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maw [93]
3 years ago
7

A population of mammals leaves an ecosystem because of increased temperatures and decreased rainfall. What’s the reason for this

shift?
A.
change in abiotic factors
B.
change in biotic factors
C.
change in food preferences
D.
change in genetic makeup
Physics
2 answers:
erastova [34]3 years ago
5 0

Answer:

The answer is question number A. Change in abiotic factors

ratelena [41]3 years ago
3 0
The answer to your question is a
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Vector A and B are given as follows:
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2i+3j+4k

Explanation:

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Grace rides her bike at a constant speed of 6 miles per hour. How far can she travel in 2 1/2 hours?
nydimaria [60]

Answer:

15~miles

Explanation:

2\frac{1}{2}*6=\frac{2*2+1}{2}*6=\frac{5}{2}*6=\frac{30}{2}=15~miles

7 0
3 years ago
What type of friction is using chalk in the summer to draw on the ground in Copley square?
Lera25 [3.4K]
The second option rolling friction
7 0
3 years ago
Water flows over a section of Niagara Falls at a rate of 2.2 × 106 kg/s and falls 65 m. What is the power wasted by the waterfal
Irina-Kira [14]

Answer:

P = 1401.4 x 10⁶ W

Explanation:

Given that

Water flow rate m = 2.2 x 10⁶ kg/s

Height ,h= 65 m

Acceleration due to gravity ,g= 9.8 m/s²

The power P is given as

P= m g h

P=Power

m=Water flow rate

h=height

Now by putting the values in the above equation

P = 2.2 x 10⁶ x 9.8 x 65 W

P = 1401.4 x 10⁶ W

Therefore the power wasted will be 1401.4 x 10⁶ W.

7 0
3 years ago
A small block of mass m = 0.032 kg can slide along the frictionless loop-the-loop, with loop radius R = 12 cm. The block is rele
Lunna [17]

Answer:

Part a)

W = 0.15 J

Part b)

W = 0.11 J

Part c)

U = 0.19 J

Part d)

U = 0.038 J

Part e)

U = 0.075 J

Part f)

It is independent of the speed of the object so all part answers will remain the same

Explanation:

Part a)

As we know that Point P is at height 5R while point Q is at height R

so the work done by gravity from P to Q is given as

W = mg(5R - R)

W = 0.032(9.8)(4)(0.12)

W = 0.15 J

Part b)

When it reaches to the top of the loop then its final height from ground is

h = 2R

so work done from P to Q is given as

W = mg(5R - 2R)

W = 3mgR

W = 0.11 J

Part c)

Potential energy at P point is given as

U = mgH

U = 0.032(9.8)(5)(0.12)

U = 0.19 J

Part d)

Potential energy at Q point is given as

U = mgH

U = 0.032(9.8)(0.12)

U = 0.038 J

Part e)

Potential energy at top point is given as

U = mgH

U = 0.032(9.8)(2)(0.12)

U = 0.075 J

Part f)

Since all the answer from part a) to part e) depends only upon the position of the object.

So here we can say that it is independent of the speed of the object so all part answers will remain the same

8 0
3 years ago
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