The answer is 45 degrees.
According to the Kinematics of projectile motion, if the purpose is to maximize range, optimum angle of landing is always 45 degrees.If the purpose is to maximize range & projection height is zero, the optimum angle of projection (and landing) is 45 degrees.
Answer:
112m/s
Explanation:
14x8=112 therefore meaning the zebra would run 112m/s
Answer:
t=37 mins -> 2220sec
We want "T" which is the pendulum time constant
Using this equation
.5A=Ae^(-t/T)
The .5A is half the amplitude
Take ln of both sides to get ride of Ae
=ln(.5)=-2220/T
Now rearrange to = T
T=-2220/ln(.5) = 3202.78sec / 60 secs = 53.38 mins -> first part of the answer.
The second part is really easy. It took 37 mins to decay half way. meaning to decay another half of 50% which equals 25% it will take an additional 37 mins!
First, let's determine the gravitational force of the Earth exerted on you. Suppose your weight is about 60 kg.
F = Gm₁m₂/d²
where
m₁ = 5.972×10²⁴ kg (mass of earth)
m₂ = 60 kg
d = 6,371,000 m (radius of Earth)
G = 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²
F = ( 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²)(60 kg)(5.972×10²⁴ kg)/(6,371,000 m )²
F = 589.18 N
Next, we find the gravitational force exerted by the Sun by replacing,
m₁ = 1.989 × 10³⁰<span> kg
Distance between centers of sun and earth = 149.6</span>×10⁹ m
Thus,
d = 149.6×10⁹ m - 6,371,000 m = 1.496×10¹¹ m
Thus,
F = ( 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²)(60 kg)(1.989 × 10³⁰ kg)/(1.496×10¹¹ m)²
F = 0.356 N
Ratio = 0.356 N/589.18 N
<em>Ratio = 6.04</em>
Answer: Mass of the planet, M= 8.53 x 10^8kg
Explanation:
Given Radius = 2.0 x 106m
Period T = 7h 11m
Using the third law of kepler's equation which states that the square of the orbital period of any planet is proportional to the cube of the semi-major axis of its orbit.
This is represented by the equation
T^2 = ( 4π^2/GM) R^3
Where T is the period in seconds
T = (7h x 60m + 11m)(60 sec)
= 25860 sec
G represents the gravitational constant
= 6.6 x 10^-11 N.m^2/kg^2 and M is the mass of the planet
Making M the subject of the formula,
M = (4π^2/G)*R^3/T^2
M = (4π^2/ 6.6 x10^-11)*(2×106m)^3(25860s)^2
Therefore Mass of the planet, M= 8.53 x 10^8kg
