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VashaNatasha [74]
3 years ago
8

When the displacement of a mass on a spring is 1/2a the half of the amplitude, what fraction of the mechanical energy is kinetic

energy?
Physics
2 answers:
Serhud [2]3 years ago
6 0

Answer:

KE : TE = 3 : 4

Explanation:

As we know that the total mechanical energy of the object which is executing SHM is given by

E_{total} = \frac{1}{2}KA^2

here we know that

A = amplitude of SHM

K = spring constant

now we know that total mechanical energy of the spring is always constant so here we can say

kinetic energy + Potential energy = total mechanical energy

we know that potential energy of the spring at any given distance is

U = \frac{1}{2}kx^2

at given position x = A/2

U = \frac{1}{2}K(\frac{A}{2})^2 = \frac{1}{8}KA^2

now we have

KE + \frac{1}{8}KA^2 = \frac{1}{2}KA^2

KE = \frac{3}{8}KA^2

now ratio of kinetic energy and total mechanical energy will be given as

KE : TE = \frac{3}{8}KA^2 : \frac{1}{2}KA^2

KE : TE = 3 : 4

USPshnik [31]3 years ago
5 0
Total energy is a spring:
E = \frac{1}{2} kx^2 +  \frac{1}{2} mv^2 =  \frac{1}{2} ka^2

At x = 0.5a:
\frac{1}{2} k \frac{a}{2} ^2 +  \frac{1}{2} mv^2 =  \frac{1}{2} ka^2 \\  \frac{1}{2} mv^2 =  \frac{1}{2} ka^2 -  \frac{1}{8} ka^2 =  \frac{3}{8} ka^2

The ration:
\frac{ \frac{3}{8}ka^2 }{ \frac{1}{2} ka^2} =  \frac{3}{4}
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A cosmic-ray proton in interstellar space has an energy of 10.0 MeV and executes a circular orbit having a radius equal to that
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Answer:

= 7.88 × 10^-12 T

Explanation:

From the above question, we are told that:

Kinetic Energy of the proton is K. E = 10.0 MeV

Step 1

We convert 10.0 MeV to Joules

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10.0 MeV = 10.0 × 1.602 × 10^-13 Joules = 1.602 × 10^-12 J

Hence, the Kinectic energy of a proton = 1.602 × 10^-12 J

Step 2

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The formula for Kinectic Energy =

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Where

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Speed of the proton = ?

1.602 × 10^-12J = 1/2 × 1.6726219 × 10^-27 × v²

1.602 × 10^-12J = 8.3631095 ×10^-28 × v²

v² = 1.602 × 10^-12/8.3631095 ×10^-28

v = √(1.602 × 10^-12/8.3631095 ×10^-28)

v = 43772331.227m/s

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Step 3

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The formula for Magnetic Field =

B = m v / q r

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mv = √2m(KE)

m = Mass of the proton = 1.6726219 × 10^-27 kg

K.E of the proton = 1.602 × 10^-12 J

v = speed of the proton = 4.4 × 10^7 m/s

q = Electric charge = 1.6 × 10^-19 C

r = radius of the orbit = 5.80Ã10^10 m

= 5.8 × 10^10m

Magnetic Field =

=√ (2 × 1.6726219 × 10^-27 kg × 1.602 × 10^-12 J) /( 1.6 × 10^-19 C × 5.80 × 10^10 m)

= 7.88 × 10^-12 T

The magnetic field in that region of space is approximately 7.88 × 10^-12 T

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