(1.6 x 10⁻¹⁹ C) x (300 N/C) = <em>4.8 x 10⁻¹⁷ Newton</em>
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Answer:
at the Equator
Explanation:
The four seasons are determined by four main positions in the Earth's orbit in its turn around the Sun (ecliptic plane), which are called solstices and equinoxes: winter solstice (Capricorn point, December 22), spring equinox (Aries point, around March 21-22), summer solstice (Cancer point, June 21) and autumn equinox (Libra point, around September 22-23).
In the equinoxes, the axis of rotation of the Earth is perpendicular to the sun's rays, which fall vertically over the equator. In solstices, the axis is inclined 23.5º, so that the sun's rays fall vertically on the Tropic of Cancer (summer in the northern hemisphere) or Capricorn (summer in the southern hemisphere).
When falling vertically on Ecuador, it generates a greater impact on the surface of the Tierre reaching a greater amount of energy and therefore UV rays.
Answer:
it is shifting from a planned a free-market economy
When analyzing inelastic collisions, we need to consider the law of conservation of momentum, which states that the total momentum, p, of the closed system is a constant. In the case of inelastic collisions, the momentum of the combined mass after the collision is equal to the sum of the momentum of each of the initial masses.
p1+p2+...=pf
In our case we only have two masses, which makes our problem fairly simple. Lets plug in the formula for momentum; p=mv.
m1v1+m2v2=(m1+m2)vf
To find the velocity of the combined mass we simply rearrange the terms.
vf=m1v1+m2v2m1+m2
Plug in the values given in the problem.
vf=(3.0kg)(1.4m/s)+(2.0kg)(0m/s)03.0kg+2.0kg
vf=.84m/s
I don't think that 4m has anything to do with the problem.
anyway. here.
A___________________B_______C
where A is the point that the train was released.
B is where the wheel started to stick
C is where it stopped
From A to B, v=2.5m/s, it takes 2s to go A to B so t=2
AB= v*t = 2.5 * 2 = 5m
The train comes to a stop 7.7 m from the point at which it was released so AC=7.7m
then BC= AC-AB = 7.7-5 = 2.7m
now consider BC
v^2=u^2+2as
where u is initial speed, in this case is 2.5m/s
v is final speed, train stop at C so final speed=0, so v=0
a is acceleration
s is displacement, which is BC=2.7m
substitute all the number into equation, we have
0^2 = 2.5^2 + 2*a*2.7
0 = 6.25 + 5.4a
a = -6.25/5.4 = -1.157
so acceleration is -1.157m/(s^2)
