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marissa [1.9K]
2 years ago
13

Which of these materials are insulators? Select the THREE (3) that apply.

Engineering
1 answer:
lubasha [3.4K]2 years ago
5 0

Answer:

wool, rubber, and plastic

Explanation:

You might be interested in
‏What is the potential energy in joules of a 12 kg ( mass ) at 25 m above a datum plane ?
Virty [35]

Answer:

E = 2940 J

Explanation:

It is given that,

Mass, m = 12 kg

Position at which the object is placed, h = 25 m

We need to find the potential energy of the mass. It is given by the formula as follows :

E = mgh

g is acceleration due to gravity

E=12\times 9.8\times 25\\\\E=2940\ J

So, the potential energy of the mass is 2940 J.

3 0
3 years ago
Mahamad Siddiqui sent false emails and letters of recommendation on behalf of individuals without their permission to nominate h
shusha [124]

Mahamad Siddiqui sent false emails and letters of recommendation on behalf of individuals without their permission to nominate himself for the Waterman Award at the National Science Foundation. His earlier emails were offered where he had solicited letters were offered as evidence. Siddiqui claimed that content of earlier emails was hearsay. Do the earlier emails come in is given below

Explanation:

1.Mohamed Siddiqui appeals his convictions for fraud and false statements to a federal agency, and obstruction in connection with a federal investigation.   Siddiqui challenges the district court's admission into evidence of e-mail and foreign depositions.

2.On February 18, 1997, Jodi Saltzman, a special agent with the NSF interviewed Siddiqui at Siddiqui's office at the University of South Alabama.   During the interview, Siddiqui signed a statement admitting that he had nominated himself for the Waterman Award, but that he had permission from Yamada and von Gunten to submit forms on their behalf.   Siddiqui also acknowledged in the statement that Westrick had recommended Siddiqui for a different award, the PECASE Award, but that Siddiqui had changed the wording of the letter to apply to the Waterman Award.   Siddiqui was indicted on April 29, 1997.

3.Siddiqui opposed the taking of the depositions on the grounds that the witnesses' personal presence at trial was necessary, and that Indian travel restrictions for its citizens residing abroad prevented him from traveling to Japan and Switzerland.   Specifically, Siddiqui asserted that because of religious persecution in India his travel to Japan or Switzerland related to the criminal action would put his family members still living in India at risk.   The magistrate judge ruled that the government had carried its burden of showing that Yamada and von Gunten would be unavailable to appear at trial, and instructed that Siddiqui's fear of obtaining a travel visa from India because of the threat of persecution of family members should not preclude the taking of the foreign depositions.

4.Yamada's deposition was taken in Japan on March 6, 1998.   At government expense, Siddiqui's counsel attended the deposition and cross-examined the witness, but was not in telephonic contact with Siddiqui during the deposition.   Yamada testified that on February 1, 1997, she received an e-mail stating that if she received a phone call from the NSF to “please tell good words about me.”   Yamada testified that she knew the e-mail was from Siddiqui because the name on the e-mail had Siddiqui's sender address, and it ended with the name “Mo” which Siddiqui had previously told her was his nickname, and which he had used in previous e-mail.

5.Yamada later admitted to Saltzman that she had not given Siddiqui permission to sign, but had made the earlier representation because she thought Siddiqui would go to jail.

6.During cross-examination of Yamada at the deposition, Siddiqui's counsel introduced an e-mail from Yamada to Siddiqui.   This e-mail contained the same e-mail address for Siddiqui as the e-mail received by Yamada and von Gunten apparently from Siddiqui.

7.Von Gunten's video deposition was taken in Switzerland.   At government expense, Siddiqui's counsel attended the deposition and cross-examined von Gunten.   During the deposition, Siddiqui was in communication with his counsel by telephone.   Von Gunten testified at the deposition that he had not submitted a letter of recommendation in favor of Siddiqui for the Waterman Award, and that he had not given Siddiqui permission to submit such a letter in his name.

8 0
3 years ago
Air enters the compressor of a simple gas turbine at 100 kPa, 300 K, with a volumetric flow rate of 5 m3/s. The compressor press
Zepler [3.9K]

Answer:

a) 3581.15067 kw

b) 95.4%

Explanation:

<u>Given data:</u>

compressor efficiency = 85%

compressor pressure ratio = 10

Air enters at:    flow rate of 5m^3/s , pressure = 100kPa, temperature = 300 K

At turbine inlet : pressure = 950 kPa, temperature = 1400k

Turbine efficiency = 88% , exit pressure of turbine = 100 kPa

A) Develop a full accounting of the exergy increase of the air passing through the gas turbine combustor in kW

attached below is a detailed solution to the given question

6 0
3 years ago
A rich industrialist was found murdered in his house. The police arrived at the scene at 11:00 PM. The temperature of the corpse
d1i1m1o1n [39]

Answer:

The dude was killed around 6:30PM

Explanation:

Newton's law of cooling states:

    T = T_m + (T_0-T_m)e^{kt}

where,

T_0 = initial temp

T_m = temp of room

T = temp after t hours

k = how fast the temp is changing

t = time (hours)

T_0 = 31     because the body was initlally 31ºC when the police found it

T_m = 22   because that was the room temp

T = 30  because the body temp drop to 30ºC after 1 hour

t = 1 because that's the time it took for the body temp to drop to 30ºC

k=???   we don't know k so we must solve for this

rearrange the equation to solve for k

T = T_m + (T_0-T_m)e^{kt}

T - T_m= (T_0-T_m)e^{kt}

\frac{T - T_m}{(T_0-T_m)}= e^{kt}

ln(\frac{T - T_m}{T_0-T_m})=kt

\frac{ln(\frac{T - T_m}{T_0-T_m})}{t}=k

plug in the numbers to solve for k

k = \frac{ln(\frac{T - T_m}{T_0-T_m})}{t}

k = \frac{ln(\frac{30 - 22}{31-22})}{1}

k=ln(\frac{8}{9})

Now that we know the value for k, we can find the moment the murder occur. A crucial information that the question left out is the temperature of a human body when they're still alive. A living human body is about 37ºC. We can use that as out initial temperature to solve this problem because we can assume that the freshly killed body will be around 37ºC.

T_0 = 37     because the body was 37ºC right after being killed

T_m = 22   because that was the room temp

T = 31  because the body temp when the police found it

k=ln(\frac{8}{9})   we solved this earlier

t = ???   we don't know how long it took from the time of the murder to when the police found the body

Rearrange the equation to solve for t

T = T_m + (T_0-T_m)e^{kt}

T - T_m= (T_0-T_m)e^{kt}

\frac{T - T_m}{(T_0-T_m)}= e^{kt}

ln(\frac{T - T_m}{T_0-T_m})=kt

\frac{ln(\frac{T - T_m}{T_0-T_m})}{k}=t

plug in the values

t=\frac{ln(\frac{T - T_m}{T_0-T_m})}{k}

t=\frac{ln(\frac{31 - 22}{37-22})}{ln(8/9)}

t=\frac{ln(3/5)}{ln(8/9)}

t=\frac{ln(3/5)}{ln(8/9)}

t ≈ 4.337 hours from the time the body was killed to when the police found it.

The police found the body at 11:00PM so subtract 4.337 from that.

11 - 4.33 = 6.66 ≈ 6:30PM

7 0
3 years ago
What is the De Broglie wavelength of an electron under 150 V acceleration?
yanalaym [24]

Answer:

0.1 nm

Explanation

Potential deference of the electron is given as V =150 V

Mass of electron m=9.1\times 10^{-31}

Let the velocity of electron = v

Charge on the electron =1.6\times 10^{-19}C

plank's constant h =6.67\times 10^{-34}

According to energy conservation eV =\frac{1}{2}mv^2

v=\sqrt{\frac{2eV}{M}}=\sqrt{\frac{2\times 1.6\times 10^{-19}\times 150}{9.1\times 10^{-31}}}=7.2627\times 10^{-6}m/sec

Now we know that De Broglie wavelength \lambda =\frac{h}{mv}=\frac{6.67\times 10^{-34}}{9.1\times 10^{-31}\times 7.2627\times10^6 }=0.100\times 10^{-9}m=0.1nm

4 0
3 years ago
Read 2 more answers
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