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Lubov Fominskaja [6]
3 years ago
7

__________ is an accurate way of drawing that shows an object's true size and shape.

Engineering
1 answer:
Bingel [31]3 years ago
6 0
ANSWER:

Detail drawing
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Technician A says that a fully charged battery is less likely to freeze than a discharged battery. Technician B says that the st
zheka24 [161]
Answer : Technician B is correct
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3 years ago
Water enters a centrifugal pump axially at atmospheric pressure at a rate of 0.12 m3
goldenfox [79]

Answer:

Water enters a centrifugal pump axially at atmospheric pressure at a rate of 0.12 m3/s and at a velocity of 7 m/s, and leaves in the normal direction along the pump casing, as shown in Fig. PI3-39. Determine the force acting on the shaft (which is also the force acting on the bearing of the shaft) in the axial direction.

Step-by-step solution:

Step 1 of 5

Given data:-

The velocity of water is .

The water flow rate is.

3 0
3 years ago
A discrete MOSFET common-source amplifier has RG = 2 MΩ, gm = 5 mA/V, ro = 100 kΩ, RD = 20kΩ, Cgs = 3pF, and Cgd = 0.5pF. The am
Papessa [141]

Answer:

a) -36.36 V/V

b) 15.17 kHz

c) 1.6 GHz

Explanation:

See attached picture.

7 0
3 years ago
A car is traveling at sea level at 78 mi/h on a 4% upgrade before the driver sees a fallen tree in the roadway 150 feet away. Th
Dmitrij [34]

Answer: V = 47.7 mi/hr

Explanation:

first we calculate elements of aero-dynamic resistance

Ka = p/2 * CD * A.f

p is the density of air(0.002378 slugs/ft^3) for zero altitude, CD is the drag coefficient(0.35) and A.f is the front region of the vehicle

so we substitute

Ka = 0.002378/2 * 0.35 * 18

Ka = 0.00749

Now we calculate the final speed of the vehicle (V2) using the relation;

S = (YbW/2gKa)In[ (UW + KaV1^2 + FriW ± Wsinθg) / (UW + KaV2^2 + FriW ± Wsinθg)

so

WE SUBSTITUTE

150 = (1.04 * 2700 / 2 * 32.2 * 0.0075) In [(0.8 * 2700 + 0.0075 *(78mil/hr * 5280ft/1min * 1hr/3600s)^2 + 0.017 * 2700 ± 2700 * 0.04) / (0.8 * 2700 + 0.0075 * V2^2 + 0.017 * 2700 ± 2700 * 0.04)]

150 = (2808/0.483) In [(2160 + 98.16 + 153.9) / ( 2160 + 0.0075V2^2 + 153.9)]

150 = 5813.66 In [ (2160 + 98.16 + 153.9) / ( 2160 + 0.0075V2^2 + 153.9)]

divide both sides by 5813.66

0.0258 = In [ (2412.06) / ( 0.0075V2^2 + 2313.9)]

take the e^ of both side

e^0.0258 = (2412.06) / ( 0.0075V2^2 + 2313.9)

1.0261 = (2412.06) / ( 0.0075V2^2 + 2313.9)]

(0.0075V2^2 + 2313.9) = 2412.06 / 1.0261

(0.0075V2^2 + 2313.9) = 2350.7

0.0075V2^2 = 2350.7 - 2313.9

0.0075V2^2 = 36.8

V2^2 = 36.8 / 0.0075

V2^2 = 4906.6666

V2 = √4906.6666

V2 = 70.0476 ft/s

converting to miles per hour

V2 = 70.0476 ft/s * 1 mil / 5280 ft * 3600s / 1hr

V = 47.7 mi/hr

8 0
3 years ago
A resistance of 30 ohms is placed in a circuit with a 90 volt battery. What current flows in the circuit?
blagie [28]

Answer:

3A

Explanation:

Using Ohms law U=I×R solve for I by I=U/R

4 0
3 years ago
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