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 <em>v</em> and <em>x</em> changes is;<u> As </u><u><em>x</em></u><u> increases, </u><u><em>v</em></u><u> increases, but the relationship is no longer linear and the values of </u><u><em>v</em></u><u> will be less for the same value of </u><u><em>x</em></u><u>.</u>
Reasons:
The energy given to the block by the spring = 
According to the principle of conservation of energy, we have;
On a flat plane, energy given to the block =
= kinetic energy of
block = 
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 = 
- The force of the weight of the block on the string,

The energy given to the block =
= The kinetic energy of block as it leaves the spring = 
Which gives;

Which is of the form;
a·x² - b = c·v²
a·x² + c·v² = b
Where;
a, b, and <em>c</em> are constants
The graph of the equation a·x² + c·v² = b is an ellipse
Therefore;
- As <em>x</em> increases, <em>v</em> increases, however, the value of <em>v</em> obtained will be lesser than the same value of <em>x</em> as when the block is on a flat plane.
<em>Please find attached a drawing related to the question obtained from a similar question online</em>
<em>The possible question options are;</em>
- <em>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</em>
- <em>The relationship is no longer linear and v will be more for the same value of x</em>
- <em>The relationship is still linear, with lesser value of v</em>
- <em>The relationship is still linear, with higher value of v</em>
- <em>The relationship is still linear, but vary inversely, such that as x increases, v decreases</em>
<em />
Learn more here:
brainly.com/question/9134528
To solve the problem it is necessary to apply the Malus Law. Malus's law indicates that the intensity of a linearly polarized beam of light, which passes through a perfect analyzer with a vertical optical axis is equivalent to:

Where,
indicates the intensity of the light before passing through the polarizer,
I is the resulting intensity, and
indicates the angle between the axis of the analyzer and the polarization axis of the incident light.
Since we have two objects the law would be,

Replacing the values,



Therefore the intesity of the light after it has passes through both polarizers is 
Answer:
The magnitude of the force of friction equals the magnitude of my push
Explanation:
Since the crate moves at a constant speed, there is no net acceleration and thus, my push is balanced by the frictional force on the crate. So, the magnitude of the force of friction equals the magnitude of my push.
Let F = push and f = frictional force and f' = net force
F - f = f' since the crate moves at constant speed, acceleration is zero and thus f' = ma = m (0) = 0
So, F - f = 0
Thus, F = f
So, the magnitude of the force of friction equals the magnitude of my push.
<span>1.0x10^3 Joules
The kinetic energy a body has is expressed as the equation
E = 0.5 M V^2
where
E = Energy
M = Mass
V = Velocity
Since the shot was at rest, the initial energy is 0. Let's calculate the energy that the shot has while in motion
E = 0.5 * 7.2 kg * (17 m/s)^2
E = 3.6 kg * 289 m^2/s^2
E = 1040.4 kg*m^2/s^2
E = 1040.4 J
So the work performed on the shot was 1040.4 Joules. Rounding the result to 2 significant figures gives 1.0x10^3 Joules</span>