Answer:
Second option 6.3 N at 162° counterclockwise from
F1->
Explanation:
Observe the attached image. We must calculate the sum of all the forces in the direction x and in the direction y and equal the sum of the forces to 0.
For the address x we have:
For the address and we have:
The forces 
and 
are known
We have 2 unknowns (
and b) and we have 2 equations.
Now we clear from the second equation and introduce it into the first equation.
Then
Then we find the value of
Finally the answer is 6.3 N at 162° counterclockwise from
F1->
<span>0.52%
First, let's convert that speed into m/s.
150 km/h * 1000 m/km / 3600 s/h = 41.667 m/s
Now let's see how much time gravity has to work on the ball. Divide the distance by the speed.
18 m / 41.667 m/s = 0.431996544 s
Now multiply that time by the gravitational acceleration to see what the vertical component to the ball's speed that gravity adds.
0.431996544 s * 9.8 m/s^2 = 4.233566131 m/s
Use the pythagorean theorem to get the new velocity of the ball.
sqrt(41.667^2 + 4.234^2) = 41.882 m/s
Finally, let's see what the difference is
(41.882 - 41.667)/41.667 = 0.005159959 = 0.5159959%
Rounding to 2 figures, gives 0.52%</span>
Answer:
Moreover, Boss says that even if Jupiter is proven to have a core, the planet still could have formed that core through disk instability. Enough dust could have collected and cemented together in the dense gas to form a core many times larger than the size of the Earth.
Explanation:
The same is true of most other objects in the solar system — except Jupiter. The gas giant is so big that it pulls the center of mass between it and the sun, also known as the barycenter, some 1.07 solar radii from the star's center — which is about 30,000 miles above the sun's surface.
69,911 km
69,911 kmJupiter/Radius
The electric field produced by a large flat plate with uniform charge density on its surface can be found by using Gauss law, and it is equal to
where
is the charge density
is the vacuum permittivity
We see that the intensity of the electric field does not depend on the distance from the plate. Therefore, the strenght of the electric field at 4 cm from the plate is equal to the strength of the electric field at 2 cm from the plate:
