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.
Answer:
568.18 N
Explanation:
From the question,
The formula for gravitational potential is given as
Ep = mgh........................ Equation 1
Where Ep = Gravitational potential, m = mass of the diver,h = Height.
But,
W = mg.................... Equation 2
Where W = weight of the diver.
Substitute equation 2 into equation 1
Ep = Wh
Make W the subject of the equation
W = Ep/h................... Equation 3
Given: Ep = 25000 J, h = 44 m
Substitute into equation 3
W = 25000/44
W = 568.18 N.
Hence the weight of the diver = 568.18 N
Answer:
Explanation:
I'm going to assume that all the givens are in m/s. If I am correct, The orange line is speeding up. It is going from 4 m/s to 7 m/s. Its slant is from lower left to upper right. It is increasing in speed.
The green line is slowing down. It is going from 4 to 0 m/s. It's slant is from upper left to lower right.
The blue line is horizontal. It is neither slowing down or speeding up.
Answer:
The force of friction acts in the direction opposite to the direction of motion. If friction would have been applied to the skier it would have resulted in a lower velocity and less kinetic energy.
Explanation:
Answer:
The arrow will leave the bow with a velocity of 10 m/s.
Explanation:
Hi there!
The potential energy stored in the bow can be calculated using the following equation:
U = 1/2 · k · d²
Where
U = elastic potential energy.
k = spring constant.
d = stretched distance of the bow
Then:
U = 1/2 · 112 N/m · (0.29 m)²
U = 4.7 J
When the bow is released, the potential energy is transformed into kinetic energy. Then, the kinetic energy of the arrow when it leaves the bow will be:
KE = 1/2 · m · v² = 4.7J
Where:
KE = kinetic energy.
m = mass of the arrow.
v = velocity of the arrow:
Then:
4.7 J = 1/2 ·0.094 kg · v²
2 · 4.7 J / 0.094 kg = v²
9.4 kg · m²/s² / 0.094 kg = v²
v = 10 m/s
The arrow will leave the bow with a velocity of 10 m/s.
