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Flauer [41]
4 years ago
13

What is simple harmonic motion​

Physics
2 answers:
juin [17]4 years ago
5 0

simple harmonic motion is a special type of periodic motion or oscillation where the restoring force is directly proportional to the displacement and acts in the direction opposite to that of displacement.

mezya [45]4 years ago
3 0

Explanation:

The motion of a particle moving along a straight line with an acceleration whose direction is always towards a fixed point on the line and whose magnitude is proportional to the distance from the fixed point is called simple harmonic motion

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An unknown charged particle passes without deflection through crossed electric and magnetic fields of strengths 187,500 V/m and
Kobotan [32]

Answer:

Explanation:

Given that

The electric fields of strengths E = 187,500 V/m and

and The magnetic  fields of strengths B = 0.1250 T

The diameter d is 25.05 cm which is converted to 0.2505m

The radius is (d/2)

= 0.2505m / 2 = 0.12525m

The given formula to find the magnetic force is F_{ma}=BqV---(i)

The given formula to find the electric force is F_{el}=qE---(ii)

The velocity of electric field and magnetic field is said to be perpendicular

Electric field is equal to magnectic field

Equate equation (i) and equation (ii)

Bqv=qE\\\\v=\frac{E}{B}

v=\frac{187500}{0.125} \\\\v=15\times10^5m/s

It is said that the particles moves in semi circle, so we are going to consider using centripetal force

F_{ce}=\frac{mv^2}{r}---(iii)

magnectic field is equal to centripetal force

Lets equate equation (i) and (iii)

Bqr=\frac{mv^2}{r} \\\\\frac{q}{m}=\frac{v}{Br}  \\\\\frac{q}{m} =\frac{15\times 10^5}{0.125\times0.12525} \\\\=\frac{15\times10^5}{0.015656} \\\\=95808383.23\\\\=958.1\times10^5C/kg

Therefore,  the particle's charge-to-mass ratio is 958.1\times10^5C/kg

b)

To identify the particle

Then 1/ 958.1 × 10⁵ C/kg

The charge to mass ratio is very close to that of a proton, which is about 1*10^8 C/kg

Therefore the particle is proton.

8 0
4 years ago
A proton of mass is released from rest just above the lower plate and reaches the top plate with speed . An electron of mass is
xenn [34]

Answer:

  v = √ 2e (V₂-V₁) / m

Explanation:

For this exercise we can use the conservation of the energy of the electron

At the highest point. Resting on the top plate

         Em₀ = U = -e V₁

At the lowest point. Just before touching the bottom plate

        Emf = K + U = ½ m v² - e V₂

Energy is conserved

         Em₀ = Emf

          -eV₁ = ½ m v² - e V₂

           v = √ 2e (V₂-V₁) / m

Where e is the charge of the electron, V₂-V₁ is the potential difference applied to the capacitor and m is the mass of the electron

3 0
4 years ago
In what order were the following energy sources<br> discovered by humans
NemiM [27]
There’s no order to go off of...
5 0
3 years ago
Potassium is a crucial element for the healthy operation of the human body. Potassium occurs naturally in our environment and th
Mariulka [41]

Answer:

a) 0.0288 grams

b) 2.6*10^{-10} J/kg

Explanation:

Given that:

A typical human  body contains about 3.0 grams of Potassium per kilogram of body mass

The abundance  for the three isotopes are:

Potassium-39, Potassium-40, and Potassium-41 with abundances are 93.26%, 0.012% and 6.728% respectively.

a)

Thus; a person with a mass of 80 kg will posses = 80 × 3 = 240 grams of potassium.

However, the amount of potassium that is present in such person is :

0.012% × 240 grams

= 0.012/100 × 240 grams

= 0.0288 grams

b)

the effective dose (in Sieverts) per year due to Potassium-40 in an 80- kg body is calculate as follows:

First the Dose in (Gy) = \frac{energy \ absorbed }{mass \ of \ the \ body}

= \frac{1.10*10^6*1.6*10^{-14}}{80}

= 2.2*10^{-10} \ J/kg

Effective dose (Sv) = RBE × Dose in Gy

Effective dose (Sv) =  1.2  *2.2*10^{-10} \ J/kg

Effective dose (Sv) = 2.6*10^{-10} J/kg

 

5 0
4 years ago
Which of the following is an example of potential energy?
Papessa [141]

Answer:

A

Explanation:

potential energy is stored energy so the ball has potential energy to bounce or roll which would then have been converted to kinetic but the rest are in motion meaning energy is no longer stored but used in motion and therefore cannot be potential energy

4 0
3 years ago
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