(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.
Answer:
Explanation:
Given
Time taken to reach ground is 
Malda initial velocity 
Let h be the height of Cliff
using 
where, u=initial velocity
t=time
In first case chirpy drop downward thus u=0


For Milada there is horizontal velocity u=95 cm/s=0.95 m/s[/tex]
time taken to reach the ground will be same so distance traveled in this time with 0.95 m/s horizontal velocity is given by

The moment of inertia of the flywheel is 2.63 kg-
It is given that,
The maximum energy stored on the flywheel is given as
E=3.7MJ= 3.7×
J
Angular velocity of the flywheel is 16000
= 1675.51
So to find the moment of inertia of the flywheel. The energy of a flywheel in rotational kinematics is given by :
E = 

By rearranging the equation:
I = 
I = 2.63 kg-
Thus the moment of inertia of the flywheel is 2.63 kg-
.
Learn more about moment of inertia here;
brainly.com/question/13449336
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The Nucleus contains Protons and Neutrons.
The Neutrons does not have a charge.
The Protons are positively charge.
Hence the charge on the Nucleus, would be the charge of the proton, which is positive.
Hence Nucleus is Positively Charged.
Answer:
E = 31.329 N/C.
Explanation:
The differential electric field
at the center of curvature of the arc is
<em>(we have a cosine because vertical components cancel, leaving only horizontal cosine components of E. )</em>
where
is the radius of curvature.
Now
,
where
is the charge per unit length, and it has the value

Thus, the electric field at the center of the curvature of the arc is:


Now, we find
and
. To do this we ask ourselves what fraction is the arc length 3.0 of the circumference of the circle:

and this is
radians.
Therefore,

evaluating the integral, and putting in the numerical values we get:

