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hoa [83]
3 years ago
10

Homeostatasis will be most affected by the romoval of the

Physics
2 answers:
ser-zykov [4K]3 years ago
4 0
My answer is the cell membrane. hope this helps.
Lady bird [3.3K]3 years ago
3 0
I would say D. Hope I helped!
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Suppose you ride your bike to the library traveling at 0.5 km/min. It takes you 25 minutes to get to the library. How far did yo
madam [21]
Answer= 12.5 km

 0.5 km      x km
_______=______
  1 min      25 min

Cross multiply
1x=12.5
divide both sides by 1
x=12.5

You traveled 12.5 km
4 0
3 years ago
Read 2 more answers
True or false. Lateral to the inguinal region is the coxal region, while superior to the scapular region is the dorsal region.
mars1129 [50]

Answer:

false

Explanation:

false

false

The right answer is false

4 0
3 years ago
Two rocks, A and B, are thrown horizontally from the top of a cliff. Rock A has an initial speed of 10 meters per second and roc
maria [59]

Answer:

i need help

Explanation:

4 0
2 years ago
What magnification would be obtained if an eyepiece with a focal length of 0.38 m was placed on telescope?
weqwewe [10]

Answer:

This question is incomplete

Explanation:

This question is incomplete because the telescope's focal length was not provided. The formula to be used here is

Magnification = telescope's focal length/eyepiece's focal length

The eyepiece's focal length was provided in the question as 0.38 m.

NOTE: Magnification can be described as the power of an instrument (in this case telescope) to enlarge an object. It has no unit and thus the two focal lengths mentioned in the formula above must be in the same unit (preferably meters since one of them is in meters already).

7 0
3 years ago
A uniform wooden plank with a mass of 75kg and length of 5m is placed on top of a brick wall so that 1.5m of plank extends beyon
jek_recluse [69]

Answer:

x₂ = 1.33 m

Explanation:

For this exercise we must use the rotational equilibrium condition, where the counterclockwise rotations are positive and the zero of the reference system is placed at the turning point on the wall

            Στ = 0

            W₁ x₁ - W₂ x₂ = 0

where W₁ is the weight of the woman, W₂ the weight of the table.

Let's find the distances.

Since the table is homogeneous, its center of mass coincides with its geometric center, measured at zero.

           x₁ = 2.5 -1.5 = 1 m

The distance of the person is x₂ measured from the turning point, at the point where the board begins to turn the girl must be on the left side so her torque must be negative

            x₂ = \frac{M_1g  }{m_2 g} \ x_1

           

let's calculate

           x₂ = \frac{100}{75}  \ 1

           x₂ = 1.33 m

7 0
3 years ago
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