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Maru [420]
3 years ago
9

The most effective way to process biodiversity is through

Physics
1 answer:
lidiya [134]3 years ago
6 0
I suppose it's C. All the other answers are everything but accurate to me
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A physicist's right eye is presbyopic (i.e., farsighted). This eye can see clearly only beyond a distance of 97 cm, which makes
Ivan

Answer:

f = 19,877 cm   and  P = 5D

Explanation:

This is a lens focal length exercise, which must be solved with the optical constructor equation

        1 / f = 1 / p + 1 / q

where f is the focal length, p is the distance to the object and q is the distance to the image.

In this case the object is placed p = 25 cm from the eye, to be able to see it clearly the image must be at q = 97 cm from the eye

let's calculate

        1 / f = 1/97 + 1/25

        1 / f = 0.05

        f = 19,877 cm

the power of a lens is defined by the inverse of the focal length in meters

         P = 1 / f

         P = 1 / 19,877 10-2

         P = 5D

5 0
3 years ago
Help me please forget about you and she was
ra1l [238]

The figure is showing a volume of 2.4 mL becuase it's feel 4 little segments.

Therefore, the answer is 2.4 mL.

5 0
1 year ago
How much work is done when a 214 newton force pushes a sleeping cow 37m across a field.
nata0808 [166]
Hello!

Answer: 
7918 J

Explanation:

We are assuming that the floor (field) is completely horizontal since there's no information about that in the statement. 

We are going to use the following formula:

W= F . Cos \alpha . D

Where:

F=214 N
\alpha =0º
D= 37m

Then, by substituting we have:

W=214N . Cos (0).37m= 7918 N.m=7918 J

8 0
3 years ago
needhelpp101 Why is the gravitational potential energy of an object 1 meter above the moon’s surface less than its potential ene
Vlad1618 [11]

If you remember the formula for potential energy,
then this question is a piece-o-cake.

  <em>Potential energy = (mass) x (<u>acceleration of gravity</u>) x (height) .</em>

-- The object's mass is the same everywhere.
-- You said that the height is the same both times.
-- How about the acceleration of gravity ? 

Compared to gravity on Earth, it's only  16.5 percent as much on the Moon. 
So naturally, from the formula, you'd expect the Potential Energy to be less
on the Moon.

4 0
3 years ago
Find the position vector of a particle that has the given acceleration and the specified initial velocity and position. a(t) = 1
kondor19780726 [428]

Answer:

Explanation:

Given

Acceleration a(t)=14t\hat{i]+\sin (t)\hat{j}+\cos (2t)\hat{k}[/tex]

and v(0)=\hat{i}

r(0)=\hat{j}

we know a=\frac{\mathrm{d} v}{\mathrm{d} t}

\int dv=\int adt

v(t)=\int (14t\hat{i}+\sin (t)\hat{j}+\cos (2t)\hat{k})dt

v(t)=7t^2\hat{i}-\cos t\hat{j}+\frac{\sin (2t)\hat{k}}{2}+c

at t=0

v(0)=0-1\cdot \hat{j}+0+c

c=\hat{i}+\hat{j}

v(t)=(7t^2+1)\hat{i}+(1-\cos t)\hat{j}+\frac{\sin (2t)\hat{k}}{2}

and \frac{\mathrm{d} r}{\mathrm{d} t}=v(t)

\int dr=\int vdt

r(t)=\int ((7t^2+1)\hat{i}+(1-\cos t)\hat{j}+\frac{\sin (2t)\hat{k}}{2})dt

r(t)=(\frac{7}{2}t^3+t)\hat{i}+(t-\sin (t))\hat{j}+\frac{1}{2}\times (-\frac{1}{2}\cos 2t)\hat{k}+c_2

at t=0

r(0)=\hat{j}

r(t)=(\frac{7}{3}t^3+t)\hat{i}+(1+t-\sin t)\hat{j}+\frac{1}{4}(1-\cos 2t)\hat{k}

       

4 0
3 years ago
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