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Luden [163]
2 years ago
5

Why aren't the electrons counted in the mass of an atom?

Physics
1 answer:
Radda [10]2 years ago
3 0

Answer:

Explanation:

The mass of an electron is small compared to the neutron and proton, so it's negligible. It's like adding zero to a number.

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A block is attached to a horizontal spring and oscillates back and forth on a frictionless horizontal surface at a frequency of
Nastasia [14]

Answer:

Amplitude = 0.058m

Frequency = 6.25Hz

Explanation:

Given

Amplitude (A) = 8.26 x 10-2 m

Frequency (f) = 4.42Hz

Conversation of energy before split

½mv² = ½KA²

Make A the subject of formula

A = v\sqrt{\frac{m}{k} }

Conversation of energy after split

½(m/2)V'² = ½(m/2)V² = ½KA'²

½(m/2)V² = ½KA'²

Make A the subject of formula

First divide both sides by ½

(m/2)V² = KA'²

Divide both sides by K

\frac{m}{2K}V² = A'²

v\sqrt{\frac{m}{2k} } = A'

Substitute v\sqrt{\frac{m}{k} } for A in the above equation

A' = A/√2

A' = 8.26 x 10^-2/√2

A' = 0.05840702012600882

Amplitude after split = 0.058 (Approximated)

Frequency (f') = f√2

f' = 4.42√2

f' = 6.25082394568908011

Frequency after split = 6.25Hz (approximated)

8 0
3 years ago
If I get a spare in one frame and then roll a 1 and 5 in the next frame, what would my score for the spare be?
bonufazy [111]
The answer is letter choice B...
8 0
2 years ago
Distinguish between directly proportional and inversely proportional with an example of each
RideAnS [48]

Consider two variables said to be "inversely proportional" to each other. If all other variables are held constant, the magnitude or absolute value of one inversely proportional variable decreases if the other variable increases, while their product (the constant of proportionality k) is always the same.

4 0
3 years ago
Kinetic energy is the energy of . It increases with and .Potential energy is the energy of or . It increases with , and
lorasvet [3.4K]

Answer:

k is the energy of motion it increase with four of load you carry

8 0
2 years ago
The speed of a certain electron is 995 km s−1 . If the uncertainty in its momentum is to be reduced to 0.0010 per cent, what unc
Tpy6a [65]

Answer:

The uncertainty in the location that must be tolerated is 1.163 * 10^{-5} m

Explanation:

From the uncertainty Principle,

Δ_{y} Δ_{p} = \frac{h}{2\pi }

The momentum P_{y} = (mass of electron)(speed of electron)

                                = (9.109 * 10^{-31}kg)(995 * 10^{3} m/s)

                                = 9.0638 * 10^{-25}kgm/s

If the uncertainty is reduced to a 0.0010%, then momentum

                              = 9.068 * 10^{-30}kgm/s

Thus the uncertainty in the position would be:

                              Δ_{y} = \frac{h}{2\pi } * \frac{1}{9.068 * 10^{-30} }

                              Δ_{y} \geq  1.163 * 10^{-5}m

5 0
3 years ago
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