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2 years ago

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We now have an algebraic expression with only one variable, which can be solved. Once we have that, we can plug it back into one

of the original equations (or the expression derived in Part A) to solve for the other variable. When this is done with the system of two equations from Parts A and B, what is the solution?

1 answer:

Len [333]2 years ago

3 0

Answer:

A=1

B=-2

Explanation:

Part A and B of the question wasn't given, however, I attached the relevant parts to solve this question as follows.

From part B as attached, it shows that the right option is C which is

2A+3B=-4

Substituting B with 3A-5 then we form the second equation as shown

2A+3(3A-5)=-4

By simplifying the above equation, we obtain

2A+9A-15=-4

Re-arranging, then

11A=-4+15

Finally

11A=11

A=1

To obtain B, we already know that 3A-5 so substituting the value of A into the above then we obtain

B=3(1)-5=-2

Therefore, required values are 1 and -2

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Answer : The correct option is, (d) $535^oC$

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 copper = $0.10cal/g^oC$

$c_2$ = specific heat of water = $1.00cal/g^oC$

$m_1$ = mass of copper = 120 g

$m_2$ = mass of water = 300 g

$T_f$ = final temperature of mixture = $35^oC$

$T_1$ = initial temperature of copper = ?

$T_2$ = initial temperature of water = $15^oC$

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

$120g\times 0.10cal/g^oC\times (35-T_1)^oC=-300g\times 1.00cal/g^oC\times (35-15)^oC$

$T_1=535^oC$

Therefore, the temperature of the kiln was, $535^oC$

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2 years ago

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the millersburg ferry (m=13000.0 kg loaded) puts its engines in full reverse and stops in 65 seconds. if the speed before brakin

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Explanation:

We can solve this problem by using the impulse theorem, which states that the impulse applied on the ferry (the product of force and time) is equal to its change in momentum:

$F \Delta t = m(v-u)$

where in this problem, we have:

F is the force applied by the brakes

$\Delta t = 65 s$ is the time interval

m = 13,000 kg is the mass of the ferry

u = 2.0 m/s is the initial velocity

v = 0 is the final velocity

And solving for F, we find the force applied by the brakes:

$F=\frac{m(v-u)}{\Delta t}=\frac{(13000)(0-2.0)}{65}=-400 N$

where the negative sign indicates that the direction is backward.

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2 years ago

On a ring road, 12 trams are spaced at regular intervals and travel at a constant speed. How many trams need to be added to the

Sphinxa [80]

3 trams must be added

Explanation:

In this problem, there are 12 trams along the ring road, spaced at regular intervals.

Calling L the length of the ring road, this means that the space between two consecutive trams is

$d=\frac{L}{12}$ (1)

In this problem, we want to add n trams such that the interval between the trams will decrease by 1/5; therefore the distance will become

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And the number of trams will become

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2 years ago

A rigid, 2.50 L bottle contains 0.458 mol He. The pressure of the gas inside the bottle is 1.83 atm. If 0.713 mol Ar is added to

stellarik [79]

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You can use PV=nRT again after Ar is added to solve for T, which will give you the final temperature. The difference between the initial and final temperatures is the change. When you're solving just be careful with the units!

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PiVi=niRTi --> Ti=(PiVi)/(niR)

PfVf=nfRTf --> Tf=(PfVf)/(nfR)

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To solve this problem it is necessary to apply the concepts related to the heat flux rate expressed in energetic terms. The rate of heat flow is the amount of heat that is transferred per unit of time in some material. Mathematically it can be expressed as:

$\frac{Q}{t} = \frac{kA}{L} (T_H - T_C)$

Where

k = 0.84 J/s⋅m⋅°C (The thermal conductivity of the material)

$A = 1m^2$ Area

$L = 5*10^{-3}m$ Length

$T_H$ = Temperature of the "hot"reservoir

$T_C$ = Temperature of the "cold"reservoir

Replacing with our values we have that,

$\frac{Q}{t} = \frac{kA}{L} (T_H - T_C)$

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