Answer:
23376 days
Explanation:
The problem can be solved using Kepler's third law of planetary motion which states that the square of the period T of a planet round the sun is directly proportional to the cube of its mean distance R from the sun.
where k is a constant.
From equation (1) we can deduce that the ratio of the square of the period of a planet to the cube of its mean distance from the sun is a constant.
Let the orbital period of the earth be 
and its mean distance of from the sun be 
.
Also let the orbital period of the planet be 
and its mean distance from the sun be 
.
Equation (2) therefore implies the following;
We make the period of the planet the subject of formula as follows;
But recall that from the problem stated, the mean distance of the planet from the sun is 16 times that of the earth, so therefore
Substituting equation (5) into (4), we obtain the following;
cancels out and we are left with the following;
Recall that the orbital period of the earth is about 365.25 days, hence;
Answer:
The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.
Explanation:
We know that,
Mass of electron 
Rest mass energy for electron = 0.511 Mev
(a). The energy required to accelerate an electron from 0.500c to 0.900c Mev
Using formula of rest,
(b). The energy required to accelerate an electron from 0.900c to 0.942c Mev
Using formula of rest,
Hence, The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.
Answer:
The circumference of the Earth is 24818.58 miles
Explanation:
Analysis conceptual : The formula of the circumference is the following:
L= π*D Formula (1)
Where:
L : is the length of the circumference in miles (mi)
π : is the constant
D : is the diameter of the circumference in miles (mi)
Known data
π = 3.1416
D= 7900 miles: Diameter of the Earth
Problem development
We apply the formula 1 to calculate the circumference of the Earth (L):
L= π*7900 miles
L= 24818.58 miles
Answer:
Option (B)
Explanation:
A lift chart usually refers to a graphical representation that is mainly used in order to improve the drawbacks of a mining model by making a comparison with any random guess, and also helps in determining the changes that occur in terms of lift scores.
It describes the binary classification of the problems associated with the mining activity. This type of chart is commonly used to differentiate the lift scores for a variety of models, and picking the best one out of all.
Thus, the correct answer is option (B).
Answer:
Tension on tendon = 1669800N
Explanation:
Detailed explanation and calculation is shown in the image below
