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gtnhenbr [62]
3 years ago
6

25) A 10 kg box is sitting on a shelf 15 feet in the air. How could you increase the potential energy of the box? (check all tha

t apply)  Raise the box to 20 feet.  Lower the box to 10 feet  Increase the mass of the box to 15 kg  Decrease the bass of the box to 5 kg  Drop the box off of the shelf.
Physics
1 answer:
irakobra [83]3 years ago
7 0
Potential energy is the energy possessed by a body or an object while at rest while kinetic energy is the energy while the body is at motion.
Potential energy is given by Mgh where m is the mass of the object, g is the gravitational acceleration (a constant) and h is the height.
Therefore, potential energy varies directly proportional to the mass of the object and the height.
Thus in this case, an increase in the mass of the box by 15 kg and raising the height of the box by 20 feet, will result to an increase in potential energy of the box.
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Why do earthquakes and volcanoes occur along continent and ocean boundaries?
jeyben [28]
Earths plates move, creating them to rub together to create earthquakes, and volcanoes occur because magma melts rocks and the pressure is so intense it causes the lava to leak out of the top of a mountain

4 0
3 years ago
Read 2 more answers
1) On the way to the moon, the Apollo astro-
kramer
(1) You must find the point of equilibrium between the two forces,

<span>G * <span><span><span>MT</span><span>ms / </span></span><span>(R−x)^2 </span></span>= G * <span><span><span>ML</span><span>ms / </span></span><span>x^2
MT / (R-x)^2 = ML / x^2

So,

x = R * sqrt(ML * MT) - ML / (MT - ML)
R = is the distance between Earth and Moon.

</span></span></span>The result should be,
x = 3.83 * 10^7m
from the center of the Moon, and 

R - x = 3.46*10^8 m
from the center of the Earth.


(2) As the distance from the center of the Earth is the number we found before,
d = R - x = 3.46*10^8m
The acceleration at this point is
g = G * MT / d^2
g = 3.33*10^-3 m/s^2
6 0
3 years ago
Consider a Hydrogen atom with the electron in the n 8 shell. What is the energy of this system? (The magnitude of the ground sta
Shtirlitz [24]

Answer:

The energy of an electron in the 8th shell is given by:  -0.2125 eV

The number of subshells is:  8

The number of orbitals is:  64

The number of electrons that fit on this shell is: 128

Explanation:

First, we find the energy of the electrons in the 8th shell. In order to do this, we recall that the energy of an electron (in the Hydrogen atom) whose principal number is n is given by:

E_{n}=-13.6\frac{1}{n^{2}}

Substituting n=8, we find that the energy is given by:

E_{8} = -13.6\frac{1}{8^{2}}=-0.2125

In order to find the number of subshells we recall that, for a given principal quantum number n, the possible values of the quantum number l, which corresponds to the number of subshells are:

0, 1, 2, ... , n-1

Since n = 8 in our problem, the possible values of l are: 0, 1, 2, 3, 4, 5, 6, 7. Therefore, the number of subshells are 8.

Now we continue with the number of orbitals. For every subshell l, we have 2l+1 possible values of m, which correspond to the orbitals. Since the possible values of l are: 0,1,2,3,4,5,6,7, therefore, we have to perform the sum:

\sum_{l=0}^{7}(2l+1) = 8^2=64

And we can conclude that the number of orbitals is equal to 64.

Finally, we know that we can fit two electrons per orbital, therefore we can have 64*2 = 128 electrons in the shell corresponding to n=8.

8 0
3 years ago
S Problem Set<br> 2.) 6.4 x 109 nm to cm
anyanavicka [17]

Answer:

6.4\cdot 10^2 cm

Explanation:

First of all, let's convert from nanometres to metres, keeping in mind that

1 nm = 10^{-9} m

So we have:

6.4\cdot 10^9 nm \cdot 10^{-9} m/nm = 6.4 m

Now we can convert from metres to centimetres, keeping in mind that

1 m = 10^2 cm

So, we find:

6.4 m \cdot 10^2 cm/m = 6.4\cdot 10^2 cm

8 0
3 years ago
List small/average stars<br><br>​
mario62 [17]

Answer:

Lol, you should do Nate, Bobby, Cindy, Joe, and Beth

Jk, if you want to be series and probably not fail go for these:

If it wants types of small/average stars, then go with

Small star names:

OGLE-TR-122B

Gliese 229 B

TRAPPIST-1

Teegarden's Star

Luyten 726-8 (A and B)

Proxima Centauri

Wolf 359 111400

Ross 248

Barnard's Star

CM Draconis B

Ross 154 167000

CM Draconis A

Kapteyn's Star

7 0
3 years ago
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