5. From the definition of work, explain why only the vertical height of the stairs is measured. Hint: Think of the data table co
1 answer:
You might be interested in
Complete Question: A charge q1 = 2.2 uC is at a distance d= 1.63m from a second charge q2= -5.67 uC. (b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as measured from q1.)
Answer:
d= 0.46 m
Explanation:
The electric potential is defined as the work needed, per unit charge, to bring a positive test charge from infinity to the point of interest.
For a point charge, the electric potential, at a distance r from it, according to Coulomb´s Law and the definition of potential, can be expressed as follows:
V =
We have two charges, q₁ and q₂, and we need to find a point between them, where the electric potential due to them, be zero.
If we call x to the distance from q₁, the distance from q₂, will be the distance between both charges, minus x.
So, we can find the value of x, adding the potentials due to q₁ and q₂, in such a way that both add to zero:
V = +
= 0
⇒:
Replacing by the values of q1, q2, and k, and solving for x, we get:
⇒ x = (2.22 μC* 1.63 m) / 7.89 μC = 0.46 m from q1.
Answer:
The maximum value of the driving force is 1044.01 N.
Explanation:
Given that,
Weight of the object, W = 50 N
Force constant of the spring, k = 210 N/m
The system is undamped and is subjected to a harmonic driving force of frequency 11.5 Hz.
Amplitude, A = 4 cm
We need to find the maximum value of the driving force. The force is given by the product of mass and maximum acceleration as :
.....(1)
A is amplitude
m is mass,
is angular frequency
Angular frequency is given by :
Equation (1) becomes :
So, the maximum value of the driving force is 1044.01 N.
<u>For the first question we use point-slope form of an equation</u>
y-
=m(x-
)
y=m(x-)+
then we plug in the known values
y=m(x-4)+7
we leave the slope as variable m since it is undefined
<u>For Question 2
</u>
A line that is parallel to the x-axis is a horizontal line and since the slope of a line is defined as the Δy/Δx (change in y/change in x) and the change in y is 0 at any 2 points observed on the line the slope is 0. (0/any number is 0)
<u>For Question 3
</u>
Following the same relationship as question 2 we can solve for the slope.
Δy/Δx
-
/-
now we plug in the known values from the two points given
(5-7)/(-3-1)
-2/-4
m=1/2 or 0.5
<u><em>and For the Final Question</em></u>
a translation left or right is done by affecting the x variable if you add 2 to x then the x value will have to be 2 less to get the same result...in other words when x is 1 the value of y is also 1...but if I wanted the whole equation translated left 2 unites then I would want the same y-value at an x-value 2 smaller... in other words, in our example x will be -1 when y is 1. For this value found on the graph to match the equation our x value must have 2 added to it in the equation....therefore the equation that translates y=|x| two units left is...
y=|x+2|
y-
y=m(x-
then we plug in the known values
y=m(x-4)+7
we leave the slope as variable m since it is undefined
<u>For Question 2
</u>
A line that is parallel to the x-axis is a horizontal line and since the slope of a line is defined as the Δy/Δx (change in y/change in x) and the change in y is 0 at any 2 points observed on the line the slope is 0. (0/any number is 0)
<u>For Question 3
</u>
Following the same relationship as question 2 we can solve for the slope.
Δy/Δx
now we plug in the known values from the two points given
(5-7)/(-3-1)
-2/-4
m=1/2 or 0.5
<u><em>and For the Final Question</em></u>
a translation left or right is done by affecting the x variable if you add 2 to x then the x value will have to be 2 less to get the same result...in other words when x is 1 the value of y is also 1...but if I wanted the whole equation translated left 2 unites then I would want the same y-value at an x-value 2 smaller... in other words, in our example x will be -1 when y is 1. For this value found on the graph to match the equation our x value must have 2 added to it in the equation....therefore the equation that translates y=|x| two units left is...
y=|x+2|