Isa ka ba sa mga kabataang nagpapahalaga at nag-iingat ng kasariang handog ng Diyos Paano ninyo ito pinahahalagahan? Sagutin ang mga tanong ng hindi bababa sa limang pangungusap.​

Answers 1

Answer:

gumawa ng nakakabuti sa iyong kapwa at sa sarili at paggalang sa nakapaligid sayo

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match the skin problems in column a with its descriptions in column b

Answer:

saan po Yung column a at column b

A 6.22-Kg piece of copper metal (s=0.385J/(g°C) is heated from 20.5°C to 324.3°C. Calculate the heat absorbed (in KJ) by the metal.
Specific Heat CapacityFormula: q = mc∆T

where in, q = heat energy

m = mass

c = specific heat capacity

∆T = change in temperature

= Tf - Ti

Given: m = 6220g Cu

c = 0.385J/g°C

Ti = 20.5°C

Tf = 324.3°C

Note: Ti = initial temperature

Tf = final temperature

Note: The unit of specific heat is in grams(g), and the mass given is in kilograms(kg), therefore the mass given must be converted first into grams(g).

kg → g

6.22kg Cu × 1000g/1kg = 6220g Cu

Unknown: q = ?

Formula: q = mc∆T

Solution:

q = mc∆T

= (6220g Cu)(0.385J/g°C)

(324.3°C - 20.5°C)

= (6220g Cu)(0.385J/g°C)(303.8°C)

= 727509.86J

Note: Units of other term are cancelled out, so the unit for heat energy will remain which is Joules(J).

Note: The problem asked for heat energy to be in kilojoules(kj)

J → kj

727509.86J × 1kj / 1000J = 727.51kj

Final Answer: q = 727.51kj

Tip: Cancelation of units is applicable only if the other same unit is on the denominator.

ฅ^•ﻌ•^ฅ

please help po sa math

sge po anong klaseng math ba?

Gumawa ng tulaplssss need ko na ​
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Answer:

kapag magtanim may aanihin

I know it is short. But I hope you can get inspiration here, I'm sorry! I'm not good in Filipino nowadays eh.

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