Answer:
7.36 × 10^22 kg
Explanation:
Mass of the man = 90kg
Weight on the moon = 146N
radius of the moon =1.74×10^6
Weight =mg
g= weight/mass
g= 146/90 = 1.62m/s^2
From the law of gravitational force
g = GM/r^2
Where G = 6.67 ×10^-11
M = gr^2/G
M= 1.62 × (1.74×10^6)^2/6.67×10^-11
= 4.904×10^12/6.67×10^-11
=0.735×10^23
M= 7.35×10^22kg. (approximately) with option c
Hello!
Recall the period of an orbit is how long it takes the satellite to make a complete orbit around the earth. Essentially, this is the same as 'time' in the distance = speed * time equation. For an orbit, we can define these quantities:
← The circumference of the orbit
speed = orbital speed, we will solve for this later
time = period
Therefore:

Where 'r' is the orbital radius of the satellite.
First, let's solve for 'v' assuming a uniform orbit using the equation:

G = Gravitational Constant (6.67 × 10⁻¹¹ Nm²/kg²)
m = mass of the earth (5.98 × 10²⁴ kg)
r = radius of orbit (1.276 × 10⁷ m)
Plug in the givens:

Now, we can solve for the period:

They were produced inside stars.
Answer:
The magnitude of electric force is 
Explanation:
Coulomb's Law:
The force of attraction or repletion is
- directly proportional to the products of charges i.e

- inversely proportional to the square of distance i.e


[ k is proportional constant=9×10⁹N m²/C²]
There are two types of force applied on Q=+2.5 μC=2.5×10⁻⁶ C
Let F₁ force be applied on Q =+2.5 μC by q₁= -5.0 μC = - 5.0×10⁻⁶ C
and F₂ force be applied on Q=+2.5 μC by q₂= 5.0 μC= 5.0×10⁻⁶ C
Since the magnitude of F₁ and F₂ are same. Therefore their y component cancel.
If we draw a line from q₁ to Q .
The it forms a triangle whose base = 4.0 cm and altitude =3.0 cm.
Let hypotenuse = r
Therefore, 
we know,


Total force 


[ r=5]
N
The magnitude of electric force is 