One-twelfth the mass of carbon-12 atom
Answer:
It's called empathy
Explanation:
All you gotta do is think about it for a minute
To solve this problem it is necessary to apply the concepts related to intensity as a function of power and area.
Intensity is defined to be the power per unit area carried by a wave. Power is the rate at which energy is transferred by the wave. In equation form, intensity I is
![I = \frac{P}{A}](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7BP%7D%7BA%7D)
The area of a sphere is given by
![A = 4\pi r^2](https://tex.z-dn.net/?f=A%20%3D%204%5Cpi%20r%5E2)
So replacing we have to
![I = \frac{P}{4\pi r^2}](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7BP%7D%7B4%5Cpi%20r%5E2%7D)
Since the question tells us to find the proportion when
![r_1 = 5r_2 \rightarrow \frac{r_2}{r_1} = \frac{1}{5}](https://tex.z-dn.net/?f=r_1%20%3D%205r_2%20%5Crightarrow%20%5Cfrac%7Br_2%7D%7Br_1%7D%20%3D%20%5Cfrac%7B1%7D%7B5%7D)
So considering the two intensities we have to
![I_1 = \frac{P_1}{4\pi r_1^2}](https://tex.z-dn.net/?f=I_1%20%3D%20%5Cfrac%7BP_1%7D%7B4%5Cpi%20r_1%5E2%7D)
![I_2 = \frac{P_2}{4\pi r_2^2}](https://tex.z-dn.net/?f=I_2%20%3D%20%5Cfrac%7BP_2%7D%7B4%5Cpi%20r_2%5E2%7D)
The ratio between the two intensities would be
![\frac{I_1}{I_2} = \frac{ \frac{P_1}{4\pi r_1^2}}{\frac{P_2}{4\pi r_2^2}}](https://tex.z-dn.net/?f=%5Cfrac%7BI_1%7D%7BI_2%7D%20%3D%20%5Cfrac%7B%20%5Cfrac%7BP_1%7D%7B4%5Cpi%20r_1%5E2%7D%7D%7B%5Cfrac%7BP_2%7D%7B4%5Cpi%20r_2%5E2%7D%7D)
The power does not change therefore it remains constant, which allows summarizing the expression to
![\frac{I_1}{I_2}=(\frac{r_2}{r_1})^2](https://tex.z-dn.net/?f=%5Cfrac%7BI_1%7D%7BI_2%7D%3D%28%5Cfrac%7Br_2%7D%7Br_1%7D%29%5E2)
Re-arrange to find ![I_2](https://tex.z-dn.net/?f=I_2)
![I_2 = I_1 (\frac{r_1}{r_2})^2](https://tex.z-dn.net/?f=I_2%20%3D%20I_1%20%28%5Cfrac%7Br_1%7D%7Br_2%7D%29%5E2)
![I_2 = 100*(\frac{1}{5})^2](https://tex.z-dn.net/?f=I_2%20%3D%20100%2A%28%5Cfrac%7B1%7D%7B5%7D%29%5E2)
![I_2 = 4W/m^2](https://tex.z-dn.net/?f=I_2%20%3D%204W%2Fm%5E2)
Therefore the intensity at five times this distance from the source is ![4W/m^2](https://tex.z-dn.net/?f=%204W%2Fm%5E2)
Answer:
490.5 J
Explanation:
m = Mass of rock = 10 kg
g = Acceleration due to gravity = 9.81 m/s²
h = Height from the ground = 5 m
The potential energy gives the energy that the body posseses due to its mass and height of the fall.
Potential energy is given by
![U=mgh](https://tex.z-dn.net/?f=U%3Dmgh)
![\Rightarrow U=10\times 9.81\times 5](https://tex.z-dn.net/?f=%5CRightarrow%20U%3D10%5Ctimes%209.81%5Ctimes%205)
![\Rightarrow U=490.5\ J](https://tex.z-dn.net/?f=%5CRightarrow%20U%3D490.5%5C%20J)
The potential energy of the rock which is at the given height is 490.5 J
Answer:
a) ![v_{max}=0.4\ m.s^{-1}](https://tex.z-dn.net/?f=v_%7Bmax%7D%3D0.4%5C%20m.s%5E%7B-1%7D)
b) ![a_{max}=1.6\ m.s^{-2}](https://tex.z-dn.net/?f=a_%7Bmax%7D%3D1.6%5C%20m.s%5E%7B-2%7D)
c) ![v_x=0.32\ m.s^{-1}](https://tex.z-dn.net/?f=v_x%3D0.32%5C%20m.s%5E%7B-1%7D)
d) ![a_x=0.96\ m.s^{-1}](https://tex.z-dn.net/?f=a_x%3D0.96%5C%20m.s%5E%7B-1%7D)
e) ![\Delta t=0.232\ s](https://tex.z-dn.net/?f=%5CDelta%20t%3D0.232%5C%20s)
Explanation:
Given:
mass of the object attached to the spring, ![m=0.5\ kg](https://tex.z-dn.net/?f=m%3D0.5%5C%20kg)
spring constant of the given spring, ![k=8\ N.m^{-1}](https://tex.z-dn.net/?f=k%3D8%5C%20N.m%5E%7B-1%7D)
amplitude of vibration, ![A=0.1\ m](https://tex.z-dn.net/?f=A%3D0.1%5C%20m)
a)
Now, maximum velocity is obtained at the maximum Kinetic energy and the maximum kinetic energy is obtained when the whole spring potential energy is transformed.
