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3 years ago

Which of the following is not an intensive physical property?

Physics

2 answers:

Nataliya [291]3 years ago

6 0

I would say freezing point im not positive about it but it sounds like it goes with the question

Kaylis [27]3 years ago

5 0

the answer is mass. hoped this helped

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EASY MULTIPLE CHOICE QUESTIONS! PLEASE HELP!

skad [1K]

1.A and 2.B there the answers

8 0

3 years ago

We know that the magnitudes of the negative charge on the electron and the positive charge on the proton are equal. Suppose, how

TEA [102]

Answer:3,45 x 10^9 N

Explanation: We have considered the total charge for each coin , this is the total atoms x 29 electrons for cooper and multiplier by electron charge, the total charge for each coin is 0,464 C

Finally we use the Coulomb law,

F=k Q/ (r)^2

6 0

3 years ago

Distinguish between linear momentum and angular momentum.

Katarina [22]

Linear momentum is in a straight line and depends on the objects mass and velocity.

Angular (rotational) momentum depends on the objects mass, velocity, and radius.

7 0

3 years ago

Assume that the function x(t) represents the length of tape that has unwound as a function of time. find θ(t), the angle through

bekas [8.4K]

We know that arc length (x(t)) is given with the following formula:
$x(t)=\theta(t) r$
Where r is the radius of the barrel. We must keep in mind that as barrel rolls its radius decreases because less and less tape is left on it.
If we say that the thickness of the tape is D then with every full circle our radius shrinks by d. We can write this down mathematically:
$r(\theta)=r_0-\frac{D\cdot \theta}{2\pi}$
When we plug this back into the first equation we get:
$x(t)=\theta(r_0-\frac{D\theta}{2\pi}})\\ \frac{D\theta^2}{2\pi}-\theta r_0+x(t)=0\\$
We must solve this quadratic equation.
The final solution is:
$\theta=\frac{\pi r_0+\sqrt{\pi \left(-2Dx(t)+\pi r_0^2\right)}}{D},\:\theta=\frac{\pi r_0-\sqrt{\pi \left(-2Dx(t)+\pi r_0^2\right)}}{D}$
It is rather complicated solution. If we asume that the tape has no thickness we get simply:
$x(t)=\theta(r_0-\frac{D\theta}{2\pi}});D=0\\
x(t)=\theta r_0\\
\theta(t)=\frac{x(t)}{r_0}$

8 0

3 years ago

An aquarium open at the top has 30-cm-deep water in it. You shine a laser pointer into the top opening so it is incident on the

Setler [38]

Answer:

You must add 8cm of water to the tank

Explanation:

In order to find how much the height is we will use the Snell Refraction law

This law relates the index of refraction of the water (n2), the index of refraction of the air (n1), the incidence angle relative to the vertical (theta1) and the refraction angle relative to the vertical (theta2) by using the next equation:

$(n1)*(sin(theta1))=(n2)*(sin(theta2))$

Then we will find the refraction angle relative to the vertical this way:

$(n1/n2)*(sin(theta1))=sin(theta2)$

$(1/1.33)*(sin(45))=sin(theta2)$

Then, theta2=32.12°

Now that we have this information we can imagine a triangle with a 30cm height and a 32.12° angle. This way we can find how much X is, this X will be the distance between the vertical line and the spot the beam hits the bottom, so we can use some trigonometry to find it, this way:

$tan(32.12)=(X/30cm)$

$X=(tan(32.12))*(30cm)$

Then, X=18.8cm, we can approximate it to 19cm

Once we have X we will add 5cm to it which is how much the beam needs to be moved, then the new X will be 24cm

Now, with the new horizontal distance we will find the new vertical distance, let´s call it Y, this way we will know how much water we must add to move the beam, then we will have a triangle with a vertical distance called Y, the same 32.12° angle will be used as we are still working with the air-water interface and a 19cm horizontal distance, then:

$tan(32.12)=(24cm/Y)$

$Y=(24cm/tan(32.12))$

Then, Y=38cm

In this case, you must add 8cm of water to the tank to move the beam on the bottom 5cm

5 0

3 years ago