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lilavasa [31]
3 years ago
15

Protons, neutrons, electrons, and a nucleus are

Physics
1 answer:
tatuchka [14]3 years ago
8 0
It would be Atoms, they’re all made up of these tiny particles
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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
How do protons, neutrons and electrons determine the properties of the atoms??? HELP! NEED SCIENCE HELP FAST!!!!
Roman55 [17]
The number of protons determines the element. Many elements also have many isotopes. Such as carbon-14 Carbon-12 and carbon-10.
4 0
3 years ago
Use these options to describe each genotype ​
Sliva [168]
A homozygous dominant
B: homozygous recessive
C: heterozygous
I hope this helps you :)
4 0
3 years ago
The transfer of energy that occurs when a force is applied over a distance is
astra-53 [7]
The transfer of energy that occurs when a force is applied over a distance is WORK.
3 0
3 years ago
A playground merry-go-round has a mass of 50 kg and a diameter of 4.0 m. There are 4 children who want to ride on it. They have
mixer [17]

Answer:

B) I1 = 1680 kg.m^2          I2 = 1120 kg.m^2

C) V = 0.84m/s      T = 29.92s

D) ω2 = 0.315 rad/s

Explanation:

The moment of inertia when they are standing on the edge:

I1 = 1/2*M*R^2 + (m1+m2+m3+m4)*R^2   where M is the mass of the merry-go-round.

I1 = 1680 kg.m^2

The moment of inertia when they are standing half way to the center:

I2 = 1/2*M*R^2 + (m1+m2+m3+m4)*(R/2)^2

I2 = 1120 kg.m^2

The tangencial velocity is given by:

V = ω1*R = 0.84m/s

Period of rotation:

T = 2π / ω1 = 29.92s

Assuming that there is no friction and their parents are not pushing anymore, we can use conservation of the angular momentum to calculate the new angular velocity:

I1*ω1 = I2*ω2    Solving for ω2:

ω2 = I1*ω1 / I2 = 0.315 rad/s

5 0
3 years ago
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