Max. spring potential energy:
![PE_s=\frac{1}{2} .k.A^2](https://tex.z-dn.net/?f=PE_s%3D%5Cfrac%7B1%7D%7B2%7D%20.k.A%5E2)
![PE_s=0.5\times 8\times 0.1^2](https://tex.z-dn.net/?f=PE_s%3D0.5%5Ctimes%208%5Ctimes%200.1%5E2)
![PE_s=0.04\ J](https://tex.z-dn.net/?f=PE_s%3D0.04%5C%20J)
When this whole spring potential is converted into kinetic energy:
![KE_{max}=0.04\ J](https://tex.z-dn.net/?f=KE_%7Bmax%7D%3D0.04%5C%20J)
![\frac{1}{2}.m.v_{max}^2=0.04](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D.m.v_%7Bmax%7D%5E2%3D0.04)
![0.5\times 0.5\times v_{max}^2=0.04](https://tex.z-dn.net/?f=0.5%5Ctimes%200.5%5Ctimes%20v_%7Bmax%7D%5E2%3D0.04)
![v_{max}=0.4\ m.s^{-1}](https://tex.z-dn.net/?f=v_%7Bmax%7D%3D0.4%5C%20m.s%5E%7B-1%7D)
b)
Max. Force of spring on the mass:
![F_{max}=k.A](https://tex.z-dn.net/?f=F_%7Bmax%7D%3Dk.A)
![F_{max}=8\times 0.1](https://tex.z-dn.net/?f=F_%7Bmax%7D%3D8%5Ctimes%200.1)
![F_{max}=0.8\ N](https://tex.z-dn.net/?f=F_%7Bmax%7D%3D0.8%5C%20N)
Now acceleration:
![a_{max}=\frac{F_{max}}{m}](https://tex.z-dn.net/?f=a_%7Bmax%7D%3D%5Cfrac%7BF_%7Bmax%7D%7D%7Bm%7D)
![a_{max}=\frac{0.8}{0.5}](https://tex.z-dn.net/?f=a_%7Bmax%7D%3D%5Cfrac%7B0.8%7D%7B0.5%7D)
![a_{max}=1.6\ m.s^{-2}](https://tex.z-dn.net/?f=a_%7Bmax%7D%3D1.6%5C%20m.s%5E%7B-2%7D)
c)
Kinetic energy when the displacement is,
:
![KE_x=PE_s-PE_x](https://tex.z-dn.net/?f=KE_x%3DPE_s-PE_x)
![\frac{1}{2} .m.v_x^2=PE_s-\frac{1}{2} .k.\Delta x^2](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20.m.v_x%5E2%3DPE_s-%5Cfrac%7B1%7D%7B2%7D%20.k.%5CDelta%20x%5E2)
![\frac{1}{2}\times 0.5\times v_x^2=0.04-\frac{1}{2} \times 8\times 0.06^2](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5Ctimes%200.5%5Ctimes%20v_x%5E2%3D0.04-%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%208%5Ctimes%200.06%5E2)
![v_x=0.32\ m.s^{-1}](https://tex.z-dn.net/?f=v_x%3D0.32%5C%20m.s%5E%7B-1%7D)
d)
Spring force on the mass at the given position,
:
![F=k.\Delta x](https://tex.z-dn.net/?f=F%3Dk.%5CDelta%20x)
![F=8\times 0.06](https://tex.z-dn.net/?f=F%3D8%5Ctimes%200.06)
![F=0.48\ N](https://tex.z-dn.net/?f=F%3D0.48%5C%20N)
therefore acceleration:
![a_x=\frac{F}{m}](https://tex.z-dn.net/?f=a_x%3D%5Cfrac%7BF%7D%7Bm%7D)
![a_x=\frac{0.48}{0.5}](https://tex.z-dn.net/?f=a_x%3D%5Cfrac%7B0.48%7D%7B0.5%7D)
![a_x=0.96\ m.s^{-1}](https://tex.z-dn.net/?f=a_x%3D0.96%5C%20m.s%5E%7B-1%7D)
e)
Frequency of oscillation:
![\omega=\sqrt{\frac{k}{m} }](https://tex.z-dn.net/?f=%5Comega%3D%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%7D%20%7D)
![\omega=\sqrt{\frac{8}{0.5} }](https://tex.z-dn.net/?f=%5Comega%3D%5Csqrt%7B%5Cfrac%7B8%7D%7B0.5%7D%20%7D)
![\omega=4\ rad.s^{-1}](https://tex.z-dn.net/?f=%5Comega%3D4%5C%20rad.s%5E%7B-1%7D)
So the wave equation is:
![x=A.\sin\ (\omega.t)](https://tex.z-dn.net/?f=x%3DA.%5Csin%5C%20%28%5Comega.t%29)
where x = position of the oscillating mass
put x=0
![0=0.1\times \sin\ (4t)](https://tex.z-dn.net/?f=0%3D0.1%5Ctimes%20%5Csin%5C%20%284t%29)
![t=0\ s](https://tex.z-dn.net/?f=t%3D0%5C%20s)
Now put x=0.08
![0.08=0.1\times \sin\ (4t)](https://tex.z-dn.net/?f=0.08%3D0.1%5Ctimes%20%5Csin%5C%20%284t%29)
![t=0.232\ s](https://tex.z-dn.net/?f=t%3D0.232%5C%20s)
So, the time taken in going from point x = 0 cm to x = 8 cm is:
![\Delta t=0.232\ s](https://tex.z-dn.net/?f=%5CDelta%20t%3D0.232%5C%20s)