A pinion and gear pair is used to transmit a power of 5000 W. The teeth numbers of pinion

An existing building is suffering from cracks in the exterior walls. The investigating engineer wants to ensure that the foundat

jek_recluse [69]

Answer:

$18 ft^{2}$

Explanation:

Soil bearing pressure= $\frac {Load}{Area}$

Since we're given pressure of 2500 psf and load of 45000 pounds

The area= $\frac {45000}{2500}=18$

Therefore, the smallest area of safe footings should not be less than $18 ft^{2}$

6 0

2 years ago

You’ve experienced convection cooling if you’ve ever extended your hand out the window of a moving vehicle or into a flowing wat

Anna35 [415]

Answer:

Condition A

Heat flux is 1400 W/M^2

Condition B

Heat flux is 12800 w/m^2

Explanation:

Given that:

$T_s$ is given as 30 degree celcius

condition A

Air temperature = - 5 degree c

convection coefficient h = 40 w/m^2. k

$heat\ flux = \frac{Q}{a}= h\Delta = 40{30 - (-5)} = 1400 w/m^2$

condition A

water temperature = 10 degree c

convection coefficient = 800 w/m^2.k

$heat\ flux = \frac{Q}{A} = H(\Delta} = 800\times (30-14) = 12800w/m^2$

7 0

2 years ago

The electricity generated by wind turbines annually in kilowatt-hours per year is given in a file. The amount of electricity is

Anika [276]

Answer:

Steps:

1. Create a text file that contains blade diameter (in feet), wind velocity (in mph) and the approximate electricity generated for the year

2. load the data file for example, in matlab, use ('fileame.txt') to load the file

3. create variables from each column of your data

for example, in matlab,

x=t{1}

y=t{2}

4. plot the wind velocity and electricity generated.

plot(x, y)

5. Label the individual axis and name the graph title.

title('Graph of wind velocity vs approximate electricity generated for the year')

xlabel('wind velocity')

ylabel('approximate electricity generated for the year')

5 0

3 years ago

Read 2 more answers

Please solve part two

Burka [1]

Answer:

Wat part 2

Explanation:

7 0

2 years ago

A 20.0 µF capacitor is charged to a potential difference of 800 V. The terminals of the charged capacitor are then connected to

Sergeu [11.5K]

Answer:

a) Q_initial = 16 * 10^-3 C

b) V_1 = V_2 = (16/3) * 10^2 V

c) E = 64/15 J

d) dE = 32/15 J of decrease

Explanation:

Given:

- Capacitor 1, C_1 = 20.0 uF

- Capacitor 2, C_2 = 10.0 uF

- Charged with P.d V = 800 V

Find:

a) the original charge of the system,

(b) the ﬁnal potential difference across each capacitor

(c) the ﬁnal energy of the system

(d) the decrease in energy when the capacitors are connected.

Solution:

a)

- The initial charge in the circuit is the one carried by the first charged capacitor.

Q_initial = C_1*V

Q_initial = 20*10^-6 * 800

Q_initial = 16 * 10^-3 C

b)

- After charging the other capacitor, we know that the total charge is conserved among two capacitor:

Q_initial = Q_1 + Q_2

- We also know that potential difference across two capacitor is also same.

V_1 = V_2 = Q_1 / C_1 = Q_2 / C_2

- Using the two equations and solve for charge Q_2:

Q_2 = Q_1*C_2/C_1

Q_2 = Q_1*10/20 = 0.5*Q_1

- using conservation of charge:

Q_initial = 1.5*Q_1

Q_1 = 16*10^-3 / 1.5 = 10.67*10^-3 C

- Hence the Voltage across each capacitor is:

V_2 = V_1 = Q_1 / C_1

= 10.67*10^-3 / 20*10^-6

= (16/3) * 10^2 V

c)

- The energy in the system is:

E = 0.5*C_eq*V^2

Where, C_eq is the equivalent capacitance of paralle circuit.

E = 0.5*(20+10)*10^-6 *((16/3) * 10^2)^2

E = 64/15 J

d)

- The decrease in energy of the capacitors is:

dE = E_initial - E_final

Where, E_initial is due to charging of the C_1 only:

dE = 0.5*10^-6*20*800^2 - (64/15)

dE = 32/5 - 64/15 = 32/15 J

5 0

3 years ago