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3 years ago

11

A wet shirt is put on a clothesline to dry on a sunny day. Do water molecules lose heat and condense, gain heat and condense or

gain heat and evaporate

1 answer:

Strike441 [17]3 years ago

8 0

Gain heat energy and evaporate

You might be interested in

the thermometer has its stem marked in millimeter instead of degree Celsius. the lower fixed point is 30mm and the upper fixed p

Lorico [155]

10°c

Explanation:

Given parameter;

Lower fixed point = 30mm

Upper fixed point = 180mm

Reading = 45mm

Unknown:

The degree celcuis temperature at 45mm = ?

Solution:

To solve this problem we simply compare the mm- scale to the celcius - scale that we know.

The upper fixed point is the boiling point of water

Lower fixed point is the freezing point of water

This shows that both the upper and lower fixed point of both thermometers are the same;

mm-scale °c scale

180mm 100°c

45mm x

30mm 0°c

Solving;

$\frac{x - 0}{100 -0} = \frac{45 - 30 }{180- 30}$

x (150) = 100 x 15

x = 10°c

learn more:

Temperature scales brainly.com/question/1603430

#learnwithBrainly

7 0

3 years ago

I have a doubt on heat and temperature. I need help with this science question and here the question: The water in a lake just r

Volgvan

Answer:

you can't go ice skating on it because if it just reached the temp then you need to wait for about 2 hours

Explanation:

5 0

3 years ago

A certain atom has atomic number Z = 25 and atomic mass number A = 52. a. What is the approximate radius of the nucleus of this

wlad13 [49]

Answer:

a)The approximate radius of the nucleus of this atom is 4.656 fermi.

b) The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527

Explanation:

$r=r_o\times A^{\frac{1}{3}}$

$r_o=1.25 \times 10^{-15} m$ = Constant for all nuclei

r = Radius of the nucleus

A = Number of nucleons

a) Given atomic number of an element = 25

Atomic mass or nucleon number = 52

$r=1.25 \times 10^{-15} m\times (52)^{\frac{1}{3}}$

$r=4.6656\times 10^{-15} m=4.6656 fm$

The approximate radius of the nucleus of this atom is 4.656 fermi.

b) $F=k\times \frac{q_1q_2}{a^2}$

k= $9\times 10^9 N m^2/C^2$ = Coulombs constant

$q_1,q_2$ = charges kept at distance 'a' from each other

F = electrostatic force between charges

$q_1=+1.602\times 10^{-19} C$

$q_2=+1.602\times 10^{-19} C$

Force of repulsion between two protons on opposite sides of the diameter

$a=2\times r=2\times 4.6656\times 10^{-15} m=9.3312\times 10^{-15} m$

$F=9\times 10^9 N m^2/C^2\times \frac{(+1.602\times 10^{-19} C)\times (+1.602\times 10^{-19} C)}{(9.3312\times 10^{-15} m)^2}$

$F=2.6527 N$

The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527

6 0

3 years ago

single goose sounds a loud warning when an intruder enters the farmyard. Some distance from the goose, you measure the sound lev

Goryan [66]

Answer:

The sound level of the 26 geese is $Z_{26}= 96.15 dB$

Explanation:

From the question we are told that

The sound level is $Z_1 = 81.0 \ dB$

The number of geese is $N = 26$

Generally the intensity level of sound is mathematically represented as

The intensity of sound level in dB for one goose is mathematically represented as

$Z_1 = 10 log [\frac{I}{I_O} ]$

Where I_o is the threshold level of intensity with value $I_o = 1*10^{-12} \ W/m^2$

$I$ is the intensity for one goose in $W/m^2$

For 26 geese the intensity would be

$I_{26} = 26 * I$

Then the intensity of 26 geese in dB is

$Z_{26} = 10 log[\frac{26 I }{I_o} ]$

$Z_{26} = 10 log (\ \ 26 * [\frac{ I }{I_o} ]\ \ )$

$Z_{26} = 10 log (\ \ 26 \ \ ) * (\ \ 10 log [\frac{ I }{I_o} ]\ \ )$

From the law of logarithm we have that

$Z_{26} = 10 log 26 + 10 log [\frac{I}{I_0} ]$

$= 14.15 + 82$

$Z_{26}= 96.15 dB$

4 0

3 years ago

What is the smallest particle that can completely represent water?

julsineya [31]

An atom would be your answer, so B!

4 0

3 years ago

Read 2 more answers