A chemical change means a new substance with new properties was formed. True False

In diving to a depth of 308 m, an elephant seal also moves 579 m due east of his starting point. What is the magnitude of the se

Kisachek [45]

Answer:

655.82m

Explanation:

By tracing the distances traveled from 308m down in the water and then 579m to the east, a right triangle can be formed and the displacement can be found by the Pythagorean theorem. This is shown in the attached image (the red line is the displacement).

One leg of the triangle will be the 308m

and the second leg of the triangle will be the 579m.

If we call the displacement $x$ , by the pythagorean theorem:

$x=\sqrt{308^2+579^2}$

$x=\sqrt{94,864+335,241}$

$x=\sqrt430,105}$

$x=655.82m$

The displacement is 655.82m

3 0

3 years ago

A car travels due east 14 miles in 18 minutes. Then, the car turns around and returns to its starting point, taking an additiona

N76 [4]

Answer:

(a) The average speed is 0.85 milles/minute

(b) The average velocity is zero

Explanation:

In order to answer part (a) and (b) you have to apply the formulas for average speed and average velocity which are:

-Average speed formula:

$v=\frac{d}{t}$

where d is the total distance traveled and t is the tota time

Replacing the given values:

$v=\frac{14+14}{18+15}\\v=\frac{28}{33} \\v=0.85$ milles/minute

Notice that you have to replace the total distance, which is 14 milles for the go plus 14 milles for the return. The same for the total time.

-Average velocity formula:

V = Δx/Δt

Where V is the velocity vector, Δx is the displacement and Δt is the change in time

V= $\frac{X2-X1}{t2-t1}$

Where X2 is the final position and X1 is the initial position

In this case X1= 0 i and X2=0 i (i is the unit vector in the x direction). So, the displacement is zero.

Therefore, the average velocity is:

V= 0 i [milles/minute]

4 0

2 years ago

A low orbit satellite at 100 km altitude had a camera that resolved a car location to within 0.3 meter. Find the minimum diamete

NeX [460]

Answer:

Minimum diameter of the camera lens is 22.4 cm

The focal length of the camera's lens is 300cm

Explanation:

y = Resolve distance = 0.3 m

h = Height of satellite = 100 km

λ = Wavelength = 550 nm

Angular resolution

$tan\theta\approx \theta =\frac{y}{h}\\\Rightarrow \theta=\frac{0.3}{100\times 10^3}=3\times 10^{-6}$

From Rayleigh criteria

$sin\theta=1.22\frac{\lambda}{D}\\\Rightarrow D=1.22\frac{\lambda}{sin\theta}\\\Rightarrow D=1.22\frac{550\times 10^{-9}}{sin3\times 10^{-6}}=0.2236\ m=22.4\ cm$