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cricket20 [7]
3 years ago
9

The students conduct experiment 2 in which the same block is connected to the same spring on a horizontal surface. The spring is

stretched a distance L2 beyond its natural length and released from rest, allowing the block-spring system to oscillate. Frictional forces are considered to be negligible. Which of the following claims is correct about how the period of oscillation for the block-spring system in experiment 2 compares with the period of oscillation for the system in experiment 1, and what evidence supports the claim?
Physics
1 answer:
sineoko [7]3 years ago
7 0

Answer:

the period does not change

Explanation:

In a system of connected spring and mass I was able to oscillate in a simple harmonic motion that is described by

          x = A cos (wt + Ф)

Where A is the amplitude of movement, w the angular velocity and Ф the initial phase.

Angular velocity is given by

         w² = k / m

The angular velocity eta related to frequency

          w = 2π f

Frequency and period are inverses

           f = 1 / T

We substitute

     4π² / T² = k / m

          T = 2π √m/k

As we ask to see the period does not depend on the amplitude, nor on the initial displacement, so the period does not change

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A neutron is confined in a one-demensional potential box of width 5.0 x 10^-15 m. Calculate the minimum kinetic energy of the ne
kvv77 [185]

Answer:

E_1=1.31\times 10^{-12}\ J

Explanation:

Given that,

Width of a one dimensional potential box, x=5\times 10^{-15}\ m

The energy of a particle in one dimensional box is given by :

E_n=\dfrac{n^2h^2}{8mx^2}

h = Planck's constant

m = the mass of the proton

For minimum kinetic energy, n = 1

E_1=\dfrac{(6.63\times 10^{-34})^2}{8\times 1.67\times 10^{-27}\times (5\times 10^{-15})^2}

E_1=1.31\times 10^{-12}\ J

So, the minimum kinetic energy of the neutron is 1.31\times 10^{-12}\ J. Hence, this is the required solution.

8 0
3 years ago
Suppose we have two planets with the same mass, but the radius of the second one is twice the size of the first one. How does th
bogdanovich [222]

The free-fall acceleration on the second planet is one-fourth the value of the first planet.

Calculation:

Consider the mass of planet A to be, M

               the mass of planet B to be, Mₓ = M

               the radius of planet A to be, R₁

               the radius of planet B to be, R₂

The acceleration due to gravity on planet A's surface is given as:

g = GM/R₁²      - (1)

Similarly, the acceleration due to gravity on planet B's surface is given as:

g' = GM/R₂²                           [where, R₂ = 2R₁]

   = GM/4R₁²    -(2)

From equation 1 & 2, we get:

g/g' = GM/R₁² ÷ GM/4R₁²

g/g' = 4/1

Thus we get,

g' = 1/4 g

Therefore, the free-fall acceleration on the second planet is one-fourth the value of the first planet.

Learn more about free-fall here:

<u>brainly.com/question/13299152</u>

#SPJ4

6 0
2 years ago
Please answerASAP
Illusion [34]

Answer:

8.9 m/s^2

Explanation:

The period of a simple pendulum is given by the equation

T=2\pi \sqrt{\frac{L}{g}}

where

L is the lenght of the pendulum

g is the acceleration due to gravity at the location of the pendulum

We notice from the formula that the period of a pendulum does not depend on the mass of the system

In this problem:

-The pendulum comes back to the point of release exactly 2.4 seconds after the release. --> this means that the period of the pendulum is

T = 2.4 s

- The length of the pendulum is

L = 1.3 m

Re-arranging the equation for g, we can find the acceleration due to gravity on the planet:

g=(\frac{2\pi}{T})^2 L=(\frac{2\pi}{2.4})^2(1.3)=8.9 m/s^2

7 0
4 years ago
The dances created and performed collectively by the ordinary people.
Readme [11.4K]

Answer:

Folk dance

Explanation:

hope i helped :)

6 0
3 years ago
How do whales bats and dolphins use echolocation?
g100num [7]
C is your answer ithink
3 0
3 years ago
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