You feel a static electric shock when free electrons build up on your body then you touch a

The critical angle for total internal reflection occurs when:___________.

Anastaziya [24]

Answer:

b) the refracted ray has an angle of 90 degrees

Explanation:

The angle of incidence is measured with respect to the normal separation of the media. The critical angle for total internal reflection occurs is:

$\theta=arcsin(\frac{n_2}{n_1})$

Here $n_2$ and $n_1$ are the refractive index of the mediums. This equation is an application of Snell's law, for the case where the refracted ray has an angle of $90^\circ$ .

5 0

3 years ago

Drag the titles to the correct boxes to complete the pairs.

PilotLPTM [1.2K]

Can you input a picture??

5 0

3 years ago

The radius of a sphere is increasing at a rate of 4 mm/s. how fast is the volume increasing when the diameter is 40 mm?

marin [14]

Using r to represent the radius and t for time, you can write the first rate as:

drdt=4mms

or

r=r(t)=4t

The formula for a solid sphere's volume is:

V=V(r)=43πr3

When you take the derivative of both sides with respect to time...

dVdt=43π(3r2)(drdt)

...remember the Chain Rule for implicit differentiation. The general format for this is:

dV(r)dt=dV(r)dr(t)⋅dr(t)dt with V=V(r) and r=r(t) .

So, when you take the derivative of the volume, it is with respect to its variable r (dV(r)dr(t)) , but we want to do it with respect to t (dV(r)dt) . Since r=r(t) and r(t) is implicitly a function of t , to make the equality work, you have to multiply by the derivative of the function r(t) with respect to t (dr(t)dt) as well. That way, you're taking a derivative along a chain of functions, so to speak (V→r→t ).

Now what you can do is simply plug in what r is (note you were given diameter) and what drdt is, because dVdt describes the rate of change of the volume over time, of a sphere.

dVdt=43π(3(20mm)2)(4mms) =6400πmm3s

Since time just increases, and the radius increases as a function of time, and the volume increases as a function of a constant times the radius cubed, the volume increases faster than the radius increases, so we can't just say the two rates are the same.

7 0

3 years ago

A box is initially sliding across a frictionless floor toward a spring which is attached to a wall. the box hits the end of the

Serjik [45]

The elastic potential energy of a spring is given by
$U= \frac{1}{2}kx^2$
where k is the spring's constant and x is the displacement with respect to the relaxed position of the spring.

The work done by the spring is the negative of the potential energy difference between the final and initial condition of the spring:
$W=-\Delta U = \frac{1}{2}kx_i^2 - \frac{1}{2}kx_f^2$

In our problem, initially the spring is uncompressed, so $x_i=0$ . Therefore, the work done by the spring when it is compressed until $x_f$ is
$W=- \frac{1}{2}kx_f^2$
And this value is actually negative, because the box is responsible for the spring's compression, so the work is done by the box.

8 0

4 years ago

The slope of the following graph represents

guajiro [1.7K]

Answer:

ok jsjajakaka you can come to me when you get home can you please send me the details of the day and I will be there at this time of year is the best way to get a hold of the guy who was the guy who was the guy who was the guy who was the guy who was the guy that was the only thing that was the case but I don't know if I can help you

4 0

3 years ago