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maxonik [38]
3 years ago
15

2. Recycled tires are frequently turned into?

Engineering
1 answer:
Leya [2.2K]3 years ago
6 0

Answer:vmmjhnv

Explanation:

b m bn

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Which option identifies the type of power system Tommy will design in the following scenario?
Sedaia [141]

Answer:

diagram of an electrical curcuit

an sketch of an HVAC system

Also 3D image of a hydrualic piston

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7 0
3 years ago
2.) A fluid moves in a steady manner between two sections in a flow
Talja [164]

Answer:

250\ \text{lbm/min}

625\ \text{ft/min}

Explanation:

A_1 = Area of section 1 = 10\ \text{ft}^2

V_1 = Velocity of water at section 1 = 100 ft/min

v_1 = Specific volume at section 1 = 4\ \text{ft}^3/\text{lbm}

\rho = Density of fluid = 0.2\ \text{lb/ft}^3

A_2 = Area of section 2 = 2\ \text{ft}^2

Mass flow rate is given by

m=\rho A_1V_1=\dfrac{A_1V_1}{v_1}\\\Rightarrow m=\dfrac{10\times 100}{4}\\\Rightarrow m=250\ \text{lbm/min}

The mass flow rate through the pipe is 250\ \text{lbm/min}

As the mass flowing through the pipe is conserved we know that the mass flow rate at section 2 will be the same as section 1

m=\rho A_2V_2\\\Rightarrow V_2=\dfrac{m}{\rho A_2}\\\Rightarrow V_2=\dfrac{250}{0.2\times 2}\\\Rightarrow V_2=625\ \text{ft/min}

The speed at section 2 is 625\ \text{ft/min}.

3 0
3 years ago
Which of the following methods of obtaining digitized images can recognize and preserve text and formatting?
Zarrin [17]
The answer is A a digital camera
6 0
3 years ago
Calculate the kinetic energy in kJ of an automobile (3,000 lb) travelling at 60 miles per hour.
AlladinOne [14]

Answer:

K = 18.26 KJ

Explanation:

See it in the pic.

4 0
3 years ago
An electronic toy is powered by three 1.58-V alkaline cells, each with an internal resistance of 0.0205 Ω, and a 1.53-V carbon-z
mina [271]

Answer:

(a) The current in amperes that flows through the toy's circuit is 0.923A

(b) The power supplied to the toy is 5.78721W

(c) The internal resistance r2 of the failed dry cell is 72Ω

Explanation:

From the circuit diagram attached. We have the electric component:

B1 = 3* 1.58 = 4.74V

B2 = 1.53V

r1 = 3*0.0205 = 0.0615Ω

r2 = 0.105Ω

R = 6.625Ω

Since the internal resistances and the resistor R are connected in series, we can calculate the total resistance RT as

RT = r1 + r2 +R = 0.0615 + 0.105 + 6.625

= 6.7915Ω

Total Voltage supplied to the circuit by both batteries V = B1 + B2 = 4.74 + 1.53 = 6.27V

(a) CIRCUIT CURRENT

The current I, flowing through the circuit is  i =\frac{V}{R_{T}} = \frac{6.27}{6.7915} =0.923A

The current in amperes that flows through the toy's circuit is 0.923A

(b) THE POWER SUPPLIED TO THE TOY

Power P = I*V =0.923*6.27 = 5.78721W

The power supplied to the toy is 5.78721W

(c) THE VALUE OF r2

Due to dry cell failure, the power supplied to the toy is reduced to 0.5W

Now Power P = \frac{V^{2} }{R} . To calculate the new total resistance of the circuit we will make R the subject of the formular

R=\frac{V^{2} }{P} = \frac{6.27^{2} }{0.5} = \frac{39.3129}{0.5} = 78.6258Ω

Remember that RT = r1 + r2 + R

r1 =RT- (R +r2)

r1 = 78.6258 - 6.6865 =71.9393Ω

The internal resistance r2 of the failed dry cell is 72Ω

6 0
3 years ago
Read 2 more answers
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