Answer:
(A) As it moves farther and farther from Q, its speed will keep increasing.
Explanation:
When a positive charge Q is fixed on a horizontal frictionless tabletop and a second charge q is released near to it then according to the Coulombs law the force acting on it decreases with the square of the distance between them.
Mathematically:
where:
r = distance between the charges
permittivity of free space
By the Newtons' second law of motion if the we know that the acceleration is directly proportional to the force applied. So as the distance between the charges increases the its acceleration also decreases therefore now the charge feels less acceleration but still continues to accelerate with a fading magnitude.
The concepts necessary to solve this problem are framed in the expression of string vibration frequency as well as the expression of the number of beats per second conditioned at two frequencies.
Mathematically, the frequency of the vibration of a string can be expressed as
Where,
L = Vibrating length string
T = Tension in the string
Linear mass density
At the same time we have the expression for the number of beats described as
Where
= First frequency
= Second frequency
From the previously given data we can directly observe that the frequency is directly proportional to the root of the mechanical Tension:
If we analyze carefully we can realize that when there is an increase in the frequency ratio on the tight string it increases. Therefore, the beats will be constituted under two waves; one from the first string and the second as a residue of the tight wave, as well
Replacing 
for n and 202Hz for 
The frequency of the tightened is 205Hz
Answer:
A.The vertical velocity is constantly increasing as the ball falls.
B.The horizontal velocity does not noticeably change as the ball falls.
G.The horizontal velocity does not affect how long it will take the ball to fall to the floor.
H.The velocity vector of the ball changes as it travels through the air.
Explanation:
As the ball is projected horizontally so here the vertical component of the velocity is zero
So the time to reach the ground is given as
so we will have
so this is the same time as the ball is dropped from H height
Since there is no force in horizontal direction so its horizontal velocity will always remain constant while vertical velocity will change at constant rate which is equal to acceleration due to gravity.
So overall the velocity vector will change due to net acceleration g
Answer:
Centripetal acceleration = 83.77m/s²
Explanation:
<u>Given the following data;</u>
Radius, r = 0.13m
Velocity, v = 3.3m/s
To find centripetal acceleration;
Centripetal acceleration is given by the formula;
Substituting into the equation, we have;
<em>Centripetal acceleration = 83.77m/s²</em>
<em>Therefore, the centripetal acceleration of the edge of the disc is 83.77 m/s². </em>
The force of friction is given by:
f = μR, where μ is the friction coefficient and R is the reaction force, which will be equal to the weight.
100 = μ x 130
μ = 0.77
