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Mariana [72]
3 years ago
8

A pipe with cross-sectional area 2.0 m^2 is joined to a second pipe with cross-sectional area 0.5 m^2. The pipes are both comple

tely filled with water. The water in the 2.0 m^2 pipe is flowing into the smaller pipe with a speed of 2 m/s. How fast is the water flowing in the second pipe?
a.) 8 m/s

b.) 0 m/s

c.) 4 m/s

d.) 2 m/s
Physics
2 answers:
docker41 [41]3 years ago
5 0

Answer:

a.) 8 m/s

Explanation:

Given that

A₁= 2 m²

V₁= 2 m/s

A₂=0.5 m²

Lets take speed in 0.5m² is V₂

As given that both pipes are connected in the series that is volume flow rate will be same

Q= A₁V₁ = A₂V₂

Now by putting the values

A₁V₁ = A₂V₂

2 x 2 = 0.5 x V₂

V₂= 8 m/s

a.) 8 m/s

IRISSAK [1]3 years ago
3 0

Answer:

v_2=8\ m/s

Explanation:

It is given that,

Area of cross section of the pipe, A_1=2\ m^2

Area of cross section of the another pipe, A_2=0.5\ m^2

Speed of water in first pipe, v_1=2\ m/s

To find,

Speed of the water flowing in the second pipe.

Solve,

Let v_2 is the speed of water flowing in the second pipe. The relation between the area of cross section and the velocity is given by the continuity equation. It is given by :

A_1v_1=A_2v_2

v_2=\dfrac{A_1v_1}{A_2}

v_2=\dfrac{2\times 2}{0.5}

v_2=8\ m/s

Therefore, the water is flowing in the second pipe at the rate of 8 m/s.

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Answer:

The final temperature of the element = 262.67°C

The power dissipated in the heating element initially = 163.21 W

The power dissipated in the heating element when the current reaches 1.23 A = 147.60 W

Explanation:

Our given parameters include;

A Nichrome heating element operates on 120 V.

Voltage (V) = 120V

Initial Current (I₁) = 1.36 A

Initial Temperature (T₁) = 28°C

Final Current (I₂) = 1.23 A

Final Temperature (T₂) = unknown ????

Temperature dependencies of resistance is given by:

R_{T(2)}=R_1[1+\alpha (T_2-T_1)]            ----------------------    (1)

in which R₁ is the resistance at temperature T₁

R_{T(2) is the resistance at temperature T₂

Given that V= IR

R = \frac{V}{I}

Therefore, the resistance at temperature 28°C is;

R_{28}= \frac{120V}{1.36A}

= 88.24Ω

R_{T(2) = \frac{120V}{1.23A}

= 97.56Ω

From (1) above;

R_{T(2)}=R_1[1+\alpha (T_2-T_1)]      

97.56 = 88.24 [ 1 + 4.5×10⁻⁴(°C)⁻¹(T₂-28°C)]

\frac{97.56}{88.24}= 1+(4.5*10^{-4})(T-28^0C)

1.1056 - 1 = 4.5×10⁻⁴(°C)⁻¹(T₂-28°C)

0.1056 = 4.5×10⁻⁴(T₂-28°C)

\frac{0.1056}{4.5*10^{-4}}= T-28^0C

T - 28° C = 234.67

T = 234.67 + 28° C

T = 262.67 ° C

(b)

What is the power dissipated in the heating element initially and when the current reaches 1.23 A

The power dissipated in the heating element initially can be calculated as:

P = I²₁R₂₈

P = (1.36A)²(88.24Ω)

P = 163.209 W

P ≅ 163.21 W

The power dissipated in the heating element when the current reaches 1.23 A can be calculated as:

P= I^2_2R_{T^0C

P = (1.23)²(97.56Ω)

P = 147.598524

P ≅ 147.60 W

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When the person is in the elevator, the scale says 77 kg. The scale is still using the same value of conversion (9.81), so the apparent weight "felt" by the scale is
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ΔT is the change in temperature.

So the law of conservation of heat tells that:

Sensible heat of Z + Sensible heat of container = Sensible heat of X

Since we have no idea what these substances are, there is no way of knowing the Cp. We can't proceed with the calculations. So, we can only assume that in the duration of 15 minutes, the whole system achieves equilibrium. Therefore, the equilibrium temperature of the system is equal to 32°C. The answer is C.
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