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tresset_1 [31]
3 years ago
6

Jaden is doing an experiment to find out if plants grow taller when given plain water or sugar water. To make sure his experimen

t is a fair test, Jaden should
Engineering
2 answers:
Paul [167]3 years ago
6 0
The answer is jaden should use two different plants and give one plain water and give the other sugar water. jaden should record their heights
Anettt [7]3 years ago
3 0
I believe he should use two separate plants. One just for tracking the growth with plain water and the other for sugar.
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In some synchronizer applications, the clock frequency f is substituted for the parameter a in metastability MTBF calculations,
Contact [7]

Answer:

Please see the attached file for the complete answer.

Explanation:

Download pdf
4 0
3 years ago
Consider a very long, cylindrical fin. The temperature of the fin at the tip and base are 25 °C and 50 °C, respectively. The dia
nekit [7.7K]

Answer:

98°C

Explanation:

Total surface area of cylindrical fin = πr² + 2πrl , r = 0.015m; l= 0.1m; π =22/7

22/7*(0.015)² + 22/7*0.015*0.1 = 7.07 X 10∧-4 + 47.1 X 10∧-4 = (54.17 X 10∧-4)m²

Temperature change, t = (50 - 25)°C = 25°C = 298K

Hence, Temperature =  150 X (54.17 X 10∧-4) X 298/123 = 242.14/124 = 2.00K =

∴ Temperature change = 2.00K

But temperature, T= (373 - 2)K = 371 K

In °C = (371 - 273)K = 98°C         

7 0
3 years ago
A 4-L pressure cooker has an operating pressure of 175 kPa. Initially, one-half of the volume is filled with liquid and the othe
vodomira [7]

Answer:

the highest rate of heat transfer allowed is 0.9306 kW

Explanation:

Given the data in the question;

Volume = 4L = 0.004 m³

V_f = V_g = 0.002 m³

Using Table ( saturated water - pressure table);

at pressure p = 175 kPa;

v_f = 0.001057 m³/kg

v_g = 1.0037 m³/kg

u_f = 486.82 kJ/kg

u_g 2524.5 kJ/kg

h_g = 2700.2 kJ/kg

So the initial mass of the water;

m₁ = V_f/v_f + V_g/v_g

we substitute

m₁ = 0.002/0.001057  + 0.002/1.0037

m₁ = 1.89414 kg

Now, the final mass will be;

m₂ = V/v_g

m₂ = 0.004 / 1.0037

m₂ = 0.003985 kg

Now, mass leaving the pressure cooker is;

m_{out = m₁ - m₂

m_{out = 1.89414  - 0.003985

m_{out = 1.890155 kg

so, Initial internal energy will be;

U₁ = m_fu_f + m_gu_g

U₁ = (V_f/v_f)u_f  + (V_g/v_g)u_g

we substitute

U₁ = (0.002/0.001057)(486.82)  + (0.002/1.0037)(2524.5)

U₁ = 921.135288 + 5.030387

U₁ = 926.165675 kJ

Now, using Energy balance;

E_{in -  E_{out = ΔE_{sys

QΔt - m_{outh_{out = m₂u₂ - U₁

QΔt - m_{outh_g = m₂u_g - U₁

given that time = 75 min = 75 × 60s = 4500 sec

so we substitute

Q(4500) - ( 1.890155 × 2700.2 ) = ( 0.003985 × 2524.5 ) - 926.165675

Q(4500) - 5103.7965 = 10.06013 - 926.165675

Q(4500) = 10.06013 - 926.165675 + 5103.7965

Q(4500) = 4187.690955

Q = 4187.690955 / 4500

Q = 0.9306 kW

Therefore, the highest rate of heat transfer allowed is 0.9306 kW

5 0
3 years ago
Provide three examples of things engineers can implement when designing environmentally sustainable projects.
Dafna1 [17]
Water supply.
Food production.
Housing and shelter.
Sanitation and waste management.
Energy development.
Transportation.
6 0
2 years ago
Read 2 more answers
Question 1 of 8.
Rasek [7]
The answer seems pretty obvious, all of the above
8 0
3 years ago
Read 2 more answers
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