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3 years ago

Why do you think the earth has fewer craters than the moon

1 answer:

4 0

The earth has an atmosphere that disintegrates most meteorites that make craters, unlike the moon, which has no atmosphere and nothing to protect it from meteorites

You might be interested in

What would happen to the two balls if one of them were kept positively charged and the charge on the other ball were slowly incr

mariarad [96]

Answer:

the balls would move closer to each other

Explanation:

8 0

2 years ago

A car traveling with constant speed travels 150 km in 7200s what is the speed of the car

Oksana_A [137]

Velocity is distance/time

so 150/7200=.0208km/s

unless you have to convert it to miles or something else. but use the formula!

5 0

3 years ago

Road rage is an aggressive driving incident where the driver has:

WARRIOR [948]

I would say that it would be A.) Been insulted. This is because you are angered by someone else. It does not necessarily mean that something has de-escelated, you've lost control of your car, or you've nothing to lose. Hopefully that helps. :)

5 0

3 years ago

11) A sled is initially given a shove up a frictionless 35º incline. It reaches a maximum height of 2.5

Andrej [43]

Answer:

7 m/s

Explanation:

To solve this problem you must use the conservation of energy.

$K1 +U1=K2+U2$

That math speak for, initial kinetic energy plus initial potential energy equals final kinetic energy plus final potential energy.

The initial PE (potential energy) is 0 because it hasn't been raised in the air yet. The final KE (kinetic energy) is 0 because it isn't moving. This gives the following:

$KE1= \frac{1}{2}mv^{2}}$

$PE2=mgh$

K1=U2

$\frac{1}{2} mv^{2} =mgh$

Solve for v

$v=\sqrt{2gh}$

Input known values and you get 7 m/s.

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 $96485.33212\; \rm C \cdot mol^{-1}$ .)

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

5 0

2 years ago