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Fiesta28 [93]
3 years ago
15

Vectors have which two properties?

Physics
1 answer:
stiks02 [169]3 years ago
3 0

Answer:

D. magnitude and direction

Explanation:

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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 <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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2 years ago
Waves move fastest in
JulijaS [17]
Liquids<span> are not </span>packed<span> as tightly as </span>solids<span>. And gases are very loosely </span>packed<span>. The spacing of the molecules enables </span>sound<span> to travel much faster through a </span>solid<span> than a gas. </span>Sound<span> travels about four times faster and farther in water than it does in air.</span>
4 0
3 years ago
Read 2 more answers
At the surface, atmospheric pressure is 1.013 × 10^5 Pa. People can normally snorkel down to a depth of roughly one meter. What
natulia [17]

Answer:

1.01 × 10⁵ Pa  

Explanation:

At the surface, atmospheric pressure is 1.013 × 10⁵ Pa.

We need to find the total pressure on the air in the lungs of a person to a depth of 1 meter.

Pressure at a depth is given by :

P=\rho gh

Where

\rho is the density of air, \rho=1.225\ kg/m^3

So,

P=1.225\times 9.8\times 1\\\\=12\ Pa

Total pressure, P = Atmospheric pressure + 12 Pa

= 1.013 × 10⁵ Pa + 12 Pa

= 1.01 × 10⁵ Pa

Hence, the total pressure is 1.01 × 10⁵ Pa.

5 0
3 years ago
A particle initially located at the origin has an acceleration of vector a = 2.00ĵ m/s2 and an initial velocity of vector v i =
natali 33 [55]

With acceleration

\mathbf a=\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)\,\mathbf j

and initial velocity

\mathbf v(0)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i

the velocity at time <em>t</em> (b) is given by

\mathbf v(t)=\mathbf v(0)+\displaystyle\int_0^t\mathbf a\,\mathrm du

\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\displaystyle\int_0^t\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)\,\mathbf j\,\mathrm du

\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf j\bigg|_{u=0}^{u=t}

\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)t\,\mathbf j

We can get the position at time <em>t</em> (a) by integrating the velocity:

\mathbf x(t)=\mathbf x(0)+\displaystyle\int_0^t\mathbf v(u)\,\mathrm du

The particle starts at the origin, so \mathbf x(0)=\mathbf0.

\mathbf x(t)=\displaystyle\int_0^t\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf j\,\mathrm du

\mathbf x(t)=\left(\left(8.00\dfrac{\rm m}{\rm s}\right)u\,\mathbf i+\dfrac12\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u^2\,\mathbf j\right)\bigg|_{u=0}^{u=t}

\mathbf x(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)t\,\mathbf i+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)t^2\,\mathbf j

Get the coordinates at <em>t</em> = 8.00 s by evaluating \mathbf x(t) at this time:

\mathbf x(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)(8.00\,\mathrm s)\,\mathbf i+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)(8.00\,\mathrm s)^2\,\mathbf j

\mathbf x(8.00\,\mathrm s)=(64.0\,\mathrm m)\,\mathbf i+(64.0\,\mathrm m)\,\mathbf j

so the particle is located at (<em>x</em>, <em>y</em>) = (64.0, 64.0).

Get the speed at <em>t</em> = 8.00 s by evaluating \mathbf v(t) at the same time:

\mathbf v(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)(8.00\,\mathrm s)\,\mathbf j

\mathbf v(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(16.0\dfrac{\rm m}{\rm s}\right)\,\mathbf j

This is the <em>velocity</em> at <em>t</em> = 8.00 s. Get the <em>speed</em> by computing the magnitude of this vector:

\|\mathbf v(8.00\,\mathrm s)\|=\sqrt{\left(8.00\dfrac{\rm m}{\rm s}\right)^2+\left(16.0\dfrac{\rm m}{\rm s}\right)^2}=8\sqrt5\dfrac{\rm m}{\rm s}\approx17.9\dfrac{\rm m}{\rm s}

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An environmental scientist gives a demonstration on composting. Roger is in the audience. He wonders how composting discarded fo
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Answer:

A

Explanation:

I just did the test

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