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LUCKY_DIMON [66]
3 years ago
9

The volume of a gas varies directly with its temperature and inversely with the pressure. If the volume of a certain gas is 15 c

ubic feet at a temperature of 300 K and a pressure of 40 pounds per square inch, what is the volume of the same gas at 320 K when the pressure is 20 pounds per square inch?
Mathematics
1 answer:
vladimir2022 [97]3 years ago
4 0

Answer:

Therefore the volume of the gas is 32 cubic feet.

Step-by-step explanation:

Given that,

The volume of a gas varies directly with its temperature and inversely with the pressure.

V\propto \frac{T}{P}

V= Volume of the gas at constant mass

T= Temperature at constant mass

P= pressure at constant mass

\frac{V_1}{V_2}=\frac{\frac{T_1}{T_2}}{\frac{P_1}{P_2}}

\therefore \frac{V_1}{V_2}=\frac{T_1P_2}{T_2P_1}.

Given that,

At constant mass, the volume of a certain gas is 15 cubic feet and a pressure of 40 pound per square inch at a temperature 300 k.

V_1=15 cubic feet, P_1=40 pound per square inch and T_1= 300 K

V_2=?, P_2=20 pound per square inch and T_2= 320k

\therefore \frac{V_1}{V_2}=\frac{T_1P_2}{T_2P_1}.

Plugging all values

\frac{15}{V_2}=\frac{300\times 20}{320\times 40}

\Rightarrow V_2=\frac{15\times 320\times 40}{300\times 20}{}

\Rightarrow V_2=32 cubic feet

Therefore the volume of the gas is 32 cubic feet.

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A dealer gains 16% by selling a mop for rs.78300.but due to competition in the market, he decides to make a profit of only 10%.
vesna_86 [32]

Answer:

The correct answer is -  72,349.2

Step-by-step explanation:

Given:

selling price = 7234.92

Initial profit = 16%

we know:

selling price = cost price + profit

Solution:

78300 = cp + 16*cp/100

profit = 78300*16/100 = 12528

cost price = 78300 - 12528 = 65772

then the profit on 10%

= 65772*10/100

= 6577.2

the new selling price = 65772+6577.2

= 72,349.2

3 0
3 years ago
Which is the value of this expression when p=-2 and q=-1?
saul85 [17]

Answer:

D. 4

Step-by-step explanation:

[(p^2) (q^{-3}) ]^{-2}.[(p)^{-3}(q)^5] ^{-2}\\\\=[(p^2) (q^{-3}) \times(p)^{-3}(q)^5 ]^{-2}\\\\=[(p^{2}) \times(p)^{-3} \times (q^{-3}) \times(q)^5 ]^{-2}\\\\=[(p^{2-3}) \times (q^{5-3}) ]^{-2}\\\\=[(p^{-1}) \times (q^{2}) ]^{-2}\\\\=(p^{-1\times (-2)}) \times (q^{2\times (-2) }) \\\\=p^{2}\times q^{-4} \\\\= \frac{p^2}{q^4}\\\\= \frac{(-2)^2}{(-1)^4}\\\\= \frac{4}{1}\\\\= 4

8 0
3 years ago
Which is the value of x is in the domain of f(x)=x-8
mrs_skeptik [129]

Answer:

8

Step-by-step explanation:

Solve for x:

x - 8 = 0

Hint: | Isolate terms with x to the left hand side.

Add 8 to both sides:

x + (8 - 8) = 8

Hint: | Look for the difference of two identical terms.

8 - 8 = 0:

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8 0
3 years ago
Write the equation of the line of best fit using the slope-intercept formula $y = mx + b$. Show all your work, including the poi
Dennis_Churaev [7]

Answer:

y=\frac{5}{7}x+\frac{135}{7}

Step-by-step explanation:

You only need two points on a line to find the equation for that line.

We are going to use 2 points that cross that line or at least come close to. You don't have to use the green points... just any point on the line will work.  You might have to approximate a little.

I see ~(67.5,67.5) and ~(64,65).

Now once you have your points, we need to find the slope.

You may use \frac{y_2-y_1}{x_2-x_1} where (x_1,y_1) \text{ and } (x_2,y_2) are points on the line.

Or you can line up the points vertically and subtract then put 2nd difference over 1st difference.

Like this:

(  64  ,   65  )

-( 67.5, 67.5 )

--------------------

-3.5        -2.5

So the slope is -2.5/-3.5=2.5/3.5=25/35=5/7.

Now use point-slope form to find the equation:

y-y_1=m(x-x_1) where m is the slope and (x_1,y_1) is a point on the line.

y-65=\frac{5}{7}(x-64)

Distribute:

y-65=\frac{5}{7}x-\frac{5}{7}\cdot 64

Simplify:

y-65=\frac{5}{7}x-\frac{320}{7}

Add 65 on both sides:

y=\frac{5}{7}x-\frac{320}{7}+65

Simplify:

y=\frac{5}{7}x+\frac{135}{7}

6 0
3 years ago
Which choices are equivalent to the quotient below √16/√8
Darya [45]

Answer:

A

Step-by-step explanation:

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Hope this helps!

8 0
3 years ago
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