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3 years ago

Why one part of the earth's surface is arid and dry and a nearby part is lush and wet is explained by the term?

1 answer:

8 0

The process that explains why one part of the earth's surface is arid and dry and a nearby part is lush and wet is areal differentiation. It is<span> an approach to geography that shows </span>the dependence of the distribution of physical and human phenomena and the relation to each other from the physical location. Areal integration on the other hand is the approach that studies how places interact with each other.

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A laser pulse takes 2.56 seconds to travel from Earth to the Moon and return. Use this to calculate how far away the Moon is. Ho

Deffense [45]

Answer:

$x = 3.84\times 10^{5}\,km$

This delay affect conversations in the sense that recipient will be receive the message just 2.56 seconds after the message was sent.

Explanation:

The laser pulse has an ondulatory nature as electromagnetic wave, which can travel in the void. Speed of light is constant and distance between Earth and the Moon is:

$x = 0.5\cdot c\cdot \Delta t$

$x = 0.5\cdot (3\times 10^{5}\,\frac{km}{s})\cdot (2.56\,s)$

$x = 3.84\times 10^{5}\,km$

This delay affect conversations in the sense that recipient will be receive the message just 2.56 seconds after the message was sent.

7 0

3 years ago

A truck is traveling 20m/s accelerates 3 m/s^2 for 4 seconds how far did it travel while it was accelerating. Using guess method

stellarik [79]

<h3> $\huge\underline\bold\blue{ƛƝƧƜЄƦ}$ </h3><h3>Given</h3>

$\blue\star$ v = 20m\s

$\blue\star$ a = 3m\s^2

$\blue\star$ t = 4sec

Firstly we have to find u

$\star$ a = $\dfrac{v - u}{t}$

$\star$ 3m\s = $\dfrac{20 - u}{4}$

$\star$ 12m\s = 20 - u

$\star$ 20 - u = 12m\s

$\star$ - u = -8

$\star$ u = 8

Now we can easily find distance by using second equation of motion

$\red\star$ s = ut + 1\2 at^2

$\red\star$ s = 8(4) + 1\2(3)(16)

$\red\star$ s = 32 + 24

$\red\star$ s = 56

So distance is 56 m\s hope it helps

5 0

3 years ago

A wildebeest and chicken participate in a race over a 2.00km long course. the wildebeest travels at a speed of 16.0m/s and chick

Nezavi [6.7K]

Answer:

(a) The distance of the chicken from the finish line is 62.5 m

(b) The stationary time of the wildebeest is 675 s

Explanation:

Given;

total distance traveled by wildebeest and chicken, d = 2 km = 2000 m

speed of the wildebeest, $v_w$ = 16 m/s

speed of the chicken, $v_c$ = 2.5 m/s

Time for wildebeest to finish the race without stopping, 2000 / 16 = 125 s

Time for chicken to finish the race without stopping, 2000/2.5 = 800 s

(b) for how long in time (in s) was the wildebeest stationary?

t(stationary) = t(chicken) - t(wildebeest)

t = 800s - 125 s

t = 675 s

(a) how far (in m) is the chicken from the Finish Line when the wildebeest resume the race?

The time taken for the wildebeest to run 1.6 km (1600 m) is given by;

t = 1600 / 16 = 100 s

The total time spent by the wildebeest before it resumed the race = stationary time + 100s

t (total) = 675 s + 100 s = 775 s

Distnace traveled by the chicken when the wildebeest resumed the race = 2.5m/s x 775s = 1937.5 m

Thus, the distance of the chicken from the finish line = 2000 m - 1937.5 m

the distance of the chicken from the finish line = 62.5 m

7 0

3 years ago

Firecrackers A and B are 600 m apart. You are standing exactly halfway between them. Your lab partner is 300 m on the other side

pishuonlain [190]

Answer:

See the explanation

Explanation:

Given:

Distance of Firecrackers A and B = 600 m

Event 1 = firecracker 1 explodes

Event 2 = firecracker 2 explodes

Distance of lab partner from cracker A = 300 m

You observe the explosions at the same time

to find:

does event 1 occur before, after, or at the same time as event 2?

Solution:

Since the lab partner is at 300 m distance from the firecracker A and Firecrackers A and B are 600 m apart

So the distance of fire cracker B from the lab partner is:

600 m + 300 m = 900 m

It takes longer for the light from the more distant firecracker to reach so

Let T1 represents the time taken for light from firecracker A to reach lab partner

T1 = 300/c

It is 300 because lab partner is 300 m on other side of firecracker A

Let T2 represents the time taken for light from firecracker B to reach lab partner

T2 = 900/c

It is 900 because lab partner is 900 m on other side of firecracker B

T2 = T1

900 = 300

900 = 3(300)

T2 = 3(T1)

Hence lab partner observes the explosion of the firecracker A before the explosion of firecracker B.

Since event 1 = firecracker 1 explodes and event 2 = firecracker 2 explodes

So this concludes that lab partner sees event 1 occur first and lab partner is smart enough to correct for the travel time of light and conclude that the events occur at the same time.

8 0

3 years ago

Name two kinds of longitudinal waves and explain how you know they are longitudinal

jasenka [17]

First of all Longitudinal waves is a matter in the medium that moves parallel to the direction of the wave travels.
1st example: sound travels parallel.
2nd example: when you talk you will here your voice again. because all the frequency bounce back to you.

In the other hand Transverse wave matter in the medium moves perpendicular to direction the wave travels.
For example: light is a good example of transverse wave.

6 0

3 years ago