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3 years ago

5

You read a research study that concludes that the higher a student's self-esteem, the better he performs in school. This sort of

relationship would best be characterized as
(1pts)

a negative correlation

a positive correlation

cause and effect

no relationship

1 answer:

LiRa [457]3 years ago

7 0

Answer:

Negative correlation

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Can I get an answer to this question please

crimeas [40]

Answer:

(i) 12 V in series with 18 Ω.

(ii) 0.4 A; 1.92 W

(iii) 1,152 J

(iv) 18Ω — maximum power transfer theorem

Explanation:

<h3>(i)</h3>

As seen by the load, the equivalent source impedance is ...

10 Ω + (24 Ω || 12 Ω) = (10 +(24·12)/(24+12)) Ω = 18 Ω

The open-circuit voltage seen by the load is ...

(36 V)(12/(24 +12)) = 12 V

The Thevenin's equivalent source seen by the load is 12 V in series with 18 Ω.

__

<h3>(ii)</h3>

The load current is ...

(12 V)/(18 Ω +12 Ω) = 12/30 A = 0.4 A . . . . load current

The load power is ...

P = I^2·R = (0.4 A)^2·(12 Ω) = 1.92 W . . . . load power

__

<h3>(iii)</h3>

10 minutes is 600 seconds. At the rate of 1.92 J/s, the electrical energy delivered is ...

(600 s)(1.92 J/s) = 1,152 J

__

<h3>(iv)</h3>

The load resistance that will draw maximum power is equal to the source resistance: 18 Ω. This is the conclusion of the Maximum Power Transfer theorem.

The power transferred to 18 Ω is ...

((12 V)/(18 Ω +18 Ω))^2·(18 Ω) = 144/72 W = 2 W

7 0

2 years ago

Compare a series circuit powered by six 1.5-volt batteries to a series circuit powered by a single 9-volt battery. Make sure the

lana [24]

Answer:

Both series circuits provide a total voltage of 9 volts to the two bulbs connected in series and the voltage will be equally divided among two bulbs and they will have same brightness. Therefore, both circuits will have same characteristics.

Explanation:

We are asked to compare two series circuits having equal number of light bulbs.

1st circuit is powered by 6 batteries each having a voltage of 1.5V

2nd circuit is powered by a single battery having a voltage of 9V.

The six batteries in the 1st circuit can be connected together in series or in parallel.

When the batteries are connected in series (positive terminal of one battery connected to negative terminal of another battery) their voltage gets added which means

Voltage of pack = number of batteries*voltage of each battery

Voltage of pack = 6*1.5

Voltage of pack = 9 volts

But the current remains same in the series connection since there is only path for the current to flow.

On the other hand, when the batteries are connected in parallel, the voltage remains same but the current increases.

Circuit 1:

In this circuit, we have 6 batteries each of 1.5 volts connected in series to provide a voltage of 9 volts.

We have connected 2 bulbs in this series circuit.

The voltage will be equally divided between two bulbs if both bulbs are identical in construction.

So there will be 4.5 volts across each bulb and both bulbs will have same brightness.

Circuit 2:

In this circuit, we have 1 battery which provide a voltage of 9 volts.

We have connected 2 bulbs in this series circuit just like in circuit 1.

The voltage will be equally divided between two bulbs if both bulbs are identical in construction.

So there will be 4.5 volts across each bulb and both bulbs will have same brightness.

Conclusion:

Both series circuits provide a total voltage of 9 volts to the two bulbs connected in series and the voltage will be equally divided among two bulbs and they will have same brightness. Therefore, both circuits will have same characteristics.

3 0

3 years ago

In the planning process of the product development life cycle what is it important to inventory

Verizon [17]

Your Answer would be A I believe.

6 0

3 years ago

assume a five layer network model. There are 700 bytes of application data. There is a 20 bye header at the transport layer, a 2

amm1812

Answer: The overhead percentage is 7.7%.

Explanation:

We call overhead, to all those bytes that are delivered to the physical layer, that don't carry real data.

We are told that we have 700 bytes of application data, so all the other bytes are simply overhead, i.e. , 58 bytes composed by the transport layer header, the network layer header, the 14 byte header at the data link layer and the 4 byte trailer at the data link layer.

So, in order to assess the overhead percentage, we divide the overhead bytes between the total quantity of bytes sent to the physical layer, as follows:

OH % = (58 / 758) * 100 = 7.7 %

4 0

3 years ago

At a point on the free surface of a stressed body, the normal stresses are 20 ksi (T) on a vertical plane and 30 ksi (C) on a ho

victus00 [196]

Answer:

The principal stresses are σp1 = 27 ksi, σp2 = -37 ksi and the shear stress is zero

Explanation:

The expression for the maximum shear stress is given:

$\tau _{M} =\sqrt{(\frac{\sigma _{x}^{2}-\sigma _{y}^{2} }{2})^{2}+\tau _{xy}^{2} }$

Where

σx = stress in vertical plane = 20 ksi

σy = stress in horizontal plane = -30 ksi

τM = 32 ksi

Replacing:

$32=\sqrt{(\frac{20-(-30)}{2} )^{2} +\tau _{xy}^{2} }$

Solving for τxy:

τxy = ±19.98 ksi

The principal stress is:

$\sigma _{x}+\sigma _{y} =\sigma _{p1}+\sigma _{p2}$

Where

σp1 = 20 ksi

σp2 = -30 ksi

$\sigma _{p1} +\sigma _{p2}=-10 ksi$ (equation 1)

$\tau _{M} =\frac{\sigma _{p1}-\sigma _{p2}}{2} \\\sigma _{p1}-\sigma _{p2}=2\tau _{M}\\\sigma _{p1}-\sigma _{p2}=32*2=64ksi$ equation 2

Solving both equations:

σp1 = 27 ksi

σp2 = -37 ksi

The shear stress on the vertical plane is zero

4 0

3 years ago