Answer:
Orbital period of the planet will be 207.06 year
Explanation:
We have given the planet have the semi major axis as 35 au
We have to find the orbital period of the planet
From Keplar's third law there is relation between the orbital period and semi major axis which is t
So
So orbital period of the planet will be 207.06 year
<u>Answer:</u>
Magnitude of resultant velocity of kayaker to the nearest tenth = 10 m/s
Direction of resultant velocity of kayaker = 49.32⁰ South of west.
<u>Explanation:</u>
Let east represents positive x- axis and north represent positive y - axis. Horizontal component is i and vertical component is j.
First kayaker paddles at 4.0 m/s in a direction 30° south of west, kayaker paddles at 4.0 m/s in a direction 210° anticlockwise from positive horizontal axis.
So velocity of kayaker = 4 cos 210 i + 4 sin 210 j = -3.46 i - 2 j
He then turns and paddles at 3.7 m/s in a direction 20° west of south, kayaker paddles at 3.7 m/s in a direction 250° anticlockwise from positive horizontal axis.
So that velocity = -1.27 i - 3.48 j
So resultant velocity of kayaker = -3.46 i - 2 j +(-1.27 i - 3.48 j) = -4.71 i - 5.48 j
Magnitude of resultant velocity of kayaker =
Magnitude of resultant velocity of kayaker to the nearest tenth = 10 m/s
Direction of resultant positive horizontal axis, θ = tan⁻¹(-5.48/-4.71) = 229.32⁰ = 49.32⁰ South of west.
The final momentum of the body is equal to 120 Kg.m/s.
<h3>What is momentum?</h3>Momentum can be described as the multiplication of the mass and velocity of an object. Momentum is a vector quantity as it carries magnitude and direction.
If m is an object's mass and v is its velocity then the object's momentum p is: . The S.I. unit of measurement of momentum is kg⋅m/s, which is equivalent to the N.s.
Given the initial momentum of the body = Pi = 20 Kg.m/s
The force acting on the body, Pf = 25 N
The time, Δt = 4-0 = 4s
The Force is equal to the change in momentum: F ×Δt = ΔP
25 × 4 = P - 20
100 = P - 20
P = 100 + 20 = 120 Kg.m/s
Therefore, the final momentum of a body is 120 Kg.m/s.
Learn more about momentum, here:
#SPJ1
Answer:
<h2>a)Explanation:
If a ball dropped from a stationary hot-air balloon that is at an altitude of 576 ft, an expression for the altitude of the ballast after t seconds can be expressed using the equation of motion;
S is the altitude of the ballest
u is the initial velocity
a is the acceleration of the body
t is the time taken to strike the ground
Since the body is dropped from a stationary air balloon, the initial velocity u will be zero i.e u = 0m/s
Also, since the ballast is dropped from a stationary hot-air balloon, the body is under the influence of gravity, the acceleration will become acceleration due to gravity i.e a = +g
Substituting this values into the equation of the motion;
a) An expression for the altitude of the ballast after t seconds is therefore
b) Given S = 576ft and g = 32ft/s², substituting this into the formula in (a);
This means that the ballast strikes the ground after 6secs
c) To get the velocity when it strikes the ground, we will use the equation of motion v = u + gt.
v = 0 + 32(6)
v = 192ft