the more the mass of a substance, the higher the specific heat needed to raise the temperature.

Specific heat is directly proportional to the mass of substance.

It also dependent on the amount of heat supplied and the nature of the substance.

The mass of a body is the amount of substance it contains.
A heavier substance has more number of atoms in it.
Since iron has a higher specific it, the atoms must be heavier or more than that of gold.

learn more:

Specific heat brainly.com/question/7210400

#learnwithBrainly

8 0

2 years ago

Read 2 more answers

Your physics textbook is sliding to the right across the table draw the vectors starting at the black dot. the location and orie

liberstina [14]

You would take a black dot at the top right of the y axis and dray it to the the far right of your x axis. hope this helps have a nice day and God bless

6 0

2 years ago

A capacitor is charged until it holds 5.0 j of energy. it is then connected across a 10-kω resistor. in 13.6 ms , the resistor d

prohojiy [21]

Answer:

The capacite is C=5.32 uF using the equations of voltage and energy in capacitance

Explanation:

The energy holds is 5 J and the resistor dissipates 2J so the energy total is 3J

Using:

$V_{t}= V_{o}e^{\frac{-t}{R*C} }$

Voltage in this case is the energy dissipated so

$E_{t}= E_{o}e^{\frac{-t}{R*C} }$

$\frac{\sqrt{E_t} }{\sqrt{E_o} } = e^{\frac{-t}{R*C} }$

$\frac{\sqrt{3 J} }{\sqrt{5J} } = e^{\frac{-13.6ms}{10kw*C} }$

Using the equation to find capacitance

$ln 0.775= e^{\frac{-13.6 x10^{3} }{10x10^{3}*C }} \\ln(0.775)= ln * e^{\frac{-13.6 x10^{3s} }{10x10^{3}*C }} \\\\ln(0.775)= {\frac{-13.6 x10^{3} }{10x10^{3}*C }} \\C= \frac{-13.6 x10^{-3} }{10x10x^{3}*ln(0.775) }$