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3 years ago

Think of a sphere with many a star-

Physics

1 answer:

ycow [4]3 years ago

3 0

Answer:

For it is a Globular Cluster

Explanation:

These Globular clusters are not considered to be a galaxies because they bond gravitationally to orbiting galaxies like the Milky Way and their masses are relatively small.

In globular clusters, the stars are closed together and they are like 70times closer to one another in their solar neighbourhood.

A globular cluster is a spherical collection of stars that orbits a galactic core. Globular clusters are very tightly bound by gravity, which gives them their spherical shapes, and relatively high stellar densities toward their centers.

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Reacting molecules must hit each other with sufficient energy in order to form products,what theory is it

7nadin3 [17]

It is called Arrhenius collision theory.

4 0

3 years ago

A golf club rotates 215 degrees and has a length (radius) equal to 29 inches. The time it took to swing the club was 0.8 seconds

vichka [17]

Answer:

The average linear velocity (inches/second) of the golf club is 136.01 inches/second

Explanation:

Given;

length of the club, L = 29 inches

rotation angle, θ = 215⁰

time of motion, t = 0.8 s

The angular speed of the club is calculated as follows;

$\omega = (\frac{\theta}{360} \times 2\pi, \ rad) \times \frac{1}{t} \\\\\omega = (\frac{215}{360} \times 2\pi, \ rad) \times \frac{1}{0.8 \ s} \\\\\omega = 4.69 \ rad/s$

The average linear velocity (inches/second) of the golf club is calculated as;

v = ωr

v = 4.69 rad/s x 29 inches

v = 136.01 inches/second

Therefore, the average linear velocity (inches/second) of the golf club is 136.01 inches/second

8 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

Question 1 of 10

ivanzaharov [21]

The answer would be 46.482 because you multiply 18.3 by 2.54 because for every inch you get 2.54 centimeters

6 0

2 years ago

Read 2 more answers

What is the change in potential energy if the distance separating the electron and proton is increased to 1.0 nm?

Vlada [557]

Answer:

$Ep=-2.3*10^{-19}J$

Explanation:

The change in potential energy can be expressed as:

$Ep=K.\frac{q1.q2}{r}$

where K is a constant with a value of $9*10^{9}\frac{N.m^{2}}{C^{2}}$ , q1 and q2 are the charges of the proton and the electron and r is the distance between them.

The charge for the proton is $+1.6*10^{-19}C$ and the charge for the electron is $-1.6*10^{-19}C$ .

Converting r=1.0nm to m:

$1.0nm*\frac{1*10^{-9}m}{1.0nm}=1*10^{-9}m$

Replacing values:

$Ep=9*10^{9}\frac{N.m^{2}}{C^{2}}.\frac{(+1.6*10^{-19}C).(-1.6*10^{-19}C)}{1*10^{-9}m}$

$Ep=-2.3*10^{-19}J$

5 0

3 years ago