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3 years ago

11

After completing a scientific investigation, a scientist will draw conclusions about their hypothesis using data they gathered d

uring their investigation.
True

False

2 answers:

marshall27 [118]3 years ago

5 0

The answer is false.

Alex777 [14]3 years ago

3 0

False, the hypothesis is before their investigation or experiment

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1+5-(-5 - 1) whats the answer

vladimir2022 [97]

Answer:

12

Explanation:

6 0

3 years ago

Read 2 more answers

At a distance D from a very long (essentially infinite) uniform line of charge, the electric field strength is 1000 N/C. At what

Alla [95]

Answer:

The correct option is (a).

Explanation:

We know that, the E is inversely proportional to the distance as follows :

$E=\dfrac{k}{d^2}$

We can write it as follows :

$\dfrac{E_1}{E_2}=(\dfrac{d_2}{d_1})^2$

Put all the values,

$\dfrac{1000}{2000}=(\dfrac{d_2}{d})^2\\\\\sqrt{\dfrac{1000}{2000}}=(\dfrac{d_2}{D})\\\\0.7071=\dfrac{d_2}{d}\\\\d_1=0.7071D\\\\d_1=\dfrac{D}{\sqrt2}$

So, the correct option is (a).

5 0

2 years ago

xConsider the following reduction potentials: Cu2+ + 2e– Cu E° = 0.339 V Pb2+ + 2e– Pb E° = –0.130 V For a galvanic cell employi

slega [8]

Answer:

Approximately $\rm 90\; kJ$ .

Explanation:

Cathode is where reduction takes place and anode is where oxidation takes place. The potential of a electrochemical reaction ( $E^{\circ}(\text{cell})$ ) is equal to

$E^{\circ}(\text{cell}) = E^{\circ}(\text{cathode}) - E^{\circ}(\text{anode})$ .

There are two half-reactions in this question. $\rm Cu^{2+} + 2\,e^{-} \rightleftharpoons Cu$ and $\rm Pb^{2+} + 2\,e^{-} \rightleftharpoons Pb$ . Either could be the cathode (while the other acts as the anode.) However, for the reaction to be spontaneous, the value of $E^{\circ}(\text{cell})$ should be positive.

In this case, $E^{\circ}(\text{cell})$ is positive only if $\rm Cu^{2+} + 2\,e^{-} \rightleftharpoons Cu$ is the reaction takes place at the cathode. The net reaction would be

$\rm Cu^{2+} + Pb \to Cu + Pb^{2+}$ .

Its cell potential would be equal to $0.339 - (-0.130) = \rm 0.469\; V$ .

The maximum amount of electrical energy possible (under standard conditions) is equal to the free energy of this reaction:

$\Delta G^{\circ} = n \cdot F \cdot E^{\circ} (\text{cell})$ ,

where

$n$ is the number moles of electrons transferred for each mole of the reaction. In this case the value of $n$ is $2$ as in the half-reactions.
$F$ is Faraday's Constant (approximately $96485.33212\; \rm C \cdot mol^{-1}$ .)

$\begin{aligned}\Delta G^{\circ} &= n \cdot F \cdot E^{\circ} (\text{cell})\cr &= 2\times 96485.33212 \times (0.339 - (-0.130)) \cr &\approx 9.0 \times 10^{4} \; \rm J \cr &= 90\; \rm kJ\end{aligned}$ .

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3 years ago

Shrinking Loop. A circular loop of flexible iron wire has an initial circumference of 168 cm , but its circumference is decreasi

DedPeter [7]

Answer:

103.1 V

Explanation:

We are given that

Initial circumference=C=168 cm

$\frac{dC}{dt}=-15cm/s$

Magnetic field,B=0.9 T

We have to find the magnitude of the emf induced in the loop after exactly time 8 s has passed since the circumference of the loop started to decrease.

Magnetic flux= $\phi=BA=B(\pi r^2)$

Circumference,C= $2\pi r$

$r=\frac{C}{2\pi}$

$r=\frac{168}{2\pi}$ cm

$\frac{dr}{dt}=\frac{1}{2\pi}\frac{dC}{dt}=\frac{1}{2\pi}(-15)=-\frac{15}{2\pi} cm/s$

$\int dr=-\int \frac{15}{2\pi}dt$

$r=-\frac{15}{2\pi}t+C$

When t=0

$r=\frac{168}{2\pi}$

$\frac{168}{2\pi}=C$

$r=-\frac{15}{2\pi}t+\frac{168}{2\pi}$

E= $-\frac{d\phi}{dt}=-\frac{d(B\pi r^2)}{dt}=-2\pi rB\frac{dr}{dt}$

$E=-2\pi(-\frac{5}{2\pi}t+\frac{168}{2\pi})B\times -\frac{15}{2\pi}$

t=8 s

B=0.9

$E=2\pi\times \frac{15}{2\pi}\times 0.9(-\frac{15}{2\pi}(8)+\frac{168}{2\pi})$

$E=103.1 V$

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3 years ago

Helppppppppppppppp asappppppppppppp

vichka [17]

The answer to the problem b.

8 0

3 years ago