Answer:
I think no.2 the answer
Because socialization and social resources are both for me
Answer:
finding Cepheid variable and measuring their periods.
Explanation:
This method is called finding Cepheid variable and measuring their periods.
Cepheid variable is actually a type of star that has a radial pulsation having a varying brightness and diameter. This change in brightness is very well defined having a period and amplitude.
A potent clear link between the luminosity and pulsation period of a Cepheid variable developed Cepheids as an important determinants of cosmic criteria for scaling galactic and extra galactic distances. Henrietta Swan Leavitt revealed this robust feature of conventional Cepheid in 1908 after observing thousands of variable stars in the Magellanic Clouds. This in fact turn, by making comparisons its established luminosity to its measured brightness, allows one to evaluate the distance to the star.
Answer:
(a) r = 1.062·R
= ![\frac{531}{500} R_E](https://tex.z-dn.net/?f=%5Cfrac%7B531%7D%7B500%7D%20R_E)
(b) r = ![\frac{33}{25} R_E](https://tex.z-dn.net/?f=%5Cfrac%7B33%7D%7B25%7D%20R_E)
(c) Zero
Explanation:
Here we have escape velocity v
given by
and the maximum height given by
![\frac{1}{2} v^2-\frac{GM}{R_E} = -\frac{GM}{r}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20v%5E2-%5Cfrac%7BGM%7D%7BR_E%7D%20%3D%20-%5Cfrac%7BGM%7D%7Br%7D)
Therefore, when the initial speed is 0.241v
we have
v =
so that;
v² =
v² = ![{\frac{0.116162\times GM}{R_E} }](https://tex.z-dn.net/?f=%7B%5Cfrac%7B0.116162%5Ctimes%20GM%7D%7BR_E%7D%20%7D)
is then
![\frac{1}{2} {\frac{0.116162\times GM}{R_E} }-\frac{GM}{R_E} = -\frac{GM}{r}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20%7B%5Cfrac%7B0.116162%5Ctimes%20GM%7D%7BR_E%7D%20%7D-%5Cfrac%7BGM%7D%7BR_E%7D%20%3D%20-%5Cfrac%7BGM%7D%7Br%7D)
Which gives
or
r = 1.062·R
(b) Here we have
![K_i = 0.241\times \frac{1}{2} \times m \times v_e^2 = 0.241\times \frac{1}{2} \times m \times \frac{2GM}{R_E} = \frac{0.241mGM}{R_E}](https://tex.z-dn.net/?f=K_i%20%3D%200.241%5Ctimes%20%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%20m%20%5Ctimes%20v_e%5E2%20%3D%200.241%5Ctimes%20%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%20m%20%20%5Ctimes%20%5Cfrac%7B2GM%7D%7BR_E%7D%20%3D%20%5Cfrac%7B0.241mGM%7D%7BR_E%7D)
Therefore we put
in the maximum height equation to get
![\frac{0.241}{R_E} -\frac{1}{R_E} =-\frac{1}{r}](https://tex.z-dn.net/?f=%5Cfrac%7B0.241%7D%7BR_E%7D%20-%5Cfrac%7B1%7D%7BR_E%7D%20%3D-%5Cfrac%7B1%7D%7Br%7D)
From which we get
r = 1.32·R
(c) The we have the least initial mechanical energy, ME given by
ME = KE - PE
Where the KE = PE required to leave the earth we have
ME = KE - KE = 0
The least initial mechanical energy to leave the earth is zero.
Answer:
h=12.41m
Explanation:
N=392
r=0.6m
w=24 rad/s
![I=0.8*m*r^{2}](https://tex.z-dn.net/?f=I%3D0.8%2Am%2Ar%5E%7B2%7D)
So the weight of the wheel is the force N divide on the gravity and also can find momentum of inertia to determine the kinetic energy at motion
![N=m*g\\m=\frac{N}{g}\\m=\frac{392N}{9.8\frac{m}{s^{2}}}](https://tex.z-dn.net/?f=N%3Dm%2Ag%5C%5Cm%3D%5Cfrac%7BN%7D%7Bg%7D%5C%5Cm%3D%5Cfrac%7B392N%7D%7B9.8%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D%7D)
![m=40kg](https://tex.z-dn.net/?f=m%3D40kg)
moment of inertia
![M_{I}=0.8*40.0kg*(0.6m} )^{2}\\M_{I}=11.5 kg*m^{2}](https://tex.z-dn.net/?f=M_%7BI%7D%3D0.8%2A40.0kg%2A%280.6m%7D%20%29%5E%7B2%7D%5C%5CM_%7BI%7D%3D11.5%20kg%2Am%5E%7B2%7D)
Kinetic energy of the rotation motion
![K_{r}=\frac{1}{2}*I*W^{2}\\K_{r}=\frac{1}{2}*11.52kg*m^{2}*(24\frac{rad}{s})^{2}\\K_{r}=3317.76J](https://tex.z-dn.net/?f=K_%7Br%7D%3D%5Cfrac%7B1%7D%7B2%7D%2AI%2AW%5E%7B2%7D%5C%5CK_%7Br%7D%3D%5Cfrac%7B1%7D%7B2%7D%2A11.52kg%2Am%5E%7B2%7D%2A%2824%5Cfrac%7Brad%7D%7Bs%7D%29%5E%7B2%7D%5C%5CK_%7Br%7D%3D3317.76J)
Kinetic energy translational
![K_{t}=\frac{1}{2}*m*v^{2}\\v=w*r\\v=24rad/s*0.6m=14.4 \frac{m}{s}\\K_{t}=\frac{1}{2}*40kg*(14.4\frac{m}{s})^{2}\\K_{t}=4147.2J](https://tex.z-dn.net/?f=K_%7Bt%7D%3D%5Cfrac%7B1%7D%7B2%7D%2Am%2Av%5E%7B2%7D%5C%5Cv%3Dw%2Ar%5C%5Cv%3D24rad%2Fs%2A0.6m%3D14.4%20%5Cfrac%7Bm%7D%7Bs%7D%5C%5CK_%7Bt%7D%3D%5Cfrac%7B1%7D%7B2%7D%2A40kg%2A%2814.4%5Cfrac%7Bm%7D%7Bs%7D%29%5E%7B2%7D%5C%5CK_%7Bt%7D%3D4147.2J)
Total kinetic energy
![K=3317.79J+4147.2J\\K=7464.99J](https://tex.z-dn.net/?f=K%3D3317.79J%2B4147.2J%5C%5CK%3D7464.99J)
Now the work done by the friction is acting at the motion so the kinetic energy and the work of motion give the potential work so there we can find height
![K-W=E_{p}\\7464.99-2600J=m*g*h\\4864.99J=m*g*h\\h=\frac{4864.99J}{m*g}\\h=\frac{4864.99J}{392N}\\h=12.41m](https://tex.z-dn.net/?f=K-W%3DE_%7Bp%7D%5C%5C7464.99-2600J%3Dm%2Ag%2Ah%5C%5C4864.99J%3Dm%2Ag%2Ah%5C%5Ch%3D%5Cfrac%7B4864.99J%7D%7Bm%2Ag%7D%5C%5Ch%3D%5Cfrac%7B4864.99J%7D%7B392N%7D%5C%5Ch%3D12.41m)
Answer: B = 1380T
Explanation: please find the attached file for the solution