Answer:
118.3 J
Explanation:
Givens:
m = 1.4 kg
V = 13 m/s
Formula for kinetic energy:
KE = (1/2)*(m)*(v)^2
KE = .5*(1.4 kg)*(13 m/s)^2
KE 118.3 J
J = Joules
Answer:
The greater the amplitude the greater the energy.
(Think of a water wave - which carries greater energy a 1 ft wave or
a 10 ft wave)
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:
(a) Magnitude of Vector = 207.73 m
(b) Direction = 65.48°
Explanation:
(a)
The formula to find out the magnitude of a resultant vector with the help of its x and y components is given as follows:
<u>Magnitude of Vector = 207.73 m</u>
(b)
For the direction of the vector we have the formula:
<u>Direction = 65.48°</u>
