the electric force decreases because the distance has an indirect relationship to the force
Explanation:
The electric force between two objects is given by
where
k is the Coulomb's constant
q1 and q2 are the charges of the two objects
r is the distance between the two objects
As we can see from the formula, the magnitude of the force is inversely proportional to the square of the distance: so, when the distance between the object increases, the magnitude of the force decreases.
Answer:
Explanation:
Two frequencies with magnitude 150 Hz and 750 Hz are given
For Pipe open at both sides
fundamental frequency is 150 Hz as it is smaller
frequency of pipe is given by
where L=length of Pipe
v=velocity of sound
for n=1
and f=750 is for n=5
thus there are three resonance frequencies for n=2,3 and 4
For Pipe closed at one end
frequency is given by
for n=0
for n=2
Thus there is one additional resonance corresponding to n=1 , between and
Answer:
From the relation above we can conclude that the as the distance between the two plate increases the electric field strength decreases
Explanation:
I cannot find any attached photo, but we can proceed anyways theoretically.
The electric field strength (E) at any point in an electric field is the force experienced by a unit positive charge (Q) at that point
i.e
But the force F
But the electric field intensity due to a point charge Q at a distance r meters away is given by
<em>From the relation above we can conclude that the as the distance between the two plate increases the electric field strength decreases</em>
This question is incomplete, the complete question is;
In a softball windmill pitch, the pitcher rotates her arm through just over half a circle, bringing the ball from a point above her shoulder and slightly forward to a release point below her shoulder and slightly forward. (Figure 1) shows smoothed data for the angular velocity of the upper arm of a college softball pitcher doing a windmill pitch; at time t = 0 her arm is vertical and already in motion. For the first 0.15 s there is a steady increase in speed, leading to a final push with a greater acceleration during the final 0.05 s before the release. In each part of the problem, determine the corresponding quantity during the first 0.15 s of the pitch.
Angular Velocity at time 0s = 12 rad/s
Angular Velocity at time 0.15s = 24 rad/s
a) What is the angular acceleration?
b) If the ball is 0.60 m from her shoulder, what is the tangential acceleration of the ball? This is the key quantity here--it's a measure of how much the ball is speeding up. Express your answer in m/s2 and in units of g
Answer:
a) the angular acceleration is 80 rad/s²
b) the tangential acceleration of the ball is;
- a = 48 m/s²
- a = 4.9 g
Explanation:
Given the data in the question;
from the graph below;
Angular Velocity at time 0s = 12 rad/s
Angular Velocity at time 0.15s = 24 rad/s
a) What is the angular acceleration;
Angular acceleration ∝ = ( -
) / dt
we substitute
Angular acceleration ∝ = ( 24 - 12 ) / 0.15
Angular acceleration ∝ = 12 / 0.15
Angular acceleration ∝ = 80 rad/s²
Therefore, the angular acceleration is 80 rad/s²
b)
If the ball is 0.60 m from her shoulder, i.e s = 0.6 m
the tangential acceleration of the ball will be;
a = ∝ × s
we substitute
a = 80 × 0.6
a = 48 m/s²
a = ( 48 / 9.8 )g
a = 4.9 g
Therefore, the tangential acceleration of the ball is;
- a = 48 m/s²
- a = 4.9 g
