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If you pair copper which has an electron affinity of 0.34 and silver which has an electron affinity of 0.80, will you make a str

Alex787 [66]

Answer:

pairing of the copper which has an electron affinity of 0.34 and the silver which has an electron affinity of 0.80 makes a strong battery.

Explanation:

All the Batteries of this world are made with two metals having different-different electron affinities. What do the phrase “electrons affinities ” mean and how do these affinities affect the voltage of the batteries?

The Electron affinities are the energy change that occurs when electrons are added to atoms. The greater the attraction of the atoms to the electrons the more energy would released. If an atom has high electron affinity, the electron will be harder to gain The greater is the difference in metal affinities, the greater the voltage. That means, if you pair Coppers with Silver, the difference between their electron affinities would be (0.80-0.34) or, 0.46 and You can make a strong battery.

3 0

2 years ago

In a distant solar system, a giant planet has

sergeinik [125]

Answer:

mass of the planet: $5.9\,10^{26}\,kg$

Explanation:

When a moon keeps a circular orbit around a planet, it is the force of gravity the one that provides the centripetal force to keep it in its circular trajectory of radius R. So if we can write that in such cases (being the mass of the planet "M" and the mass of the moon "m"), we can form an equation by making the centripetal force on the moon equal the force of gravity (using the Newton's Universal Law of Gravity):

$m\frac{v^2}{R}=G\frac{M\,m}{R^2}$

where we used here the tangential velocity (v) of the moon around the planet. This equation can be further simplified by dividing both sides by "m" and multiplying both sides by the orbital radius R:

$m\frac{v^2}{R}=G\frac{M\,m}{R^2}\\v^2=G\frac{M}{R}$

Notice that the mass of the moon has actually disappeared from the equation, which tells us that the orbiting velocity and period do not depend on the mass of the moon, but on the mass of the actual planet.

We know the orbital radius R ( $5.32\,10^5\,km=5.32\,10^8\,m$ , the value of the Universal Gravitational constant G, and we can estimate the value of the tangential velocity of the moon since we know it period: 36.3 hrs = 388800 seconds.

We know that the moon makes a full circumference ( $2\,\pi\,R$ ) in 388800 seconds, therefore its tangential velocity is:

$v=\frac{2\,\pi\,5.32\,10^8}{388800} \frac{m}{s} \\v=8.6\,10^3\,\frac{m}{s}$

where we rounded the velocity to one decimal.

Notice that we have converted all units to the SI system, so when using the formula to solve for the mass of the planet, the answer comes directly in kg.

Now we use this value for the tangential velocity to estimate the mass of the planet in the first equation we made and simplified:

$v^2=G\frac{M}{R}\\M=\frac{v^2\,R}{G} \\M=\frac{(8.6\,10^3)^2\,5.32\,10^8}{6.67\,10^{-11}}kg\\M=5.9\,10^{26}\,kg$

8 0

2 years ago

How does energy from the sun affect the motion of molecules in a gas compared to molecules in a liquid?

AnnZ [28]

Molecules in a gas move faster than in a liquid.

hope it helps.

3 0

1 year ago

Read 2 more answers

Identical guns fire identical bullets horizontally at the same speed from the same height above level planes, one on the Earth a

Natasha_Volkova [10]

Answer:

I. The horizontal distance traveled by the bullet is greater for the Moon.

II. The flight time is less for the bullet on the Earth.

Explanation:

Horizontal distance depends on the initial speed, height and gravity. Bullets have the same initial speed and are shot from the same height. In these conditions horizontal distance only depends on gravity, which is inversely proportional. Therefore, the less gravity the greater the horizontal distance. Gravity slows bullet and causes its impact on the ground. Since gravity is greater in Earth, the bullet hits faster on the earth.

3 0

1 year ago

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A cannon is mounted on a tower above a wide, level field. The barrel of the cannon is 20 m above the ground below. A cannonball

OLga [1]

Answer:

Cannonball will be in flight before it hits the ground for 2.02 seconds

Explanation:

Initial height from ground = 20 meter.

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.