Select the correct answer.

1. A team of students have designed a battery-powered cooler, which promises to keep beverages at a high-drinkability temperatur

Anit [1.1K]

Answer:

Minimum electrical power required = 3.784 Watts

Minimum battery size needed = 3.03 Amp-hr

Explanation:

Temperature of the beverages, $T_L = 36^0 F = 275.372 K$

Outside temperature, $T_H = 100^0F = 310.928 K$

rate of insulation, $Q = 100 Btu/h$

To get the minimum electrical power required, use the relation below:

$\frac{T_L}{T_H - T_L} = \frac{Q}{W} \\W = \frac{Q(T_H - T_L)}{T_L}\\W = \frac{100(310.928 - 275.372)}{275.372}\\W = 12.91 Btu/h\\1 Btu/h = 0.293071 W\\W = 12.91 * 0.293071\\W_{min} = 3.784 Watt$

V = 5 V

Power = IV

$W_{min} = I_{min} V\\3.784 = 5I_{min}\\I_{min} = \frac{3.784}{5} \\I_{min} = 0.7568 A$

If the cooler is supposed to work for 4 hours, t = 4 hours

$I_{min} = 0.7568 * 4\\I_{min} = 3.03 Amp-hr$

Minimum battery size needed = 3.03 Amp-hr

6 0

2 years ago

A ball bearing has been selected with the bore size specified in the catolog as 35.000 mm to 35.020 mm. Specify appropriate mini

Fofino [41]

Answer:

the minimum shaft diameter is 35.026 mm

the maximum shaft diameter is 35.042mm

Explanation:

Given data;

D-maximum = 35.020mm and d-minimum = 35.000mm

we have to go through Tables "Descriptions of preferred Fits using the Basic Hole System" so from the table, locational interference fits H7/p6

so From table, Selection of International Trade Grades metric series

the grade tolerance are;

ΔD = IT7(0.025 mm)

Δd = IT6(0.016 mm)

Also from Table "Fundamental Deviations for Shafts" metric series

Sf = 0.026

so

D-maximum

Dmax = d + Sf + Δd

we substitute

Dmax = 35 + 0.026 + 0.016

Dmax = 35.042 mm

therefore the maximum diameter of shaft is 35.042mm

d-minimum

Dmin = d + Sf

Dmin = 35 + 0.026

Dmin = 35.026 mm

therefore the minimum diameter of shaft is 35.026 mm

8 0

2 years ago

Calculate the current in the 8-W resistor of Figure below by using Thevenin’s theorem. What will be its value of connections of

kvv77 [185]

solution from c hegg. hope it helps. see photos explanation

7 0

3 years ago

An aircraft is flying at 300 mph true airspeed has a 50 mph tailwind. What is its ground speed?

Free_Kalibri [48]

Answer:

304.13 mph

Explanation:

Data provided in the question :

The Speed of the flying aircraft = 300 mph

Tailwind of the true airspeed = 50 mph

Now,

The ground speed will be calculated as:

ground speed = $\sqrt{300^2+50^2}$

or

The ground speed = $\sqrt{92500}$

or

The ground speed = 304.13 mph

Hence, the ground speed is 304.13 mph

8 0

3 years ago

Two uniformly charged conducting plates are parallel to each other. They each have area A. Plate #1 has a positive charge Q whil

Karo-lina-s [1.5K]

Answer:

$E = \frac{3Q}{2A\epsilon_0}$

Explanation:

By Gauss Law for electric field:

$E = \frac{\sigma}{2\epsilon_0}$

Where $\sigma$ is the charge density Q/A. Since we have 2 parallel plates with different charge, the electric field at point P in the gap would be the sum of 2 field

$E = E_1 + E_2$

$E = \frac{Q}{2A\epsilon_0} + \frac{2Q}{2A\epsilon_0}$

$E = \frac{3Q}{2A\epsilon_0}$

5 0

2 years ago