Answer:
In physics, the kinetic energy (KE) of an object is the energy that it possesses due to its motion
In classical mechanics, the gravitational potential at a location is equal to the work (energy transferred) per unit mass that would be needed to move an object to that location from a fixed reference location. It is analogous to the electric potential with mass playing the role of charge. The reference location, where the potential is zero, is by convention infinitely far away from any mass, resulting in a negative potential at any finite distance.
In mathematics, the gravitational potential is also known as the Newtonian potential and is fundamental in the study of potential theory. It may also be used for solving the electrostatic and magnetostatic fields generated by uniformly charged or polarized ellipsoidal bodies
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.
<span>Objective Lenses: Usually you will find 3 or 4 objective lenses on a microscope. They almost always consist of 4X, 10X, 40X and 100X powers. When coupled with a10X (most common) eyepiece lens, we get total magnifications of 40X (4X times10X), 100X , 400X and 1000X.</span>
A. Ocean winds bring rain and fog and often bring warm water that keeps the climate mild
Answer:
1) 3.07kgm/s
2) 5.56kgm/s
3) 76.16N
4) 4.33kgm/s
5) 0.57s
6) -8.66J
Explanation:
Given
m = 0.221kg
v = 13.9m/s
θ = 25°
t = 0.073s
1) to get the magnitude of the initial momentum is the racquet ball. We use the formula,
P(i) = mv(i)
P(i) = 0.221 * 13.9
P(i) = 3.07kgm/s
2) Magnitude of the change in momentum of the ball,
P(i,x) = P(i) cos θ
P(i,x) = 3.07 * cos25
P(i,x) = 3.07 * 0.9063
P(i,x) = 2.78
ΔP = 2P(i,x)
ΔP = 2 * 2.78 = 5.56kgm/s
3) magnitude of the average force exerted by the wall,
F(ave) = ΔP/Δt
F(ave) = 5.56/0.073
F(ave) = 76.16N
4) ΔP(z) = mv(f) - mv(i)
ΔP(z) = 0.221*-7.8 - 0.221*11.8
ΔP(z) = -1.72 - 2.61
ΔP(z) = 4.33kgm/s
5) F(ave) = ΔP/Δt
Δt = ΔP/F(ave)
Δt = 4.33 / 76.16
Δt = 0.57s
6) KE(i) = 0.5mv(i)²
KE(f) = 0.5mv(f)²
ΔKE = 0.5m[v(f)² - v(i)²]
ΔKE = 0.5 * 0.221 [(-7.8)² - 11.8²]
ΔKE = 0.1105 ( 60.84 - 139.24 )
ΔKE = 0.1105 * -78.4
ΔKE = -8.66J
