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3 years ago

Please help for 21! will mark brainliest

Physics

1 answer:

Katyanochek1 [597]3 years ago

5 0

The answer for this question is D

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What is the percentage by weight of a solution prepared by dissolving 10 g of NaCl in 100 g of H2O?

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Explanation:

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3 years ago

guys pls help which one is it

Crazy boy [7]

Answer:

a mixture

since it had the power to separate as the question says

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Data that consists of numbers and measurements is called __________ data.

Vsevolod [243]

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3 years ago

Read 2 more answers

A 90 kg firefighter needs to climb the stairs of a 20-m-tall building while carrying a 40 kg backpack filled with gear. How much

Alexxx [7]

The power required will be equal to 463.3 watt

Power is defined as the rate of work

$P=\frac{W}{t}$

here P is power , W is work done , t is time taken

work done is defined as force times displacement

here s is displacement

here total force will be the weight of the firefighter plus the weight of gear

therefore F=(90+40)×9.8 N

so substituting value in equation of power we get

P= $\frac{25480}{55}$

p=463.3 watt

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brainly.com/question/12282215

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2 years ago

To calibrate the calorimeter electrically, a constant voltage of 3.6 V is applied and a current of 2.6 A flows for a period of 3

hodyreva [135]

Answer : The correct option is, (c) $3.7\times 10^2J/^oC$

Explanation :

First we have to calculate the energy or heat.

Formula used :

$E=V\times I\times t$

where,

E = energy (in joules)

V = voltage (in volt)

I = current (in ampere)

t = time (in seconds)

Now put all the given values in the above formula, we get:

$E=(3.6V)\times (2.6A)\times (350s)$

$E=3276J$

Now we have to calculate the heat capacity of the calorimeter.

Formula used :

$C=\frac{E}{\Delta T}=\frac{E}{T_{final}-T_{initial}}$

where,

C = heat capacity of the calorimeter

$T_{initial}$ = initial temperature = $20.3^oC$

$T_{final}$ = final temperature = $29.1^oC$

Now put all the given values in this formula, we get:

$C=\frac{3276J}{(29.1-20.3)^oC}$

$C=372.27J/^oC=3.7\times 10^2J/^oC$

Therefore, the heat capacity of the calorimeter is, $3.7\times 10^2J/^oC$

7 0

3 years ago