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algol [13]
2 years ago
10

How much heat is required to raise the temperature of

Mathematics
1 answer:
Elina [12.6K]2 years ago
4 0

Answer:

The amount of heat required to raise the temperature of liquid water is 9605 kilo joule .

Step-by-step explanation:

Given as :

The mass of liquid water = 50 g

The initial temperature = T_1 = 15°c

The final temperature = T_2  = 100°c

The latent heat of vaporization of water = 2260.0 J/g

Let The amount of heat required to raise temperature = Q Joule

Now, From method

Heat = mass × latent heat × change in temperature

Or, Q = m × s × ΔT

or, Q =  m × s × ( T_2 - T_1 )

So, Q = 50 g × 2260.0 J/g × ( 100°c - 15°c )

Or, Q =  50 g × 2260.0 J/g × 85°c

∴   Q = 9,605,000  joule

Or, Q =  9,605 × 10³ joule

Or, Q = 9605 kilo joule

Hence The amount of heat required to raise the temperature of liquid water is 9605 kilo joule . Answer

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Your friend is trying to grow her hair as long as her cousin’s hair. The table shows their hair lengths y (in inches) in differe
Effectus [21]
<span>Given the table that shows the hair lengths y (in inches) of your friend and her cousin in different months x.

Month      Friends Hair(in)      Cousins Hair(in)
    3                     4                              7
    8                    6.5                            9.

To solve for the cousins hair, recall that the equation of a line is given by

y = mx + c

From the table,

7 = 3m + c . . . (1)
9 = 8m + c . . . (2)

(1) - (2) ⇒ -2 = -5m
\Rightarrow m= \frac{2}{5} =0.4

Substituting for m into equation (1) gives:

7=3(0.4)+c \\  \\ \Rightarrow c=7-1.2=5.8

Therefore, the equation representing the growth of the cousin's hair is given by y = 1.2x + 5.8
</span>
7 0
2 years ago
Sasha has 3.20 in U.S. coins. She has the same number of quarters and nickels. What is the greatest number of quarters she could
RSB [31]

Answer:

10 quarters = $2.50

10 nickels = $0.50

that leaves $0.20 for other coins (dimes / pennies)

Step-by-step explanation:

First, suppose she has only quarters and nickels and no other coins.  Then if C is the identical number of coins of each type, then 5C + 25C = 320, so 30C = 320 and 3C = 32, but there is no integer solution to this.  So she must have at least one other type of coin.

Assume she has only quarters, nickels, and dimes.  Then if D is the number of dimes, 5C +  25C + 10D = 320, which means 30C + 10D = 320, or 3C + D = 32.  The smallest D can be is 2, leaving 3C = 30 and thus C = 10.  So in this scenario she would have 10 quarters, 10 nickels, and two dimes to make $2.50 + $0.50 + $0.20 = $3.20.

This has to be the highest number, because if she had 11 quarters and 11 nickels, that alone would add up to 11(0.25) + 11(0.05) = $3.30, which would already be too much.

8 0
2 years ago
Read 2 more answers
If the ratio of the sides of two squares is 3:1, what is the ratio of their perimeters?
Anvisha [2.4K]

Answer:

Ratio of the perimeters =3:1

Step-by-step explanation:

We have given that : Ratio of the sides of two squares is 3:1

To find : Ratio of their perimeters

Solution : Let the length of the sides are 3:1 = 3x : x

                  Formula of perimeter of square = 4(side)

Using the formula ,

Perimeter of 1 square = 4×3x= 12x

Perimeter of 2 square = 4×x= 4x

Ratio of the perimeter of 1 square and 2 square = 12x : 4x

                                                                                 = 3 : 1


8 0
2 years ago
How do i write 12.009 in a word form?
tia_tia [17]
Your answer will be twelve and nine thousandths because in place value decimals are like tenths, hundredths, thousandths. In place value for regular numbers are ones, tens, hundreds and so on. So, when you have 12.009, you are going to separate the decimals with the numbers. Then, write down the number in words which would be twelve. Now, go to the decimal and write that down, which is 9 thousandths. Finally, combine the both of them to get: twelve and nine thousandths.

Hope this helps :) 
and good luck

5 0
2 years ago
Roberto wants to display his 18 sports cards in an album. Some pages hold 2 cards and others hold 3 cards. How many different wa
WITCHER [35]
<span>65 = number of different arrangements of 2 and 3 card pages such that the total number of card slots equals 18. 416,154,290,872,320,000 = number of different ways of arranging 18 cards on the above 65 different arrangements of page sizes. ===== This is a rather badly worded question in that some assumptions aren't mentioned. The assumptions being: 1. The card's are not interchangeable. So number of possible permutations of the 18 cards is 18!. 2. That all of the pages must be filled. Since the least common multiple of 2 and 3 is 6, that means that 2 pages of 3 cards can only be interchanged with 3 pages of 2 cards. So with that said, we have the following configurations. 6x3 card pages. Only 1 possible configuration. 4x3 cards and 3x2 cards. These pages can be arranged in 7!/4!3! = 35 different ways. 2x3 cards and 6x2 cards. These pages can be arranged in 8!/2!6! = 28 ways 9x2 card pages. These can only be arranged in 1 way. So the total number of possible pages and the orders in which that they can be arranged is 1+35+28+1 = 65 possible combinations. Now for each of those 65 possible ways of placing 2 and 3 card pages such that the total number of card spaces is 18 has to be multiplied by the number of possible ways to arrange 18 cards which is 18! = 6402373705728000. So the total amount of arranging those cards is 6402373705728000 * 65 = 416,154,290,872,320,000</span>
6 0
2 years ago
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