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Mechanical waves move energy through a medium by vibrating particles. Mechanical waves can't move energy through a vacuum because there is no matter inside of a vacuum. The three types of mechanical waves are transverse waves, surface waves, and longitudinal waves.
Answer:
Mass can never be negative. Everything has mass. Just like how they ask you to find area under the graph in maths. If the area is in the 3rd and 4th quadrant, when calculated, you would get negative answer.However, area can not be negative because it is a place/ location. It's exactly the same as mass.
Answer:
d. We can calculate it by applying Newton's version of Kepler's third law
Explanation:
The measurements of a Star like the Sun have several problems, the first one is distance, but the most important is the temperature since as we get closer all the instruments will melt. This is why all measurements must be indirect because of the effects that these variables create on nearby bodies.
Kepler's laws are deduced from Newton's law of universal gravitation, in these laws the mass of the Sun affects the orbit of the planets since it creates a force of attraction, if measured the orbit and the time it takes to travel it we can know the centripetal acceleration and with it knows the force, from where we clear the mass of the son.
Let's review the statements of the exercise
.a) False. We don't have good enough models for this calculation
.b) False. The size of the sun is very difficult to measure because it is a mass of gas, in addition the density changes strongly with depth
.c) False. The amount of light that comes out of the sun is not all the light produced and is due to quantum effects where the mass of the sun is not taken into account
.d) True. This method has been used to calculate the mass of the sun and the other planets since the variable distance and time are easily measured from Earth
Correct answer is D
Answer:
a) T = 2.26 N, b) v = 1.68 m / s
Explanation:
We use Newton's second law
Let's set a reference system where the x-axis is radial and the y-axis is vertical, let's decompose the tension of the string
sin 30 =
cos 30 =
Tₓ = T sin 30
T_y = T cos 30
Y axis
T_y -W = 0
T cos 30 = mg (1)
X axis
Tₓ = m a
they relate it is centripetal
a = v² / r
we substitute
T sin 30 = m
(2)
a) we substitute in 1
T =
T =
T = 2.26 N
b) from equation 2
v² =
If we know the length of the string
sin 30 = r / L
r = L sin 30
we substitute
v² =
v² =
For the problem let us take L = 1 m
let's calculate
v =
v = 1.68 m / s
Answer:
Explanation:
Calculate the volume of the lead
![V=\frac{m}{d}\\\\=\frac{10g}{11.3g'cm^3}](https://tex.z-dn.net/?f=V%3D%5Cfrac%7Bm%7D%7Bd%7D%5C%5C%5C%5C%3D%5Cfrac%7B10g%7D%7B11.3g%27cm%5E3%7D)
Now calculate the bouyant force acting on the lead
![F_L = Vpg](https://tex.z-dn.net/?f=F_L%20%3D%20Vpg)
![F_L=(\frac{10g}{11.3g/cm^3} )(1g/cm^3)(9.8m/s^2)\\\\=8.673\times 10^{-3}N](https://tex.z-dn.net/?f=F_L%3D%28%5Cfrac%7B10g%7D%7B11.3g%2Fcm%5E3%7D%20%29%281g%2Fcm%5E3%29%289.8m%2Fs%5E2%29%5C%5C%5C%5C%3D8.673%5Ctimes%2010%5E%7B-3%7DN)
This force will act in upward direction
Gravitational force on the lead due to its mass will act in downward direction
Hence the difference of this two force
![T=mg-F_L\\\\=(10\times10^{-3}kg(9.8m/s^2)-8.673\times 10^{-3}\\\\=8.933\times10^{-3}N](https://tex.z-dn.net/?f=T%3Dmg-F_L%5C%5C%5C%5C%3D%2810%5Ctimes10%5E%7B-3%7Dkg%289.8m%2Fs%5E2%29-8.673%5Ctimes%2010%5E%7B-3%7D%5C%5C%5C%5C%3D8.933%5Ctimes10%5E%7B-3%7DN)
If V is the volume submerged in the water then bouyant force on the bobber is
![F_B=V'pg](https://tex.z-dn.net/?f=F_B%3DV%27pg)
Equate bouyant force with the tension and gravitational force
![F_B=T_mg\\\\V'pg=\frac{(8.933\times10^{-2}N)+mg}{pg} \\\\V'=\frac{(8.933\times10^{-2}N)+mg}{pg}](https://tex.z-dn.net/?f=F_B%3DT_mg%5C%5C%5C%5CV%27pg%3D%5Cfrac%7B%288.933%5Ctimes10%5E%7B-2%7DN%29%2Bmg%7D%7Bpg%7D%20%5C%5C%5C%5CV%27%3D%5Cfrac%7B%288.933%5Ctimes10%5E%7B-2%7DN%29%2Bmg%7D%7Bpg%7D)
Now Total volume of bobble is
![\frac{V'}{V^B} =\frac{\frac{(8.933\times10^{-2})+Mg}{pg} }{\frac{4}{3} \pi R^3 }\times100\\\\=\frac{\frac{(8.933\times10^{-2})+(3)(9.8)}{(1000)(9.8)} }{\frac{4}{3} \pi (4.0\times10^{-2})^3 }\times100\\\\](https://tex.z-dn.net/?f=%5Cfrac%7BV%27%7D%7BV%5EB%7D%20%3D%5Cfrac%7B%5Cfrac%7B%288.933%5Ctimes10%5E%7B-2%7D%29%2BMg%7D%7Bpg%7D%20%7D%7B%5Cfrac%7B4%7D%7B3%7D%20%5Cpi%20R%5E3%20%7D%5Ctimes100%5C%5C%5C%5C%3D%5Cfrac%7B%5Cfrac%7B%288.933%5Ctimes10%5E%7B-2%7D%29%2B%283%29%289.8%29%7D%7B%281000%29%289.8%29%7D%20%7D%7B%5Cfrac%7B4%7D%7B3%7D%20%5Cpi%20%284.0%5Ctimes10%5E%7B-2%7D%29%5E3%20%7D%5Ctimes100%5C%5C%5C%5C)
=![\large\boxed{4.52 \%}](https://tex.z-dn.net/?f=%5Clarge%5Cboxed%7B4.52%20%5C%25%7D)