Be easy-going and be yourself.
Answer:
The value is 
Explanation:
From the question we are told that
The period of the asteroid is 
Generally the average distance of the asteroid from the sun is mathematically represented as
![R = \sqrt[3]{ \frac{G M * T^2 }{4 \pi} }](https://tex.z-dn.net/?f=R%20%3D%20%5Csqrt%5B3%5D%7B%20%5Cfrac%7BG%20M%20%2A%20T%5E2%20%7D%7B4%20%5Cpi%7D%20%7D)
Here M is the mass of the sun with a value

G is the gravitational constant with value 
![R = \sqrt[3]{ \frac{6.67 *10^{-11} * 1.99*10^{30} * [5.55 *10^{9}]^2 }{4 * 3.142 } }](https://tex.z-dn.net/?f=R%20%3D%20%5Csqrt%5B3%5D%7B%20%5Cfrac%7B6.67%20%2A10%5E%7B-11%7D%20%20%2A%201.99%2A10%5E%7B30%7D%20%2A%20%5B5.55%20%2A10%5E%7B9%7D%5D%5E2%20%7D%7B4%20%2A%203.142%20%7D%20%7D)
=> 
Generally

So

=> 
=> 
Answer:
176.58 m
Explanation:
t = Time taken = 6 seconds
u = Initial velocity = 0
v = Final velocity
s = Displacement
g = Acceleration due to gravity = 9.81 m/s² = a
Equation of motion

The object travels 176.58 m from the cliff in 6 seconds.
Answer:
Weight of the dog on surface of earth is 140.14 Newton.
Given:
mass of the dog = 14.3 kg
To find:
Weight of the dog = ?
Formula used:
Weight of the dog is given by,
W = mg
Where, W = weight of the dog
m = mass of the dog
g = acceleration due to gravity
Solution:
Weight of the dog is given by,
W = mg
Where, W = weight of the dog
m = mass of the dog = 14.3 kg
g = acceleration due to gravity
W = 14.3 × 9.8
W = 140.14 Newton
Weight of the dog on surface of earth is 140.14 Newton.
Explanation:
The protons determines the atomic number which helps in identifying an atom of an element