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spayn [35]
2 years ago
9

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

horizontal surface of negligible friction, releasing the block, and measuring the velocity v of the block as it leaves the spring, as shown in Figure 1. The experiments indicate that as x increases, so does v in a linear relationship. The surface is now lifted so that the surface is at an angle θ above the horizontal. Which of the following indicates how the relationship between v and x changes?
Physics
1 answer:
Verizon [17]2 years ago
6 0

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 = \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 <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

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An older camera has a lens with a focal length of 60mm and uses 34-mm-wide film to record its images. Using this camera, a photo
lesya692 [45]

Answer:

24.71 mm

Explanation:

Distance is proportional to focal length, so

d∝f

which means

\frac{d'_1}{d'_2}=\frac{f_1}{f_2}

Magnification of first lens

M_2=-\frac{d'_1}{d_1}

                   and

M_2=\frac{h'_1}{h_1}

Similarly, magnification of second lens

M_2=-\frac{d'_2}{d_1}

                   and

M_2=\frac{h'_2}{h_1}

From the above equations we get

\frac{M_1}{M_2}=\frac{d'_1}{d_2'}

                   and

\frac{M_1}{M_2}=\frac{h'_1}{h_2'}

which means,

\frac{d'_1}{d_2'}=\frac{h'_1}{h_2'}

and

\frac{d'_1}{d_2'}=\frac{f_1}{f_2}

So, we get

\frac{f_1}{f_2}=\frac{h'_1}{h_2'}\\\Rightarrow f_2=f_1\times\frac{h_2'}{h'_1}\\\Rightarrow f_2=60\times\frac{14}{34}=24.71\ mm

∴ Focal length should this camera's lens is 24.71 mm

6 0
3 years ago
A soup company wants to manufacture a can in the shape of a right circular cylinder that will hold 500 cm^3 of liquid. The mater
Rom4ik [11]
The area of the top and bottom:
2πr²
Cost for top and bottom:
2πr²  x 0.02
= 0.04πr²

Area for side:
2πrh
Cost for side:
2πrh x 0.01
= 0.02πrh

Total cost:
C = 0.04πr² + 0.02πrh

We know that the volume of the can is:
V = πr²h
h = 500/πr²

Substituting this into the cost equation to get a cost function of radius:
C(r) = 0.04πr² + 0.02πr(500/πr²)
C(r) = 0.04πr² + 10/r

Now, we differentiate with respect to r and equate to 0 to obtain the minimum value:

0 = 0.08πr - 10/r²
10/r² = 0.08πr
r³ = 125/π

r = 3.41 cm
4 0
3 years ago
A small glider is coasting horizontally when suddenly a very heavy piece of cargo falls out of the bottom of the plane.
myrzilka [38]

Answer:

a. The plane speeds up but the cargo does not change speed.

Explanation:

Just to make it clear, the question is as follows from what I understand.

A small glider is coasting horizontally when suddenly a very heavy piece of cargo falls out of the bottom of the plane.  You can neglect air resistance.

Just after the cargo has fallen out:

a. The plane speeds up but the cargo does not change speed.

b. The cargo slows down but the plane does not change speed.

c. Neither the cargo nor the plane change speed.

d. The plane speeds up and the cargo slows down.

e. Both the cargo and the plane speed up.

And we are requested to choose the right answer under the given conditions. We know the glider has no motor, then it must be in free fall movement, then it is experiencing some force that pulls it to the from due to the gravity effect on it, and a force in general is calculated by

F=m*a, m:= mass of the object, a:= acceleration.

Here we are only considering the horizontal effect of the forces, then since the mass is reduced the acceleration must increase to compensate and maintain  the equilibrium of the forces, then the glider being lighter can travel faster due to the acceleration. On the other hand by the time the cargo left the glider there was no acceleration and the speed it had at the moment he left the plane continues, then the cargo does not change its speed, then horizontally speaking the answer would be a. The plane speeds up but the cargo does not change speed.

5 0
3 years ago
When we learn new information, we often change
EastWind [94]
The data from our experiment.

3 0
3 years ago
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You have a gold nugget and find its mass to be 25 grams. How much volume does this gold nugget occupy?
Kay [80]

According to the solving, the volume that this gold nugget occupies is 1.29 cm³.

<h3>What is volume?</h3>

Volume is a standard measure of three-dimensional space that is occupied. It is typically mathematically represented using standard or SI-derived units. Volume and the concept of length are connected. It is sometimes referred to as the object's capacity.

d= m/ V

m=  dV

V= m/d

<h3>So according to the given data,</h3>

The mass of the gold nugget is 25 grams.

The density of gold is 19.32 g/cm³.

We know that,

m= Vd

To find out the Volume(V),

V= m/d

V= (25 g)/(19.32 g/cm³)

V= 1.29 cm³.

According to the solving, the volume this gold nugget occupies is 1.29cm³.

To know more about questions of volume visit,

brainly.com/question/1578538

#SPJ1

8 0
1 year ago
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