Answer:
A) x _electron = 0.66 10² m
, B) x _Eart = 1.13 10² m
, C) d_sphere = 1.37 10⁻² mm
Explanation:
A) Let's use a ball for the nucleus, the electron is at a farther distance the sphere for the electron must be at a distance of
Let's use proportions rule
x_ electron = 0.529 10⁻¹⁰ /1.2 10⁻¹⁵ 1.5
x _electron = 0.66 10⁵ mm = 0.66 10² m
B) the radii of the Earth and the sun are
= 6.37 10⁶ m
tex]R_{Sum}[/tex] = 6.96 10⁸ m
Distance = 1.5 10¹¹ m
x_Earth = 1.5 10¹¹ / 6.96 10⁸ 1.5
x _Eart = 1.13 10² m
C) The radius of a sphere that represents the earth, if the sphere that represents the sun is 1.5 mm, let's use another rule of proportions
d_sphere = 1.5 / 6.96 10⁸ 6.37 10⁶
d_sphere = 1.37 10⁻² mm
Answer:
the length of the now stationary spacecraft = 89.65m
Explanation:
In contraction equation, Length contraction L is the shortening of the measured length of an object moving relative to the observer’s frame.
Thus, it has a formula;
L = L_o(√(1 - (v²/c²))
Where in this question;
L = 71m and v = 0.610 c
Thus;
71 = L_o (√(1 - ((0.61c)²/c²))
c² will cancel out to give;
71 = L_o (√(1 - 0.61²)
71 = L_o (√(1 - 0.61²)
71 = 0.792L_o
L_o = 71/0.792
L_o = 89.65m
Answer: Mass of the planet, M= 8.53 x 10^8kg
Explanation:
Given Radius = 2.0 x 106m
Period T = 7h 11m
Using the third law of kepler's equation which states that the square of the orbital period of any planet is proportional to the cube of the semi-major axis of its orbit.
This is represented by the equation
T^2 = ( 4π^2/GM) R^3
Where T is the period in seconds
T = (7h x 60m + 11m)(60 sec)
= 25860 sec
G represents the gravitational constant
= 6.6 x 10^-11 N.m^2/kg^2 and M is the mass of the planet
Making M the subject of the formula,
M = (4π^2/G)*R^3/T^2
M = (4π^2/ 6.6 x10^-11)*(2×106m)^3(25860s)^2
Therefore Mass of the planet, M= 8.53 x 10^8kg
Answer:
a) E=228391.8 N/C
b) E=-59345.91N/C
Explanation:
You can use Gauss law to find the net electric field produced by both line of charges.
Where E1 and E2 are the electric field generated at a distance of r1 and r2 respectively from the line of charges.
The net electric field at point r will be:
a) for y=0.200m, r1=0.200m and r2=0.200m:
![E=\frac{1}{2\pi(8.85*10^{-12}C^2/Nm^2)}[\frac{4.80*10^{-6}C}{0.200m}-\frac{2.26*10^{-6}C}{0.200m}}]=228391.8N/C](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B1%7D%7B2%5Cpi%288.85%2A10%5E%7B-12%7DC%5E2%2FNm%5E2%29%7D%5B%5Cfrac%7B4.80%2A10%5E%7B-6%7DC%7D%7B0.200m%7D-%5Cfrac%7B2.26%2A10%5E%7B-6%7DC%7D%7B0.200m%7D%7D%5D%3D228391.8N%2FC)
b) for y=0.600m, r1=0.600m, r2=0.200m:
![E=\frac{1}{2\pi(8.85*10^{-12}C^2/Nm^2)}[\frac{4.80*10^{-6}C}{0.600m}-\frac{2.26*10^{-6}C}{0.200m}}]=-59345.91N/C](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B1%7D%7B2%5Cpi%288.85%2A10%5E%7B-12%7DC%5E2%2FNm%5E2%29%7D%5B%5Cfrac%7B4.80%2A10%5E%7B-6%7DC%7D%7B0.600m%7D-%5Cfrac%7B2.26%2A10%5E%7B-6%7DC%7D%7B0.200m%7D%7D%5D%3D-59345.91N%2FC)
Answer:
A. plot an H-R diagram for the stars in the cluster.
Explanation:
A star cluster can be defined as a constellation of stars, due to gravitational force, which has the same origin.
The astronomy student would have to plot an H-R diagram for the stars in the cluster and determine the age of the cluster by observing the turn-off point. The turn-off is majorly as a result of gradual depletion of the source of energy of the star. Thus, it projects off the constellation.
