1). Calculate how long it takes an object to fall 4,000 m after it's dropped. (Use D = (1/2) (g) (T²) . D is 4,000 m. g = 9.8 m/s². Find T .)
2). Calculate how far the object will move HORIZONTALLY in that length of time, if it's moving at 75 m/s. (Distance = (75 m/s) x (time) . )
Answer:
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PART A) Yes, the fact that there is a frictional force acting on the satellite generates a loss of energy due to friction. What causes satellite to diminish its orbit during its tour. In fact, many satellites have rectifier systems that allow them to position themselves and remain in their orbit for a long time to avoid being trapped by the Earth's gravity Force and fall into the atmosphere where they would probably be torn apart.
PART B) As a similarity, one could start by mentioning the structure of the two equations are similar and have their own constants who were responsible for supporting them. While the law of gravity speaks of the masses of the bodies the electrostatic law speaks of the charges of the bodies. For both the force is inversely proportional to the square of the distance that separates them.
However, the most notable difference between them is basically their statement. While one of the equations speaks about greavedad the other reflects the electromagnetic phenomena. It should be noted that the force of gravity is much weaker than the electromagnetic force and that the latter has the capacity of attraction and repulsion. While the gravitational force only that of attraction.
A pebbled, uneven road would be easier to see at night because it minimizes the reflection of light from car’s light coming in the opposite direction. It is difficult to see when driving on the rainy day because the roadway reflects light from cars coming in the opposite <span>directions.</span>
Answer:
a) Eₓ = - A y + 2B x
, b) Ey = -Ax –C
, c) Ez = 0
, d) The correct answer is 3
Explanation:
The electric field and the electric power are related
E = - dV / ds
a) Let's find the electric field on the x axis
Eₓ = - dV / dx
dV / dx = A y - B 2x
Eₓ = - A y + 2B x
b) calculate the electric field on the y-axis
Ey = - dV / dy
dV / dy = A x + C
Ey = -Ax –C
c) the electric field on the z axis
dv / dz = 0
Ez = 0
.d) at which point the electric field is zero
Since the electric field is a vector quantity all components must be zero
X axis
0 = = - A y + 2B x
y = 2B / A x
Axis y
0 = -Ax –C
.x = -C / A
We substitute this value in the previous equation
.y = 2B / A (-C / A)
.y = 2 B C / A2
The correct answer is 3
