If the frequency of a periodic wave is doubled, the period of the wave will be

Lake Baikal in Siberia has a maximum depth of 1642 m. What is the water

Veseljchak [2.6K]

Explanation:

Fluid gauge pressure is:

P = ρgh

where ρ is the fluid density and h is the depth of the fluid.

P = (1000 kg/m³) (9.8 m/s²) (1642 m)

P = 16,091,600 Pa

Rounded to four significant figures, the gauge pressure is 16.09 MPa.

3 0

3 years ago

Find the absolute value of the change of the gravitational potential energy (GPE) of the puck-Earth system from the moment the p

jekas [21]

Answer:

0.16joules

Explanation:

Using the relation for The gravitational potential energy

E= Mgh

Where,

E= Potential energy

h = Vertical Height

M = mass

g = Gravitational Field Strength

To find the vertical component of angle of launch Where the angle is 22°

h= sin theta

So E = mghsintheta

= 0.18 x 0.98 x 0.253 sin22

=0.16joules

Explanation:

7 0

3 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

3 years ago

What and where is the asteroid belt?<br><br> Please ANSWER THIS

Rudiy27

Answer:

The asteroid belt is a region of our solar system, between the orbits of Mars and Jupiter, in which many small bodies orbit our sun.

Explanation:

Hope this helps!

3 0

3 years ago

Read 2 more answers

I NEED HELP PLEASE, THANKS! :)

Luden [163]

Answer:

The ball will be attracted to the negatively charged plate. It'll touch and pick up some electrons from the plate so that the ball becomes negatively charged. Immediately the ball is repelled by the negative plate and is attracted to the positive plate. The ball gives up electrons to the positive plate so that it is positively charged and suddenly attracts to the negative plate again, flies over to it and picks up enough electrons to be repulsed by negative plate and again to the positive plate and that continues.

8 0

3 years ago