which of the following is not an example of force? A. magnetism B. Electric force C. energy D. friction

Students perform a set of experiments by placing a block of mass m against a spring, compressing the spring a distance x along a

Verizon [17]

Increasing the angle of inclination of the plane decreases the velocity of the block as it leaves the spring.

The statement that indicates how the relationship between v and x changes is; As x increases, v increases, but the relationship is no longer linear and the values of v will be less for the same value of x.

Reasons:

The energy given to the block by the spring = $\mathbf{0.5 \cdot k \cdot x^2}$

According to the principle of conservation of energy, we have;

On a flat plane, energy given to the block = $0.5 \cdot k \cdot x^2$ = kinetic energy of

block = $0.5 \cdot m \cdot v^2$

Therefore;

0.5·k·x² = 0.5·m·v²

Which gives;

x² ∝ v²

x ∝ v

On a plane inclined at an angle θ, we have;

The energy of the spring = $\mathbf{0.5 \cdot k \cdot x^2}$

The force of the weight of the block on the string, $F = m \cdot g \cdot sin(\theta)$

The energy given to the block = $0.5 \cdot k \cdot x^2 - m \cdot g \cdot sin(\theta)$ = The kinetic energy of block as it leaves the spring = $\mathbf{0.5 \cdot m \cdot v^2}$

Which gives;

$0.5 \cdot k \cdot x^2 - m \cdot g \cdot sin(\theta) = 0.5 \cdot m \cdot v^2$

Which is of the form;

a·x² - b = c·v²

a·x² + c·v² = b

Where;

a, b, and c are constants

The graph of the equation a·x² + c·v² = b is an ellipse

Therefore;

As x increases, v increases, however, the value of v obtained will be lesser than the same value of x as when the block is on a flat plane.

Please find attached a drawing related to the question obtained from a similar question online

The possible question options are;

As x increases, v increases, but the relationship is no longer linear and the values of v will be less for the same value of x

The relationship is no longer linear and v will be more for the same value of x

The relationship is still linear, with lesser value of v

The relationship is still linear, with higher value of v

The relationship is still linear, but vary inversely, such that as x increases, v decreases

Learn more here:

brainly.com/question/9134528

6 0

2 years ago

Along the line connecting the two charges, at what distance from the charge q1 is the total electric field from the two charges

Nostrana [21]

Answer:

$r = d (\frac{\sqrt {q_1}}{\sqrt{q_1} + \sqrt{q_2}})$

Explanation:

Here two charges are placed at distance "d" apart

now the net value of electric field at some position between two charges will be ZERO

so we will have

electric field due to charge 1 = electric field due to charge 2

$E_1 = E_2$

Let the position where net field is zero will lie at distance "r" from q1

$\frac{kq_1}{r^2} = \frac{kq_2}{(d-r)^2}$

now we will have

$\frac{(d - r)^2}{r^2} = \frac{q_2}{q_1}$

now square root both sides

$\frac{d}{r} - 1 = \sqrt{\frac{q_2}{q_1}}$

now we have

$\frac{d}{r} = \sqrt{\frac{q_2}{q_1}} + 1$

so we have

$r = d (\frac{\sqrt {q_1}}{\sqrt{q_1} + \sqrt{q_2}})$

8 0

3 years ago

Not in book

umka2103 [35]

Answer:

$x=2.4365\ m$

and

$x=-1.4365\ m$

Explanation:

Given:

first charge, $q_1=5\times 10^{-3}\ C$

second charge, $q_2=3\times 10^{-3}\ C$

position of first charge, $x_1=-2\ m$

position of second charge, $x_2=-1\ m$

Now since there are only 2 charges and of the same sign so they repel each other. This repulsion will be zero at some point on the line joining the charges.

Now, according to the condition, electric field will be zero where the effects of field due to both the charges is equal.

$E_1=E_2$

since first charge is greater than the second charge so we may get a point to the right of the second charge and the distance between the two charges is 1 meter.

$\frac{1}{4\pi.\epsilon_0} \frac{q_1}{(r+1)^2} =\frac{1}{4\pi.\epsilon_0} \frac{q_2}{(r)^2}$

$\frac{5\times 10^{-3}}{(r+1)^2} = \frac{3\times 10^{-3}}{(r)^2}$

$3(r^2+1+2r)=5r^2$

$2r^2-6r-3=0$

$r=3.4365 \&\ r=-0.4365$

Since we have assumed that the we may get a point to the right of second charge so we calculate with respect to the origin.

$x=-1+3.4365=2.4365\ m$

and

$x=-1-0.4365=-1.4365\ m$

6 0

3 years ago

A lamp consumes 1000J of ekectrical energy in 10s. Calculate its power.

Mice21 [21]

You do 1000 divide it by 10 which equals 100 W

4 0

3 years ago

Select the option that best summarizes the reason for the use of position-time graphs when studying physical science. (1 point)

Sedaia [141]

The position-time graphs show the relationship between the position of an object (shown on the y-axis) and the time (shown on the x-axis) to show velocity.

<h3>What is velocity?</h3>

Velocity is a vector quantity that tells the distance an object has traveled over a period of time.

Displacement is a vector quality showing total length of an area traveled by a particular object.

Imagine a time-position graph where the velocity of an object is constant. What will be observed on the graph concerning the slope of the line segment as well as the velocity of the object?

The slope of the line is equal to zero and the object will be stationary.

The position-time graphs show the relationship between the position of an object (shown on the y-axis) and the time (shown on the x-axis) to show velocity.

To learn more about velocity refer to the link

brainly.com/question/18084516

#SPJ2

4 0

1 year ago

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which of the following is not an example of force? A. magnetism B. Electric force C. energy D. friction​

which of the following is not an example of force? A. magnetism B. Electric force C. energy D. friction