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hram777 [196]
3 years ago
12

A small piece of aluminum (atomic number 13) contains 10^15 atoms (The atomic number is the number of protons; it determines the

(positive) electric charge of the nucleus and, thus, the number of electrons in a neutral atom.) If the piece of aluminum has a net positive charge of 3.0 uc what fraction of the electrons that the aluminum had when it was neutral would have had to be lost?
Physics
1 answer:
Otrada [13]3 years ago
8 0

Answer:

3 micro coulombs = 3 * 10E-6 coulombs     charge of aluminum

13 * 10E15 * 1.6 * 10E-19 = 2.08 E-3 Coulombs - charge of atoms in Al

3 * 10E-6 / 2.08 * E-3 = 1.44 * E-3 = .00144 = .144 %

.00144 of the original electrons would have to be lost

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8 0
4 years ago
Read 2 more answers
If the kinetic energy of the 40kg box is 784 J, what is the velocity before it strikes the ground?
mamaluj [8]

Answer:

Explanation:

KE=\frac{1}{2}mv^2

784=\frac{1}{2}(40)v^2

784=20v^2

39.2=v^2

v=6.26m/s

4 0
3 years ago
Which example best represents translational kenetic energy
Mila [183]

Answer:

an apple falling off a tree

Explanation:

5 0
3 years ago
The electric potential at the origin of an xy-coordinate system is 40 V. A -8.0-μC charge is brought from x = +∞ to that point.
vredina [299]

Answer:

-320 μJ.

Explanation:

Consider a point with an electrical charge of q. Assume that V is the electrical potential at the position of that charge. The electrical potential of that point charge will be equal to:

\text{Potential Energy} = q \cdot V.

Keep in mind that since both q and V might not be positive, the size of the electrical potential energy might not be positive, either.

For this point charge,

  • q = \rm -8.0\; \mu C; (that's -8.0 microjoules, which equals to \rm -8.0\times 10^{-6}\; J)
  • V = \rm 40\; V.

Hence its electrical potential energy:

\text{Potential Energy} = q\cdot V = \rm (-8.0\; \mu C) \times 40\; V = -320\; \mu J.

Why is this value negative? The electrical potential energy of a charge is equal to the work needed to bring that charge from infinitely far away all the way to its current position. Also, negative charges are attracted towards regions of high electrical potential. Bringing this \rm -8.0\; \mu C negative charge to the origin will not require any external work. Instead, this process will release 320 μJ of energy. As a result, the electrical potential energy is a negative value.

7 0
3 years ago
A car accelerates from rest at 3.6 m/s 2 . How much time does it need to attain a speed of 5 m/s?
Olenka [21]

car starts from rest

v_i = 0

final speed attained by the car is

v_f = 5 m/s

acceleration of the car will be

a = 3.6 m/s^2

now the time to reach this final speed will be

t = \frac{v_f - v_i}{a}

t = \frac{5 - 0}{3.6}

t = 1.39 s

so it required 1.39 s to reach this final speed

6 0
4 years ago
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