The force on charge Y is the same as the force on charge X
Explanation:
We can answer this problem by applying Newton's third law of motion, which states that:
"When an object A exerts a force on object B (action force), then object B exerts an equal and opposite force on object A (reaction force)"
In this problem, we can identify object A as charge X and object B as charge Y. The magnitude of the electrostatic force between them is given by
(1)
where:
is the Coulomb's constant
are the two charges
r is the separation between the two charges
According to Newton's third law, therefore, the magnitude of the force exerted by charge X on charge Y is the same as the force exerted by charge Y on charge X (and it is given by eq.(1)), however their directions are opposite.
Learn more about Newton's third law:
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Answer: option 1 : the electric potential will decrease with an increase in y
Explanation: The electric potential (V) is related to distance (in this case y) by the formulae below
V = kq/y
Where k = 1/4πε0
Where V = electric potential,
k = electric constant = 9×10^9,
y = distance of potential relative to a reference point, ε0 = permittivity of free space
q = magnitude of electronic charge = 1.609×10^-19 c
From the formulae, we can see that q and k are constants, only potential (V) and distance (y) are variables.
We have that
V = k/y
We see the potential(V) is inversely proportional to distance (y).
This implies that an increase in distance results to a decreasing potential and a decrease in distance results to an increase in potential.
This fact makes option 1 the correct answer
At t =0, the velocity of A is greater than the velocity of B.
We are told in the question that the spacecrafts fly parallel to each other and that for the both spacecrafts, the velocities are described as follows;
A: vA (t) = ť^2 – 5t + 20
B: vB (t) = t^2+ 3t + 10
Given that t = 0 in both cases;
vA (0) = 0^2 – 5(0) + 20
vA = 20 m/s
For vB
vB (0) = 0^2+ 3(0) + 10
vB = 10 m/s
We can see that at t =0, the velocity of A is greater than the velocity of B.
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Read each question carefully. Show all your work for each part of the question. The parts within the question may not have equal weight. Spacecrafts A and B are flying parallel to each other through space and are next to each other at time t= 0. For the interval 0 <t< 6 s, spacecraft A's velocity v A and spacecraft B's velocity vB as functions of t are given by the equations va (t) = ť^2 – 5t + 20 and VB (t) = t^2+ 3t + 10, respectively, where both velocities are in units of meters per second. At t = 6 s, the spacecrafts both turn off their engines and travel at a constant speed. (a) At t = 0, is the speed of spacecraft A greater than, less than, or equal to the speed of spacecraft B?
<h3>Answer</h3>
1104 km/hour
<h3>Explanation</h3>
Distance between Dallas Texas to New York = 2760 km
Time the plane took from Dallas to New York = 2 hours
Time the plane took from New York back to Dallas = 2.5 hours
Formula to use
<h3>distance = speed x time </h3>
Speed the plane took from Dallas to New York
2760 = 2 x speed
speed = 2760 / 2
= 1380 km/hour
Speed the plane took from New York to Dallas (ROUND TRIP)
2760 = 2.5 x speed
speed = 2760 / 2.5
= 1104 km/hour
The gravitational potential energy of the object is 100 J.
Gravitational potential energy stored in an object is the work done in raising the object to a height <em>h</em> against the gravitational force acting on it.
The gravitational force acting on a body is its weight mg, where m is its mass and g, the acceleration due to gravity.
Work done by a force is equal to the product of the force and the displacement made by the point of application of the force.
The weight of the object is given as 20 J and it is raised to a height of 5 m.
The gravitational potential energy of the object is 100 J.
