Which is correct help asap

Can someone please help me with science.

pav-90 [236]

On the dog's return trip (between t = 10 and t = 12.5 seconds), the slope of the position function is steeper than during the first 5 seconds, which means the dog ran home faster. The only option that captures this is D.

You can check to make sure that the dog indeed runs twice as fast on the return trip. The slope of the position function during the first 5 seconds is

(change in position) / (change in time) = (5 - 0) / (5 - 0) = 5/5 = 1

while during the return trip, it is

(0 - 5) / (12.5 - 10) = -5/2.5 = -2

Ignoring the sign (which only indicates the direction in which the dog was running), we see that the dog's speed on the return trip was indeed twice as high as during the first 5 s.

8 0

3 years ago

An airplane flies northwest for 250 mi and then west for 150 mi. What is the resultant displacement of the plane after this time

Kazeer [188]

Answer:371.564 mi

Explanation:

Given

Airplane flies northwest for 250 mi and then travels west 150 mi

That is first it travels 250cos45 in - ve x direction and simultaneously 250sin45 in y direction

after that it travels 150 mi in -ve x direction

So its position vector is given by

$r=-250cos45\hat{i}-150\hat{i}+250sin45\hat{j}$

$r=-\left ( 250cos45+150\right )\hat{i}+250sin45\hat{j}$

so magnitude of displacement is

$|r|=\sqrt{\left ( \frac{250}{\sqrt{2}}+150\right )^2+\left ( \frac{250}{\sqrt{2}}\right )^2}$

|r|=371.564 mi

7 0

3 years ago

At the very end of Wagner's series of operas The Ring of Nibelung, Brunnhilde takes the golden ring form the finger of the dead

Blababa [14]

Answer:

a) 404 m² b) apparent height = 7.5 m

Explanation:

This question is about refraction and total internal refraction.

Here I will take refractive index of air and water

$n_{air}=1\\ n_{water}=1.33=4/3$

Now let's look at the diagram I have attached here

At some angle A, the light from the ring (yellow point) under water will be totally internally refracted (B = 90°), which means that rays of light (yellow arrow) that make large enough angle A will not be able to escape from the water. Since we assumed that the ring is a point, there will be a critical cone of angle A with the ring at its apex which traces a circle of radius R on the surface of water, which, beyond this radius, no light could escape.

According to snell's law

$\frac{sin(B)}{sin(A)} = \frac{n_{water}}{n_{air}} = 4/3$

At critical angle B = 90°

$\frac{3)}{4}sin(B) = [tex]\frac{3}{4} sin(90^\circ ) = 0.75 = sin(A)$

Therefore

$A = 48.6^\circ$

With this, we can find the radius of the circle (refer to my diagram)

$h* tan (A) = R\\R =11.3 m$

And with that we can find the area

$A = \pi R^2=404\ m^2$

Additional Problem

For apparent depth from above, we can think that, since we are accustomed to seeing light at the speed of c in air, our brain interpret light from any source to be traveling at c. This causes light that originated under water, which has the speed of

$v_{water} = \frac{c}{n_{water}} = 0.75c$