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MakcuM [25]
3 years ago
12

A 23.5 g piece of aluminum metal is initially at 100.0°C. It is dropped into a coffee cup-calorimeter containing 130.0 g of wate

r at a temperature of 23.0°C. After stirring, the final temperature of both copper and water is 26.0°C. Assuming no heat losses, and that the specific heat capacity of water is 4.184 J/(g·°C), what is the molar heat capacity of aluminum, Cm(Al)?
Physics
1 answer:
vivado [14]3 years ago
5 0

Answer: The molar heat capacity of aluminum is 25.3J/mol^0C

Explanation:

heat_{absorbed}=heat_{released}

As we know that,  

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]         .................(1)

where,

q = heat absorbed or released

m_1 = mass of water = 130.0 g

m_2 = mass of aluminiunm = 23.5 g

T_{final} = final temperature = 26.0^oC=(273+26)K=299K

T_1 = temperature of water = 23^oC=(273+23)K=296K

T_2 = temperature of aluminium = 100^oC=273+100=373K

c_1 = specific heat of water= 4.184J/g^0C

c_2 = specific heat of aluminium= ?

Now put all the given values in equation (1), we get

130.0\times 4.184\times (299-296)=-[23.5\times c_2\times (299-373)]

c_2=0.938J/g^0C

Molar mass of Aluminium = 27 g/mol

Thus molar heat capacity =0.938J/g^0C\times 27g/mol=25.3J/mol^0C

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Tamiku [17]

Answer:

the angle is about 67.79 degrees

Explanation:

We know that at its maximum height, the vertical component of the projectile's launching (initial) velocity (Vyi) is zero, so at that point it total velocity equals the horizontal component of the initial velocity (Vxi = 0.5 m/s)

We also know that the maximum height of the projectile is given by the square of its initial vertical component of the velocity (Vyi) divided by 2g, therefore half of such distance is :

half\,\,max-height = \frac{v_{yi}^2}{4\,g}

we can use this information to find the y component of the velocity at that height via the formula:

v_{yf}^2-v_{yi}^2=-2\,g\,\Delta y\\\\v_{yf}^2-v_{yi}^2=-2\,g\,(\frac{v_{yi}^2}{4\,g} )\\v_{yf}^2=v_{yi}^2-\frac{v_{yi}^2}{2} \\v_{yf}^2=\frac{v_{yi}^2}{2}

Now we use the information that tells us the speed of the projectile at this height to be 1 m/s. That should be the result of the vector addition of the vertical and horizontal components:

1=\sqrt{v_{yf}^2+0.5^2} \\1=\sqrt{\frac{v_{yi}^2}{2} +0.5^2}\\1^2=\frac{v_{yi}^2}{2} +0.5^2}\\1-0.5^2=\frac{v_{yi}^2}{2} \\2(1-0.5^2)=v_{yi}^2\\1.5 = v_{yi}^2\\v_{yi}=\sqrt{1.5} \\

Now we can use the arc-tangent to calculate the launching angle, since we know the two initial component of the velocity vector:

tan(\theta)=\frac{v_{yi}}{v_{xi}} =\frac{\sqrt{1.5} }{0.5} \\\theta= arctan(\frac{\sqrt{1.5} }{0.5})=67.79^o

3 0
3 years ago
You are standing on a moving bus, facing forward, and you suddenly fall forward. You can imply form this that the bus’s a. Veloc
UkoKoshka [18]

ANSWER:

a. Velocity decreased

STEP-BY-STEP EXPLANATION:

If the person inside the bus experiences a forward movement, this means that the bus is braking, since by action-reaction to compensate for the movement, the body moves forward.

When braking, it means that there is a decrease in speed, therefore, the correct answer is a. Velocity decreased

3 0
1 year ago
An aeroplane accelerates from rest to 65 m/s for take-off. It travels for
Virty [35]

Answer:

1.76m/s²

Explanation:

Given parameters:

Initial velocity  = 0m/s

Final velocity  = 65m/s

Distance traveled  = 1200m

Unknown:

Acceleration  = ?

Solution:

This is linear velocity and we apply the appropriate motion equation to solve this problem.

      V²  = U² + 2as

S is the distance

u is the initial velocity

V is the final velocity

a is the acceleration

Now, insert the parameters and solve;

             65²  =  0² + 2 x a x 1200

             4225 = 2400a

                  a  = 1.76m/s²

5 0
3 years ago
The state of strain at a point is plane strain with εx = ε0, εy = –2ε0, γxy = 0, where ε0 is a positive constant. What is the no
Marat540 [252]

Answer:

The normal strain along an axis oriented 45° from the positive x axis in the clockwise direction is -ε₀/2

Explanation:

Given that

\epsilon_{x}=\epsilon_{o}\\\\\epsilon_{y}=-2\epsilon_{o}\\\\\gamma_{xy}=0\\\\\theta=-45^{o}\\\\\epsilon_{x_{1}}=?

From equation of normal strain in x direction:

\epsilon_{x_{1}}=\epsilon_{x}cos^{2}\theta+\epsilon_{y}sin^{2}\theta+\gamma_{xy{ sin\theta cos\theta

Substituting the values:

\epsilon_{x_{1}}=\epsilon_{o}cos^{2}(-45)-2\epsilon_{o}sin^{2}(-45)+0\\\\\epsilon_{x_{1}}=\frac{\epsilon_{o}}{2}-2\frac{\epsilon_{o}}{2}\\\\\epsilon_{x_{1}}=-\frac{\epsilon_{o}}{2}

6 0
2 years ago
Constants A capacitor is connected across an ac source that has voltage amplitude 59 0 V and frequency 77 0 Hz Part C What is th
Vladimir79 [104]

Answer:

C = 1.77 \times 10^{-4} F

Explanation:

As we know that in AC circuit we have

V = i x_c

here we have

V = 59 V

i = 5.05 A

so we will have

x_c = \frac{59}{5.05}

x_c = 11.68 ohm

also we know that

x_c = \frac{1}{\omega C}

here we will have

11.68 = \frac{1}{(2\pi 77)C}

C = 1.77 \times 10^{-4} F

7 0
2 years ago
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