Answer:
Final temperature, 
Explanation:
Given that,
Mass of silver ring, m = 4 g
Initial temperature, 
Heat released, Q = -18 J (as heat is released)
Specific heat capacity of silver, 
To find,
Final temperature
Solution,
The expression for the specific heat is given by :





So, the final temperature of silver is 21.85 degrees Celsius.
First let’s pick one at random
Less than 5 percentile
That means they lined 100 kids up and less than 5% of them are that certain weight.
So if you apply this to every option you will see the answer is
A). Less than 5 percentile
The formula is=1/2(m x v^2)
so = 1/2*(0.05)*(310)^2
ans is =2402.5 joules
Answer:
Tp/Te = 2
Therefore, the orbital period of the planet is twice that of the earth's orbital period.
Explanation:
The orbital period of a planet around a star can be expressed mathematically as;
T = 2π√(r^3)/(Gm)
Where;
r = radius of orbit
G = gravitational constant
m = mass of the star
Given;
Let R represent radius of earth orbit and r the radius of planet orbit,
Let M represent the mass of sun and m the mass of the star.
r = 4R
m = 16M
For earth;
Te = 2π√(R^3)/(GM)
For planet;
Tp = 2π√(r^3)/(Gm)
Substituting the given values;
Tp = 2π√((4R)^3)/(16GM) = 2π√(64R^3)/(16GM)
Tp = 2π√(4R^3)/(GM)
Tp = 2 × 2π√(R^3)/(GM)
So,
Tp/Te = (2 × 2π√(R^3)/(GM))/( 2π√(R^3)/(GM))
Tp/Te = 2
Therefore, the orbital period of the planet is twice that of the earth's orbital period.