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Eddi Din [679]
2 years ago
4

In a heating system, cold outdoor air at 7°C flowing at a rate of 4 kg/min is mixed adiabatically with heated air at 66°C flowin

g at a rate of 5 kg/min. Determine the exit temperature of the mixture. Solve using appropriate software.
Physics
1 answer:
Sindrei [870]2 years ago
8 0

Answer:

The exit temperature of the mixture is 39.7°.

Explanation:

Given that,

Outdoor temperature = 7°C

Flowing rate = 4 kg/min

Temperature = 66°C

Flowing rate = 5 kg/min

We need to calculate the exit temperature of the mixture

Using  balance equation for the system

m_{1}T_{1}+m_{2}T_{2}=(m_{1}+m_{2})T_{3}

T_{3}=\dfrac{m_{1}T_{1}+m_{2}T_{2}}{(m_{1}+m_{2})}

Put the value into the formula

T_{3}=\dfrac{4\times 7+5\times 66}{4+5}

T_{3}=39.7^{\circ}

Hence, The exit temperature of the mixture is 39.7°.

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Answer:

1.) Magnitude = 5596 N

2.) Direction = 60 degrees

Explanation: You are given that the breakdown vehicle A is exerting a force of 4000 N at angle 45 degree to the vertical and breakdown vehicle B is exerting a force of 2000 N

Let us resolve the two forces into X and Y component

Sum of the forces in the X - component will be 4000 × cos 45 = 2828.43 N

Sum of the forces in the Y - component will be 2000 + ( 4000 × sin 45 )

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The resultant force R will be

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Substitutes the forces at X component and Y component into the formula

R = sqrt ( 2828.43^2 + 4828.43^2 )

R = sqrt ( 31313752.53 )

R = 5595.87 N

The direction will be

Tan Ø = Y/X

Substitute Y and X into the formula

Tan Ø = 4828.43 / 2828.43

Tan Ø = 1.707106

Ø = tan^-1( 1.707106 )

Ø = 59.64 degree

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8 0
2 years ago
In a certain clock, a pendulum of length L1 has a period T1 = 0.95s. The length of the pendulum
gulaghasi [49]

Answer:

Ratio of length will be \frac{L_2}{L_1}=1.108

Explanation:

We have given time period of the pendulum when length is L_1 is T_1=0.95sec

And when length is L_2 time period T_2=1sec

We know that time period is given by

T=2\pi \sqrt{\frac{L}{g}}

So 0.95=2\pi \sqrt{\frac{L_1}{g}}----eqn 1

And 1=2\pi \sqrt{\frac{L_2}{g}}-------eqn 2

Dividing eqn 2 by eqn 1

\frac{1}{0.95}=\sqrt{\frac{L_2}{L_1}}

Squaring both side

\frac{L_2}{L_1}=1.108

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2 years ago
A 12-volt battery causes 0.60 ampere to flow through a circuit that contains a lamp and a resistor connected in parallel. The la
san4es73 [151]
The answer to this question is A
5 0
1 year ago
Electric field lines always begin at _______ charges (or at infinity) and end at _______ charges (or at infinity). One could als
lesantik [10]

Answer:

b)

Explanation:

By convention, the electric field lines (which are tangent to the direction of the electric field at a given point) always begin at positive charges, and finish at negative charges.

This is a consequence of the convention that states that the electric field has the direction of the trajectory of a positive test charge when released from rest in an electric field.

(As the positive charge would move away from positive charges and would  be attracted by negative ones).

So, the combination of answers that is true is b) (positive, negative, positive).

3 0
2 years ago
A magnetic field of 0.080 T is in the y-direction. The velocity of wire segment S has a magnitude of 78 m/s and components of 18
Annette [7]

Answer:

Part a)

Induced EMF when length vector is along Z direction is 0.72 V

Part b)

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As we know that the motional EMF induced in the wire is given as

E = (v\times B). L

1)

As we know that

v = 18\hat i + 24\hat j + 72\hat k

B = 0.080 \hat j

L = 0.50 \hat k

now we have

\vec v \times \vec B = 1.44\hat k - 5.76 \hat i

so we have

E = 1.44 (0.50) = 0.72 V

2)

If the length vector is along Y direction then we have

L = 0.50 \hat j

so again we have

\vec v \times \vec B = 1.44\hat k - 5.76 \hat i

so we have

EMF = 0

6 0
2 years ago
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