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4 years ago

7

Can someone plz help me Thank you in advance:)

1 answer:

yulyashka [42]4 years ago

4 0

Answer : The specific heat of the metal is, $2178.67J/kg^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of metal = ?

$c_2$ = specific heat of ice = $2000J/kg^oC$

$m_1$ = mass of metal = 1.00 kg

$m_2$ = mass of ice = 1.00 kg

$T_f$ = final temperature of mixture = $-8.88^oC$

$T_1$ = initial temperature of metal = $5.00^oC$

$T_2$ = initial temperature of ice = $-24.0^oC$

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

$(1.00kg)\times c_1\times (-8.88-5.00)^oC=-[(1.00kg)\times 2000J/kg^oC\times (-8.88-(-24.0))^oC]$

$c_1=2178.67J/kg^oC$

Therefore, the specific heat of the metal is, $2178.67J/kg^oC$

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Which of the following could be points on the unit circle: A. (4/3, 4/5) B. (5/13, 12/13)

IRINA_888 [86]

Answer:

Option (2) and (4) are correct.

$(\frac{5}{13},\frac{12}{13} )$ and $(\frac{6}{7},\frac{\sqrt{13}}{7} )$ are points on the unit circle.

Step-by-step explanation:

Given : Some points of circles.

We have to choose which points could be points on the unit circle.

We know, the equation of circle is

$x^2+y^2=r^2$ ..........(1)

where r is radius of circle.

We check each point for x and y values on the equation of circle and see which point gives radius = 1

Thus,

1) $(\frac{4}{3},\frac{4}{5} )$

Put in LHS of (1) , we have,

$(\frac{4}{3})^2+(\frac{4}{5})^2$

Simplify, we have,

$=\frac{16}{9}+\frac{16}{25}$

$=\frac{544}{225}\neq 1$

Thus, $(\frac{4}{3},\frac{4}{5} )$ is not a point on the unit circle.

2) $(\frac{5}{13},\frac{12}{13} )$

Put in LHS of (1) , we have,

$(\frac{5}{13})^2+(\frac{12}{13})^2$

Simplify, we have,

$=\frac{25}{169}+\frac{144}{169}$

$=\frac{169}{169}= 1$

Thus, $(\frac{5}{13},\frac{12}{13} )$ is a point on the unit circle.

3) $(\frac{1}{3},\frac{2}{3} )$

Put in LHS of (1) , we have,

$(\frac{1}{3})^2+(\frac{2}{3})^2$

Simplify, we have,

$=\frac{1}{9}+\frac{4}{9}$

$=\frac{5}{9}\neq 1$

Thus, $(\frac{1}{3},\frac{2}{3} )$ is not a point on the unit circle.

4) $(\frac{6}{7},\frac{\sqrt{13}}{7} )$

Put in LHS of (1) , we have,

$(\frac{6}{7})^2+(\frac{\sqrt{13}}{7})^2$

Simplify, we have,

$=\frac{36}{49}+\frac{13}{49}$

$=\frac{49}{49}= 1$

Thus, $(\frac{6}{7},\frac{\sqrt{13}}{7} )$ is a point on the unit circle.

Thus, $(\frac{5}{13},\frac{12}{13} )$ and $(\frac{6}{7},\frac{\sqrt{13}}{7} )$ are points on the unit circle.

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The probability that he winds up wearing the white shirt and tan pants is then calculated as:

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