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3 years ago

9

.What is the momentum of a car with a mass of 1,300 kg traveling north at a speed of 28 m/

s?the momentum of the car

1 answer:

Maksim231197 [3]3 years ago

3 0

44,563 m/s
hopoe this helps

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an object weighting 100g is thrown upwards from the ground at a speed of 100 m/s.where will the potential energy of the object b

Kay [80]

Answer:

333.3 m

Explanation:

Given

$m =100g\ =\ 0.1kg\\v = 100 m/s\\g = 10 m/s ^2$

Potential energy = $\frac{2}{3}\ of\ Kinetic\ energy$ ......Equation(1)

We know that

Potential energy=mgh

Kinetic energy = $\frac{1}{2} mv^{2}$

Now From the Equation(1)

$mgh=\frac{2}{3}*\frac{1}{2} mv^{2}\\ gh=\frac{v^{2} }{3} \\10 * h=\ \frac{10000}{3}\\ h=\ \frac{1000}{3} \\h=333.3\ m$

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2 years ago

What is a statement that’s summarizes a pattern found in nature

sergey [27]

The answer to your question is,

A scientific law.

-Mabel <3

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2 years ago

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escribe the differences between the terrestrial and jovian planets and classify each of the 8 planets to the proper type

Radda [10]

Answer:

*******

Explanation:

is obviously the correct answer they are both so different yet so alike

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2 years ago

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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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2 years ago

The sun rose at 6:54 this morning.<br>is this an observation or inference?

jeka57 [31]

Observation: You saw/ observed something and you are stating that.

Inference: You are predicting something to be true. If so and so, then this will occur.

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2 years ago

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