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>
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brainly.com/question/9134528
Answer:
The energy entering, reflecting, absorbed, and emitted by the earth system are the components of the Earth's radiation budget.
Explanation:
I hope this helps also I hope you have a great day and a new year.
The light refects ofthe sides
The minimum value of the coefficient of static friction between the block and the slope is 0.53.
<h3>Minimum coefficient of static friction</h3>
Apply Newton's second law of motion;
F - μFs = 0
μFs = F
where;
- μ is coefficient of static friction
- Fs is frictional force
- F is applied force
μ = F/Fs
μ = F/(mgcosθ)
μ = (250)/(50 x 9.8 x cos15)
μ = 0.53
Thus, the minimum value of the coefficient of static friction between the block and the slope is 0.53.
Learn more about coefficient of friction here: brainly.com/question/20241845
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Answer:
The launching point is at a distance D = 962.2m and H = 39.2m
Explanation:
It would have been easier with the drawing. This problem is a projectile launching exercise, as they give us data after the window passes and the wall collides, let's calculate with this data the speeds at the point of contact with the window.
X axis
x = Vox t
t = x / vox
t = 7.1 / 340
t = 2.09 10-2 s
In this same time the height of the window fell
Y = Voy t - ½ g t²
Let's calculate the initial vertical speed, this speed is in the window
Voy = (Y + ½ g t²) / t
Voy = [0.6 + ½ 9.8 (2.09 10⁻²)²] /2.09 10⁻² = 0.579 / 0.0209
Voy = 27.7 m / s
We already have the speed at the point of contact with the window. Now let's calculate the distance (D) and height (H) to the launch point, for this we calculate the time it takes to get from the launch point to the window; at this point the vertical speed is Vy2 = 27.7 m / s
Vy = Voy - gt₂
Vy = 0 -g t₂
t₂ = Vy / g
t₂ = 27.7 / 9.8
t₂ = 2.83 s
This is the time it also takes to travel the horizontal and vertical distance
X = Vox t₂
D = 340 2.83
D = 962.2 m
Y = Voy₂– ½ g t₂²
Y = 0 - ½ g t2
H = Y = - ½ 9.8 2.83 2
H = 39.2 m
The launching point is at a distance D = 962.2m and H = 39.2m