To solve the problem, use Kepler's 3rd law :
T² = 4π²r³ / GM
Solved for r :
r = [GMT² / 4π²]⅓
but first covert 6.00 years to seconds :
6.00years = 6.00years(365days/year)(24.0hours/day)(6...
= 1.89 x 10^8s
The radius of the orbit then is :
r = [(6.67 x 10^-11N∙m²/kg²)(1.99 x 10^30kg)(1.89 x 10^8s)² / 4π²]⅓
= 6.23 x 10^11m
Answer:
1,700feet
Explanation:
If an object in free fall travels a distance s that is directly proportional to the square of the time t, this can be represented mathematically as;
S = kt²where;
k is the proportionality constant
K = s/t²
s1/t1²= s2/t2²= Sn/tn²= k for values of the distance and time. Using the formula
s1/t1² = s2/t2² where;
s1 is the falling distance in time t1 s2 is the falling distance in time t2
Given s1 = 1088feet, t1 = 8secs, s2 = ? t2 = 10secs
Substituting this value in the formula to get s2, we have;
1088/8²= s2/10²
64s2= 108800
s2 = 108800/64
s2 = 1,700feet
This means the object will fall a distance of 1,700feet in 10seconds
The speed of a beam of light will appear the same to all observers.
<h3>Speed of light:</h3>
The exact distance that light travels in a vacuum in a second is 299,792,458 meters (983,571,056 feet). This is roughly 186,282 miles per second, or "c," the symbol for light speed in mathematics.
According to Einstein's theory of special relativity, an object (such as a single photon of light) cannot appear to move faster than the speed of light. This physical rule is not broken by Nemiroff's experiment since the laser pointer emits a stream of photons rather than a single photon.
Nothing can move faster than light in a vacuum, according to one of physics' most revered rules. However, in a recent experiment, this speed restriction was broken when a laser pulse traveled at more than 300 times the speed of light.
Learn more about speed here:
brainly.com/question/13202563
#SPJ4
Answer:
P = n P₀ 4.9 10¹⁴ Pa
Explanation:
The radiation pressure for full absorption is
P = S / c
Where S the pointing vector, which is equal to the intensity of the beam that is defined as the energy per unit area per unit time
The energy of the protons can be calculated
Em = K = ½ m v²
Area
A = π r²
Intensity is
I = n ½ m v² / π r²
I = ½ n m /π v² / r²
We replace
S = U / t A
S = ½ n m /π v² / r² Δt
The pressure is
P = 1/c (½ n m /π (v / r Δt)²2
Δt = 45 10⁻⁹ s
P = n [½ m /πc (v/r)²] 4.9 10¹⁴
The amount in square brackets is the pressure that a proton creates, which is why it is useful
P = n P₀ 4.9 10¹⁴ Pa
Where Po is the pressure created by a proton
