To contrast inner and outer planets we will start with the climate of the planets and then move on to there lighting. To start the planets closet to the sun, mercury, venus, earth and mars, are all hot compared to the further one, jupiter, saturn, uranus, neptune. This distance also makes the farthe away planets darker than the ones closer. Now to compare all the planets vary from either gass or solid, rocky or icy. All of them spin around the sun and all have objects spinning around them, moons.
Force , F = ma
F = m(v - u)/t
Where m = mass in kg, v= final velocity in m/s, u = initial velocity in m/s
t = time, Force is in Newton.
m= 1.2*10³ kg, u = 10 m/s, v = 20 m/s, t = 5s
F = 1.2*10³(20 - 10)/5
F = 2.4*10³ N = 2400 N
Answer:
-320 μJ.
Explanation:
Consider a point with an electrical charge of 
. Assume that 
is the electrical potential at the position of that charge. The electrical potential of that point charge will be equal to:
.
Keep in mind that since both and might not be positive, the size of the electrical potential energy might not be positive, either.
For this point charge,
; (that's -8.0 microjoules, which equals to 
)
.
Hence its electrical potential energy:
.
Why is this value negative? The electrical potential energy of a charge is equal to the work needed to bring that charge from infinitely far away all the way to its current position. Also, negative charges are attracted towards regions of high electrical potential. Bringing this 
negative charge to the origin will not require any external work. Instead, this process will release 320 μJ of energy. As a result, the electrical potential energy is a negative value.
Answer:
<u>The magnitude of the friction force is 8197.60 N</u>
Explanation:
Using the definition of the centripetal force we have:
Where:
- m is the mass of the car
- v is the speed
- R is the radius of the curvature
Now, the force acting in the motion is just the friction force, so we have:
<u>Therefore the magnitude of the friction force is 8197.60 N</u>
I hope it helps you!
The net force on particle particle q1 is 13.06 N towards the left.
<h3>
Force on q1 due to q2</h3>
F(12) = kq₁q₂/r₂
F(12) = (9 x 10⁹ x 13 x 10⁻⁶ x 7.7 x 10⁻⁶)/(0.25²)
F(12) = -14.41 N (towards left)
<h3>Force
on q1 due to q3</h3>
F(13) = (9 x 10⁹ x 7.7 x 10⁻⁶ x 5.9 x 10⁻⁶)/(0.55²)
F(13) = 1.352 N (towards right)
<h3>Net force on q1</h3>
F(net) = 1.352 N - 14.41 N
F(net) = -13.06 N
Thus, the net force on particle particle q1 is 13.06 N towards the left.
Learn more about force here: brainly.com/question/12970081
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