What is the most common way to organize data in a experiment

Which of the following best illustrates a pair of sentences that are joined by an understood relationship?

zubka84 [21]

Answer : Option C) Detective Smiley scanned the dim hallway. He pulled his pistol from its holster.

Explanation : Amongst the given other choices there seems to be no relationship between two consecutive sentences. The only sentence which seemed to have a connection between previous and later sentence was option C. Where it is clearly stated that the detective named as Smiley was walking through the hallway which was dimly lit. The second sentence has a co-relation to the previous one as it extends the sentence. Detective smiley then pulled out his pistol from his holster after walking through the hallway.

This confirms that the correct answer for this question is Option C.

8 0

3 years ago

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A 191 191 kg sculpture hangs from a horizontal rod that serves as a pivot about which the sculpture can oscillate. The sculpture

Studentka2010 [4]

Answer:

r = 0.31 m

Explanation:

Given that,

Mass of the sculpture, m = 191 kg

The sculpture's moment of inertia with respect to the pivot is, $I=17.2\ kg-m^2$

Frequency of oscillation, f = 0.925 Hz

Let r is the distance of the the pivot from the sculpture's center of mass. The frequency of oscillation is given by :

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{mgr}{I}}$

$r=\dfrac{4\pi^2f^2I}{mg}$

$r=\dfrac{4\pi^2\times 0.925^2\times 17.2}{191\times 9.8}$

r = 0.31 m

So, the pivot is 0.31 meters from the sculpture's center of mass. Hence, this is the required solution.

4 0

3 years ago

The unit called the____<br> is defined based on the amount of<br> work a horse can do in 1 minute.

Kruka [31]

Answer:

horse power

Explanation:

8 0

3 years ago

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Emotional Intelligence includes:

vazorg [7]

Answer: True

Explanation

5 0

3 years ago

. An expedition in the arctic sea to study the e ect of climate change on deep-ocean ecosystemrequires precise measurement of te

Sergio039 [100]

Answer:

The complete question contains 2 parts such that

part a: Where the probe is submerged in water. In this case the wattage of heater is 90.47 W

part b: Where the probe is on a vessel. In this case the wattage of heater is So 121.34 W

Explanation:

Part a

For the first part when it is assumed that the probe is submerged in the water. In this case the heat loss is only due to the convective heat transfer which is given as

$Q=h \times A \times (T_{body}-T_{surr})$

Here

Q is the heat loss which is to be calculated

A is the area of the body which is given as $4 \pi r^2 \\$ . It is calculated as

$A=4 \pi r^2 \\A=4 \times 3.14 \times (0.2)^2 \\A=0.5026 m^2$

Tbody is 10 C
Tsurr is 0C
h is the convection coefficient of water which is 18 W/m^2K

$Q=h \times A \times (T_{body}-T_{surr})\\Q=18 \times 0.5026 \times (10-0)\\Q=90.47 W\\$

So the wattage of heater is 90.47 W

Part b

Now when the probe is above sea water now, the losses are both because of convection and radiation now the loss is given as

$Q_{con}=h_{air} \times A \times (T_{body}-T_{air})$

Here

Q_con is the heat loss due to convection which is to be calculated

A is the area of the body which is given as 0.5026 m^2

Tbody is 10 C
Tair is -10C
h is the convection coefficient of water which is 10 W/m^2K

$Q_{con}=h_{air} \times A \times (T_{body}-T_{air})\\Q_{con}=10 \times 0.5026 \times (10+10)\\Q_{con}=100.52 Watts$

Radiation loss is given as

$Q_{rad}=E_{emission}-E_{radiation}\\Q_{rad}=\epsilon A \tau (T_{body}^4-T_{sky}^4)\\$

Here

Q_rad is the heat loss due to radiation which is to be calculated

A is the area of the body which is given as 0.5026 m^2

Tbody is 10+273=283K
Tsky is 0+273=273K
ε is the emissivity of the shell which is 0.85

τ is the coefficient which is $5.67 \times 10^{-8}$

$Q_{rad}=E_{emission}-E_{radiation}\\Q_{rad}=\epsilon A \tau (T_{body}^4-T_{sky}^4)\\Q_{rad}=0.85 \times 0.5026 \times 5.67 \times 10^{-8} ((283)^4-(273)^4)\\Q_{rad}=20.823 Watts$

So the total value of heat loss is

$Q_{total}=Q_{convective}+Q_{radiation}\\Q_{total}=100.52+20.823 W\\Q_{total}=121.341 W$

So the wattage of heater is 121.34 W

3 0

3 years ago