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3 years ago

12

In what climates would you expect soil to form fastest? Why?

1 answer:

enyata [817]3 years ago

5 0

Answer:

Soils develop faster in warm, moist climates and slowest in cold or arid ones. Rainfall is one of the most important climate factors in soil formation.

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Hope this helped God bless :)

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Ira is concerned about his posture. He tests it by seeing how long he can hold good posture and finds that he feels the most dis

larisa [96]

Answer: C

Explanation:

JUST TO SIMPLE

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3 years ago

s=1/2(u+v)t check the dimensional consistency for this? where s is displacement, t is time, u is initial velocity, v is final ve

Anettt [7]

Answer:

See the explanation below.

Explanation:

The units of work are consistent since if we work in the international system of measures we have the following dimensional quantities for velocity, distance and time.

s = displacement [m]

v and u = velocity [m/s]

t = time [s]

Now using these units in the given equation.

$s = 0.5*([m/s]+[m/s])*[s]\\s=0.5*[m/s]*[s]\\s = 0.5*[m]$

So the expression is good, and dimensional has consistency.

8 0

3 years ago

A ball is thrown straight up into the air with an initial speed of 8.0 m/s. How long does it take for the rocket to reach its hi

andrezito [222]

What is the highest point??

3 0

3 years ago

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(NEED HELP PLEASE) A physics student goes to the roof of the school, 24.15 m above the ground, and drops a pumpkin straight down

slavikrds [6]

Answer:

t = 2.2 s

Explanation:

Given that,

Height of the roof, h = 24.15 m

The initial velocity of the pumpkin, u = 0

We need to find the time taken for the pumpkin to hit the ground. Let the time be t. Using second equation of kinematics to find it as follows :

$h=ut+\dfrac{1}{2}at^2$

Here, u = 0 and a = g

$h=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2h}{g}} \\\\t=\sqrt{\dfrac{2\times 24.15}{9.8}} \\\\t=2.22\ s$

So, it will take 2.22 s for the pumpkin to hit the ground.

7 0

3 years ago

Water falls without splashing at a rate of 0.200 L/s from a height of 3.60 m into a 0.730 kg bucket on a scale. If the bucket is

dimaraw [331]

Answer:

15.106 N

Explanation:

From the given information,

The weight of the bucket can be calculated as:

$W_b = m_bg = \\ \\ W_b = (0.730 \ kg) ( 9.80 \ m/s^2) \\ \\ W_b = 7.154 \ N$

The mass of the water accumulated in the bucket after 3.20s is:

$m_w= (0.20 \ L/s) ( 3.20)s$

$m _w=0.64 \ kg$

To determine the weight of the water accumulated in the bucket, we have:

$W_w = m_w g$

$W_w = ( 0.64 \ kg )(9.80\ m \ /s^2)$

$W_w = 6.272 \ N$

For the speed of the water before hitting the bucket; we have:

$v = \sqrt{2gh}$

$v = \sqrt{2*9.80 \ m/s^2 * 3.60 \ m}$

v = 8.4 m/s

Now, the force required to stop the water later when it already hit the bucket is:

$F = v ( \dfrac {dm}{dt} )$

$F = (8.4 \ m/s)( 0.200 \ L/s)$

F = 1.68 N

Finally, the reading scale is:

$F_{scale$ = 7.154 N + 6.272 N + 1.68 N

= 15.106 N

6 0

2 years ago

Read 2 more answers