The inner planets are usually rocky because the gravitational pull is stronger closer to the star or in this case the sun. The dust and rocky particles that are left over after a super nova or in a nebula will tend to orbit closer to a proto-star when a solar system is in its early days. In our solar system these planets are Mercury, Venus, Earth and Mars. Gases are less dense and will be less affected by the pull of gravity because rocky particles have more mass. The outer planets are gas giants formed from clouds of gas that would be further out in the spinning disk around a proto-star.
The precision (relative error) of the centripetal force is 1%.
<h3>Relative error</h3>
This is the error in measurement of a variable obtained in comparison with other variables.
F = mv²/r
where;
- F is centripetal force
- m is mass
- v is velocity
- r is radius
F/m = v²/r
F/m = (0.01v)²/(0.01r)
F/m = 0.01v²/r
F/m = 1%(v²/r)
Thus, the precision (relative error) of the centripetal force is 1%.
Learn more about relative error here: brainly.com/question/13370015
Answer:
2
Explanation:
We know that in the Fraunhofer single-slit pattern,
maxima is given by

Given values
θ=2.12°
slit width a= 0.110 mm.
wavelength λ= 582 nm
Now plugging values to calculate N we get

Solving the above equation we get
we N= 2.313≅ 2
The force the escaping gas exerts of the rocket is 10.42 N.
<h3>
Force escaping gas exerts</h3>
The force the escaping gas exerts of the rocket is calculated as follows;
F = m(v - u)/t
where;
- m is mass of the rocket
- v is the final velocity of the rocket
- u is the initial velocity of the rocket
- t is time of motion
F = (0.25)(40 - 15)/0.6
F = 10.42 N
Thus, the force the escaping gas exerts of the rocket is 10.42 N.
Learn more about force here: brainly.com/question/12970081
#SPJ1
Answer:
The center of mass of the two-ball system is 7.05 m above ground.
Explanation:
<u>Motion of 0.50 kg ball:</u>
Initial speed, u = 0 m/s
Time = 2 s
Acceleration = 9.81 m/s²
Initial height = 25 m
Substituting in equation s = ut + 0.5 at²
s = 0 x 2 + 0.5 x 9.81 x 2² = 19.62 m
Height above ground = 25 - 19.62 = 5.38 m
<u>Motion of 0.25 kg ball:</u>
Initial speed, u = 15 m/s
Time = 2 s
Acceleration = -9.81 m/s²
Substituting in equation s = ut + 0.5 at²
s = 15 x 2 - 0.5 x 9.81 x 2² = 10.38 m
Height above ground = 10.38 m
We have equation for center of gravity

m₁ = 0.50 kg
x₁ = 5.38 m
m₂ = 0.25 kg
x₂ = 10.38 m
Substituting

The center of mass of the two-ball system is 7.05 m above ground.