Answer:
please the answer below
Explanation:
(a) If we assume that our origin of coordinates is at the position of charge q1, we have that the potential in both points is
![V_1=k\frac{q_1}{r-1.0}-k\frac{q_2}{1.0}=0\\\\V_2=k\frac{q_1}{r+5.2}-k\frac{q_2}{5.2}=0\\\\](https://tex.z-dn.net/?f=V_1%3Dk%5Cfrac%7Bq_1%7D%7Br-1.0%7D-k%5Cfrac%7Bq_2%7D%7B1.0%7D%3D0%5C%5C%5C%5CV_2%3Dk%5Cfrac%7Bq_1%7D%7Br%2B5.2%7D-k%5Cfrac%7Bq_2%7D%7B5.2%7D%3D0%5C%5C%5C%5C)
k=8.89*10^9
For both cases we have
![k\frac{q_1}{r-1.0}=k\frac{q_2}{1.0}\\\\q_1(1.0)=q_2(r-1.0)\\\\r=\frac{q_1+q_2}{q_2}\\\\k\frac{q_1}{r+5.2}=k\frac{q_2}{5.2}\\\\q_1(5.2)=q_2(r+5.2)\\\\r=\frac{5.2q_1-5.2q_2}{q_2}](https://tex.z-dn.net/?f=k%5Cfrac%7Bq_1%7D%7Br-1.0%7D%3Dk%5Cfrac%7Bq_2%7D%7B1.0%7D%5C%5C%5C%5Cq_1%281.0%29%3Dq_2%28r-1.0%29%5C%5C%5C%5Cr%3D%5Cfrac%7Bq_1%2Bq_2%7D%7Bq_2%7D%5C%5C%5C%5Ck%5Cfrac%7Bq_1%7D%7Br%2B5.2%7D%3Dk%5Cfrac%7Bq_2%7D%7B5.2%7D%5C%5C%5C%5Cq_1%285.2%29%3Dq_2%28r%2B5.2%29%5C%5C%5C%5Cr%3D%5Cfrac%7B5.2q_1-5.2q_2%7D%7Bq_2%7D)
(b) by replacing this values of r in the expression for V we obtain
![k\frac{q_1}{\frac{5.2(q_1-q_2)}{q_2}+5.2}=k\frac{q_2}{5.2}\\\\\frac{q_1}{q_2}=\frac{(q_1-q_2)}{q_2}-1.0=\frac{q_1-q_2-q_2}{q_2}=\frac{q_1-2q_2}{q_2}](https://tex.z-dn.net/?f=k%5Cfrac%7Bq_1%7D%7B%5Cfrac%7B5.2%28q_1-q_2%29%7D%7Bq_2%7D%2B5.2%7D%3Dk%5Cfrac%7Bq_2%7D%7B5.2%7D%5C%5C%5C%5C%5Cfrac%7Bq_1%7D%7Bq_2%7D%3D%5Cfrac%7B%28q_1-q_2%29%7D%7Bq_2%7D-1.0%3D%5Cfrac%7Bq_1-q_2-q_2%7D%7Bq_2%7D%3D%5Cfrac%7Bq_1-2q_2%7D%7Bq_2%7D)
hope this helps!!
They are fused in the core of the star due to great pressures and temperatures. They are made all the way through iron. At that point the star dies. If it is a really large star it will become a supernova when it dies, creating all of the elements beyond iron as well, but only in its death. No star can create anything beyond iron in its life cycle
Answer:
the time interval that an earth observer measures is 4 seconds
Explanation:
Given the data in the question;
speed of the spacecraft as it moves past the is 0.6 times the speed of light
we know that speed of light c = 3 × 10⁸ m/s
so speed of spacecraft v = 0.6 × c = 0.6c
time interval between ticks of the spacecraft clock Δt₀ = 3.2 seconds
Now, from time dilation;
t = Δt₀ / √( 1 - ( v² / c² ) )
t = Δt₀ / √( 1 - ( v/c )² )
we substitute
t = 3.2 / √( 1 - ( 0.6c / c )² )
t = 3.2 / √( 1 - ( 0.6 )² )
t = 3.2 / √( 1 - 0.36 )
t = 3.2 / √0.64
t = 3.2 / 0.8
t = 4 seconds
Therefore, the time interval that an earth observer measures is 4 seconds
Answer: 2561.7 pounds
Explanation:
If we assume the total weight of an airplane (in pounds units) as a <u>linear function</u> of the amount of fuel in its tank (in gallons) and we make a Weight vs amount of fuel graph, which resulting slope is 5.7, we can use the slope equation of the line:
(1)
Where:
is the slope of the line
is the airplane weight with 51 gallons of fuel in its tank (assuming we chose the Y axis for the airplane weight in the graph)
is the fuel in airplane's tank for a total weigth of 2390.7 pounds (assuming we chose the X axis for the a,ount of fuel in the tank in the graph)
This means we already have one point of the graph, which coordinate is:
![(X_{1},Y_{1})=(51,2390.7)](https://tex.z-dn.net/?f=%28X_%7B1%7D%2CY_%7B1%7D%29%3D%2851%2C2390.7%29)
Rewritting (1):
(2)
As Y is a function of X:
(3)
Substituting the known values:
(4)
(5)
(6)
Now, evaluating this function when X=81 (talking about the 81 gallons of fuel in the tank):
(7)
(8) This means the weight of the plane when it has 81 gallons of fuel in its tank is 2561.7 pounds.
To solve this problem we will apply the definitions given in Newtonian theory about the Force of gravity, and the Force caused by weight. Both will be defined below, and in equal equilibrium condition to clear the variable concerning acceleration due to gravity. Finally, with the values provided in the statement, it will be replaced.
The equation for the gravitational force between the Earth and the object on the surface of the Earth is
![F_g = \frac{Gm_em_o}{r^2_e}](https://tex.z-dn.net/?f=F_g%20%3D%20%5Cfrac%7BGm_em_o%7D%7Br%5E2_e%7D)
Where,
G = Universal gravitational constant
= Mass of Earth
= Distance between object and center of earth
= Mass of Object
The equation for the gravitational pulling force on the object due to gravitational acceleration is
![F_g = m_o g](https://tex.z-dn.net/?f=F_g%20%3D%20m_o%20g)
Equation the two expression we have
![m_o g = \frac{Gm_em_o}{r_e^2}](https://tex.z-dn.net/?f=m_o%20g%20%3D%20%5Cfrac%7BGm_em_o%7D%7Br_e%5E2%7D)
![g = \frac{Gm_e}{r_e^2}](https://tex.z-dn.net/?f=g%20%3D%20%5Cfrac%7BGm_e%7D%7Br_e%5E2%7D)
This the acceleration due to gravity which is composite constant.
Replacing with our values we have then
![g = \frac{(6.67*10^{-11}N\cdot m^2/kg^2)(5.98*10^{24}kg)}{6378km(\frac{10^3m}{1km})^2}](https://tex.z-dn.net/?f=g%20%3D%20%5Cfrac%7B%286.67%2A10%5E%7B-11%7DN%5Ccdot%20m%5E2%2Fkg%5E2%29%285.98%2A10%5E%7B24%7Dkg%29%7D%7B6378km%28%5Cfrac%7B10%5E3m%7D%7B1km%7D%29%5E2%7D)
![g = 9.8m/s^2](https://tex.z-dn.net/?f=g%20%3D%209.8m%2Fs%5E2)
The value of composite constant is
. Here, the composite constant is nothing but the acceleration due to gravity which is constant always.