PHYSICS COLLEGE

A block of mass 0.221 kg is placed on top of a light, vertical spring of force constant 5365 N/m and pushed downward so that the spring is compressed by 0.097 m. After the block is released from rest, it travels upward and then leaves the spring. To what maximum height above the point of release does it rise?

Answers

Answer 1
Answer:

Answer:

The maximum height above the point of release is 11.653 m.

Explanation:

Given that,

Mass of block = 0.221 kg

Spring constant k = 5365 N/m

Distance x = 0.097 m

We need to calculate the height

Using stored energy in spring

U=(1)/(2)kx^2...(I)

Using gravitational potential energy

U' =mgh....(II)

Using energy of conservation

E_(i)=E_(f)

U_(i)+U'_(i)=U_(f)+U'_(f)

(1)/(2)kx^2+0=0+mgh

h=(kx^2)/(2mg)

Where, k = spring constant

m = mass of the block

x = distance

g = acceleration due to gravity

Put the value in the equation

h=(5365*(0.097)^2)/(2*0.221*9.8)

h=11.653\ m

Hence, The maximum height above the point of release is 11.653 m.

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Answers

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Explanation:

See attachment

If the speed of light in a medium is 2 x 10^8 m/s, the medium's index of refraction is?

Answers

speed of light in the air is 3 x 10^8
so index of refractions would be speed of light divided by speed in the medium
3/2 = 1.5

Answer: n=1.5

by the way it is glass :) 

The most soaring vocal melody is in Johann Sebastian Bach's Mass in B minor. In one section, the basses, tenors, altos, and sopranos carry the melody from a low D to a high A. In concert pitch, these notes are now assigned frequencies of 146.8 Hz and 880.0 Hz. (Use 343 m/s as the speed of sound, and 1.20 kg/m3 as the density of air.)a. Find the wavelength of the initial note.
b. Find the wavelength of the final note.
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d. Find the pressure amplitude of the final note.
e. Find the displacement amplitude of the initial note.
f. Find the displacement amplitude of the final note.

Answers

Answer:

Detailed step wise solution is attached below

Explanation:

(a) wavelength of the initial note 2.34 meters

(b) wavelength of the final note 0.389 meters

(d) pressure amplitude of the final note 0.09 Pa

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Consider position [x] = L, time [t] = T, velocity [v] = L/T and acceleration [a] = L/T 2 . Find the exponent A in the equation v = a^2 t^ A /x

Answers

Answer:

The exponent A in the equation is 3.

Explanation:

v = a^2 t^ A /x

v = (a^2t^A)/(x) \n\nvx = a^2t^A\n\n((L)/(T))(L) = ((L)/(T^2))^2(T)^A\n\n (L^2)/(T)= ((L^2)/(T^4))(T)^A\n\n (L^2)/(T) *(T^4)/(L^2) = (T)^A\n\nT^3 = T^A\n\n(T^3)/(T^3) = (T^A)/(T^3)\n\nT^(3-3) = T^(A-3)\n\n3-3 = A-3\n\n0 = A-3\n\nA = 3

Therefore, the exponent A in the equation is 3.

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Answers

Answer:

electric flux is 280  Nm²/C  

so correct option is D 280  Nm²/C

Explanation:

radius r = 0.50 m

angle = 30 degree

field strength = 713 N/C

to find out

the electric flux through the surface

solution

we find here electric flux by given formula that is

electric flux = field strength × area× cos∅   .......1

here area = πr² = π(0.50)²

put here all value in equation  1

electric flux = field strength × area× cos∅  

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Answers

Answer:

14.7 psi is equal to 19051.2 pounds per square yard.

Explanation:

Dimensionally speaking, a square yard equals 1296 square inches. Therefore, we need to multiply the atmospheric pressure by 1296 to obtain its equivalent in pounds per square yard. That is:

p = 14.7\,(lbf)/(in^(2))* 1296\,(in^(2))/(yd^(2))

p = 19051.2\,(lbf)/(yd^(2))

14.7 psi is equal to 19051.2 pounds per square yard.

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