(a) Determine the circumference of the Earth through the equation,
C = 2πr
Substituting the known values,
C = 2π(1.50 x 10¹¹ m)
C = 9.424 x 10¹¹ m
Then, divide the answer by time which is given to a year which is equal to 31536000 s.
orbital speed = (9.424 x 10¹¹ m)/31536000 s
orbital speed = 29883.307 m/s
Hence, the orbital speed of the Earth is ~29883.307 m/s.
(b) The mass of the sun is ~1.9891 x 10³⁰ kg.
Well first of all, when it comes to orbits of the planets around
the sun, there's no such thing as "orbital paths", in the sense
of definite ("quantized") distances that the planets can occupy
but not in between. That's the case with the electrons in an atom,
but a planet's orbit can be any old distance from the sun at all.
If Mercury, or any planet, were somehow moved to an orbit closer
to the sun, then ...
-- its speed in orbit would be greater,
-- the distance around its orbit would be shorter,
-- its orbital period ("year") would be shorter,
-- the temperature everywhere on its surface would be higher,
-- if it has an atmosphere now, then its atmosphere would become
less dense, and might soon disappear entirely,
-- the intensity of x-rays, charged particles, and other forms of
solar radiation arriving at its surface would be greater.
Answer:
8.756 rad/s²
Explanation:
Given that:
A motorcycle accelerates uniformly from rest, then initial velocity v_i = 0 m/s
It final velocity v_f = 24.8 m/s
time (t) = 9.87 s
radius (r) of each tire = 0.287 m
Firstly; the linear acceleration of the motor cycle is determined as follows:
=(V_f - v_i)/t
=(24.8-0)/9.87
=2.513 m/s²
Then; the magnitude of angular acceleration
α = /r
=2.513/0.287
=8.756 rad/s²
The answer is C is the velocity is the same then the acceleration is also the same.
