The gravitational force the sun experiences from the earth is 3.48×10²²N, which is exactly the same as the force the sun experiences from the earth.
- Gravity is a force that develops as a result of the attraction between mass-containing objects. The mass of the object has a direct relationship to the strength of this attraction. r equals the separation of two objects.
F = G (M₁M₂)/r²
Where, F the gravitational force
G=6.67×10⁻¹¹Nm²kg⁻² gravitational constant
M₁=5.98×10²⁴kg mass of earth
M₂= 1.99×10³⁰ kg the mass of the sun
r =15×10¹⁰ m is the distance between sun and earth
Putting all the values in above equation,
F = 6.67×10⁻¹¹Nm²kg⁻²(5.98×10²⁴kg 1.99×10³⁰ kg)/15×10¹⁰ m
On solving the above equation we get,
F = 3.48×10²²N
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The information that could be gathered about a star whose light curve has multiple symmetrical depths is ; The shape and surface variegation of the star
The light curves of a KBO ( moons and stars ) are measured as a rate of the brightness of a star in relation to time. therefore the study of the light curve having multiple symmetrical depths ( depth of brightness ) will give an information about the shape/size and the surface variegation of the star
Hence we can conclude that The information that could be gathered about a star whose light curve has multiple symmetrical depths is ; The shape and surface variegation of the star
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Answer:
Explanation:
In linear motion, the directions of all the vectors describing the system are equal and constant which means the objects move along the same axis and do not change direction.
Answer:
c. V = k Q1 * Q2 / R1 potential energy of Q1 and Q2 separated by R
V2 / V1 = (R1 / R2) = 1/4
V2 = V1 / 4
Answer:
The energy stored in the solenoid is 7.078 x 10⁻⁵ J
Explanation:
Given;
diameter of the solenoid, d = 2.80 cm
radius of the solenoid, r = d/2 = 1.4 cm
length of the solenoid, L = 14 cm = 0.14 m
number of turns, N = 200 turns
current in the solenoid, I = 0.8 A
The cross sectional area of the solenoid is given as;

The inductance of the solenoid is given by;

The energy stored in the solenoid is given by;
E = ¹/₂LI²
E = ¹/₂(2.212 x 10⁻⁴)(0.8)²
E = 7.078 x 10⁻⁵ J
Therefore, the energy stored in the solenoid is 7.078 x 10⁻⁵ J