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: Acceleration will have 2 components, vertical and horizontal.
Net-vertical component can be positive, zero or negative depending upon the magnitude of the upward component of the applied acceleration.
Net-horizontal acceleration will be equal to the horizontal component of the applied acceleration.
Explanation:
Since acceleration is a vector quantity and the cart is being pushed up the ramp, the ramp would be at some angle to the horizontal and hence there will be vertical and horizontal components of acceleration.
<u>For vertical acceleration:</u>
If the magnitude of the upward component of the applied acceleration is greater than the value of the acceleration due to gravity then the net vertical acceleration will be upward because it will overtake the value of acceleration due to gravity.
In case the upward component of the applied acceleration is lesser than the value of the acceleration due to gravity then the net vertical acceleration will be downward.
<u>For horizontal acceleration:</u>
This component remains unaffected and is equal to the horizontal component of the applied acceleration because there is no other acceleration acting in the horizontal direction.
But the net acceleration will not be solely in the vertical or horizontal direction because the block has to move forward on the inclined ramp so there will always exist a horizontal and a vertical component making the net acceleration to parallel to the ramp in upward direction if the body is going up the ramp.
Draw a diagram to illustrate the problem as shown below.
The vertical component of the launch velocity is
v = (8.5 m/s)*sin30° = 4.25 m/s
The horizontal component of the launch velocity is
8.5*cos30° = 7.361 m/s
Assume that aerodynamic resistance may be ignored.
Because the horizontal distance traveled is 19 m, the time of travel is
t = 19/7.361 = 2.581 s
The downward vertical travel is modeled by
h = (-4.25 m/s)*(2.581 s) + 0.5*(9.8 m/s²)*(2.581 s)²
= 21.675 m
Answer: The height is 21.7 m (nearest tenth)
Answer:
position as a function of time is y = 0.05 × cos(9.9)t
Explanation:
given data
mass = 5 kg
length = 10 cm = 0.1 m
displaced = 5 cm
to find out
position as a function of time
solution
we will apply here equilibrium that is
mass × g = k × length
put here value and find k
k =
k = 490 N/m
and ω is
ω =
ω =
ω = 9.9
so here position w.r.t time is
y = 0.05 × cosωt
y = 0.05 × cos(9.9)t
so position as a function of time is y = 0.05 × cos(9.9)t