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WARRIOR [948]
2 years ago
7

In grassland regions, rainy seasons and drought seasons determine, in part, the _____. kinds of resident organisms spread of fir

es average temperature location of fresh water streams
Physics
2 answers:
qaws [65]2 years ago
6 0

kinds of resident organisms
trasher [3.6K]2 years ago
6 0

The correct answer is option A

In grassland regions, rainy seasons the drought seasons determine, in part the kinds of resident organism.

The adaptions of the organism depends on the climatic and environmental conditions in which they live.

By knowing the environmental conditions of the place we get an idea of the organisms that survive there.

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Step 8: Observe How Changes in the Speed of the Bottle Affect Beanbag Height
lina2011 [118]

Answer:

When the speed of the bottle is 2 m/s, the average maximum height of the beanbag is <u>0.10</u>  m.

When the speed of the bottle is 3 m/s, the average maximum height of the beanbag is<u> 0.43</u>  m.

When the speed of the bottle is 4 m/s, the average maximum height of the beanbag is  <u>0.87</u> m.

When the speed of the bottle is 5 m/s, the average maximum height of the beanbag is  <u>1.25</u> m.

When the speed of the bottle is 6 m/s, the average maximum height of the beanbag is  <u>1.86</u> m.

Sorry for not answering early on! If anyone in the future needs help, I got these answers from 2020 egenuity, though I can't post the picture for proof. Stay Safe!

6 0
2 years ago
Read 2 more answers
One beaker contains 100 mL of pure water and second beaker contains 100 mL of seawater. The two beakers are left side by side on
Harman [31]

Answer: The beaker containing pure water has decreased more.

Explanation:

In both cases, the decrease of water level is due to evaporation. We know that evaporation is a surface phenomenon. In the case of salt water, the salt molecules somewhat hinders the evaporation process of the water molecules and hence the salt water evaporates at a slower rate than pure water.

Hence, pure water level falls more.

7 0
2 years ago
Read 2 more answers
A large crate sits on the floor of a warehouse. Paul and Bob apply constant horizontal forces to the crate. The force applied by
Delicious77 [7]

Answer:

W = -510.98J

Explanation:

Force = 43N, 61° SW

Displacement = 12m, 22° NE

Work done is given as:

W = F*d*cosA

where A = angle between force and displacement.

Angle between force and displacement, A = 61 + 90 + 22 = 172°

W = 43 * 12 * cos172

W = -510.98J

The negative sign shows that the work done is in the opposite direction of the force applied to it.

6 0
2 years ago
As shown in the figure below, a bullet is fired at and passes through a piece of target paper suspended by a massless string. Th
NikAS [45]

Answer:

M = 0.730*m

V = 0.663*v

Explanation:

Data Given:

v_{bullet, initial} = v\\v_{bullet, final} = 0.516*v\\v_{paper, initial} = 0\\v_{paper, final} = V\\mass_{bullet} = m\\mass_{paper} = M\\Loss Ek = 0.413 Ek

Conservation of Momentum:

P_{initial} = P_{final}\\m*v_{i} = m*0.516v_{i} + M*V\\0.484m*v_{i} = M*V .... Eq1

Energy Balance:

\frac{1}{2}*m*v^2_{i} = \frac{1}{2}*m*(0.516v_{i})^2 + \frac{1}{2}*M*V^2 + 0.413*\frac{1}{2}*m*v^2_{i}\\\\0.320744*m*v^2_{i} = M*V^2\\\\M = \frac{0.320744*m*v^2_{i} }{V^2}  ....... Eq 2

Substitute Eq 2 into Eq 1

0.484*m*v_{i} = \frac{0.320744*m*v^2_{i} }{V^2} *V  \\0.484 = 0.320744*\frac{v_{i} }{V} \\\\V = 0.663*v_{i}

Using Eq 1

0.484m*v_{i} = M* 0.663v_{i}\\\\M = 0.730*m

7 0
2 years ago
A toy rocket is launched vertically from ground level (y = 0 m), at time t = 0.0 s. The rocket engine provides constant upward a
Dvinal [7]

Answer:

2.13 s

Explanation:

Hi!

At t = 0s the rocket is at rest in its platform, so the intial speed is zero. I f the acceleration is A, then the height Y, and the speed V are:

Y=\frac{A}{2}t^2

V=At

We nedd to find time T during  which the rocket engine provides upward acceleration. We know that:

64\;m=\frac{A}{2}T^2\\ 60\frac{m}{s} =AT\\

With these 2 equations we can find A and T (dropping units for simplicity):

A=\frac{60}{T} \\64 =\frac{30}{T} T^2=30T\\T=\frac{64}{30}\approx 2.13\;s

4 0
2 years ago
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