2 years ago

I feel so pressured..

Engineering

2 answers:

topjm [15]2 years ago

7 0

Answer: Thats alright bro keep yo head up

aleksley [76]2 years ago

3 0

Answer:

Well, Hope you feel better..

Explanation:

You might be interested in

Lois is in high school and enjoys studying math and science subjects. She aspires to get into the robotics industry after comple

Inessa [10]

Answer:

d.partivipate in a robotic club

hope helps

mark me please

4 0

3 years ago

A sample of wastewater is diluted 10 times. The diluted solution has an ultimate biochemical oxygen demand (BOD), Lo, of 30 mg/L

zzz [600]

Answer:

474.59 mg/L

Explanation:

Given that

BOD = 30 mg/L

Original BOD = 30 mg/L × dilution factor

Original BOD = 30 mg/L × 10 = 300 mg/L

$L_o = \frac{BOD}{1-e^{-5t}}$

here $L_o$ is the ultimate BOD ; BOD is the biochemical oxygen demand ; t = 0.20 /day

$L_o = \frac{300}{1-e^{-5(0.20)}}$

$L_o = 474.59 \ mg/L$

3 0

4 years ago

How many volts of electricity would it take to power up an entire city? Take Tokyo for example. Please explain!

jek_recluse [69]

Answer:

maybe a couple million volts

Explanation:

because there are outlets you need to charge your phone maybe you car if you have a tesla and skyscrapers

3 0

3 years ago

What possible scenarios may happen if you do the task without using PPE?

ale4655 [162]

Without PPE, employees are at risk of Cuts and punctures. Chemical burns. Electric shocks. Exposure to excessive noise or vibration.

8 0

3 years ago

Consider a Carnot refrigeration cycle executed in a closed system in the saturated liquid–vapor mixture region using 1.06 kg of

Alexxandr [17]

Answer:

$P_m_i_n = 442KPA$

Explanation:

We are given:

m = 1.06Kg

$T_H = 1.2T_L$

T = 22kj

Therefore we need to find coefficient performance or the cycle

$COP_R = \frac {1}{(T_R/T_l) -1}$

$= \frac {1 }{1.2-1}$

= 5

For the amount of heat absorbed:

$Q_l = COP_R Wm$

= 5 × 22 = 110KJ

For the amount of heat rejected:

$Q_H = Q_L + W_m$

= 110 + 22 = 132KJ

[tex[ q_H = \frac{Q_L}{m} [/tex];

= $= \frac{132}{1.06}$

= 124.5KJ

Using refrigerant table at hfg = 124.5KJ/Kg we have 69.5°c

Convert 69.5°c to K we have 342.5K

To find the minimum temperature:

$T_L = \frac{T_H}{1.2}$ ;

$T_L = \frac{342.5}{1.2}$

= 285.4K

Convert to °C we have 12.4°C

From the refrigerant R -134a table at $T_L$ = 12.4°c we have 442KPa

6 0

3 years ago