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Andre45 [30]
3 years ago
7

if a forest exosystem is removed through clear-cutting, many species of organisms that lived in that ecosystem dsappear. Their l

ost is due to?
Physics
1 answer:
Ksju [112]3 years ago
4 0

The loss of species of organisms is due to loss of their habitat.
If you went through a city and cut all the houses down to their
foundations, the people who lived in them would either die or
go somewhere else if they could.

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Select the statement that shows the correct relationship between current, voltage, and power.
bogdanovich [222]
 Best Answer:  <span>Power = Current * Voltage 

Therefore (B) is correct.

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Reducing friction in a machine
SVETLANKA909090 [29]

Isn't the correct answer it increases efficiency, so C?

Sorry if I'm wrong but pretty sure that's right!

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3 years ago
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if the car goes from 0 to 60 mph in 2.3 seconds, using p.l.i.m., find the average net force acting on the car.
motikmotik

The net force acting on the car is 11.66m newtons.

We need to know about force to solve this problem. Force is proportional to mass and acceleration. It can be written as

F = m . a

where F is force, m is mass and a is acceleration.

The acceleration can be determined by using uniform motion using this equation

vt = vo + a . t

a = (vt - vo) / t

where vt is final speed, vo is initial speed, and t is time interval.

From the question above, we know that:

vo = 0 mph = 0 m/s

vt = 60 mph = 26.82 m/s

t = 2.3 s

m = m

By combining the following equation, we can get the net force

F = m . a

F = m . (vt - vo) / t

F = m . (26.82 - 0) / 2.3

F = 11.66m newtons

where m is mass of the car

Hence, the net force acting on the car is 11.66m newtons

For more on force at: brainly.com/question/25239010

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5 0
2 years ago
A Thomson's gazelle can run at very high speeds, but its acceleration is relatively modest. A reasonable model for the sprint of
frosja888 [35]

1. 27.3 m/s

The velocity of the gazelle at any time is given by:

v=u+at

where

u is the initial velocity

a is the acceleration

t is the time

Here we have:

u = 0 (the gazelle starts from rest)

a=4.2 m/s^2

t = 6.5 s

Substituting the data, we find the gazelle's top speed:

v=0+(4.2)(6.5)=27.3 m/s

2. 3.8 s

The distance covered by the gazelle is

d = 30 m

We know that the gazelle accelerates during the first part of the motion and then it continues at constant speed. We need to find first if the gazelle completes the race during the first part of its motion (accelerated motion); to do this, we can calculate what would be the distance covered by the gazelle before reaching the top speed, after t = 6.5 s:

d'=\frac{1}{2}at^2 = \frac{1}{2}(4.2)(6.5)^2=88.7 m

Which is larger than 30 m: this means that the gazelle covers the 30 m during its accelerated motion. Therefore, we can use again the equation:

d=\frac{1}{2}at^2

And substituting d = 30 m, we find the time:

t=\sqrt{\frac{2d}{a}}=\sqrt{\frac{2(30)}{4.2}}=3.8 s

3. 10.6 s

In this case, the  distance the gazelle must cover is 200 m.

We know that in the first 6.5 s, the gazelle covers a distance of 88.7 m.

In the second part of the motion, the gazelle continues at its top speed, which is:

v = 27.3 m/s

The gazelle still have to cover a distance of

d' = 200-88.7 =111.3 m

Therefore, the time taken to cover this distance is

t'=\frac{d'}{v}=\frac{111.3}{27.3}=4.1 s

So, the total time the gazelle needs to cover 200 m is

t = 6.5 + 4.1 = 10.6 s

6 0
4 years ago
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