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oee [108]
3 years ago
13

Why is cell called structual unit of life​

Physics
1 answer:
Anika [276]3 years ago
4 0

Answer:

The cells are called the structural unit of life as all living organisms are made up of cells .......

Explanation:

please mark me as Brainliest and follow me

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Object 1 and 2 attract each other with a electrostatic force of 36.0 units. If the charge of object 1 is changed to four times t
Mrac [35]
The answer would be 6
8 0
3 years ago
What are tiny sacs at the end of the bronchioles filled with air called?
WARRIOR [948]

Answer:

Alveoli

The bronchioles end in tiny air sacs called alveoli, where oxygen is transferred from the inhaled air to the blood.

Hope this helps :)

7 0
4 years ago
The near point of an eye is 56.0 cm. A corrective lens is to be used to allow this eye to focus clearly on objects at the distan
irakobra [83]

Answer:

Explanation:

Near point = 56 cm .

near point of healthy person = 25 cm

person suffers from long sightedness

convex lens will be required .

object distance u  = 25 cm

image distance   v = 56 cm

both will be negative as both are in front of the lens.

lens formula

I/v - 1 / u = 1/f

- 1/56 +1/25 = 1/f

- .01785 + .04 = 1/f

1/f  = .02215

f = 45.15 cm .

4 0
4 years ago
Gravity is defined as: Select one: a. an apple falling from a tree toward the ground. b. ability to accomplish a job with the le
Scorpion4ik [409]

Answer:

c. natural force or pull toward the earth

Explanation:

Gravity of the earth is the force of attraction that it naturally possesses to attract any mass.

An apple falls on the earth due to this force of gravity.

The force of gravity near to the surface of the earth is given as:

F=m.g

where:

m= mass of the body

g= acceleration due to gravity

The variation of the gravitational force with height is given as:

F'=m.g'

where:

g'=g\times (1+\frac{h}{R} )^{-2}

where

R = radius of the earth \approx64000\ km

4 0
3 years ago
A particle initially located at the origin has an acceleration of vector a = 2.00ĵ m/s2 and an initial velocity of vector v i =
natali 33 [55]

With acceleration

\mathbf a=\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)\,\mathbf j

and initial velocity

\mathbf v(0)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i

the velocity at time <em>t</em> (b) is given by

\mathbf v(t)=\mathbf v(0)+\displaystyle\int_0^t\mathbf a\,\mathrm du

\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\displaystyle\int_0^t\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)\,\mathbf j\,\mathrm du

\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf j\bigg|_{u=0}^{u=t}

\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)t\,\mathbf j

We can get the position at time <em>t</em> (a) by integrating the velocity:

\mathbf x(t)=\mathbf x(0)+\displaystyle\int_0^t\mathbf v(u)\,\mathrm du

The particle starts at the origin, so \mathbf x(0)=\mathbf0.

\mathbf x(t)=\displaystyle\int_0^t\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf j\,\mathrm du

\mathbf x(t)=\left(\left(8.00\dfrac{\rm m}{\rm s}\right)u\,\mathbf i+\dfrac12\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u^2\,\mathbf j\right)\bigg|_{u=0}^{u=t}

\mathbf x(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)t\,\mathbf i+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)t^2\,\mathbf j

Get the coordinates at <em>t</em> = 8.00 s by evaluating \mathbf x(t) at this time:

\mathbf x(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)(8.00\,\mathrm s)\,\mathbf i+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)(8.00\,\mathrm s)^2\,\mathbf j

\mathbf x(8.00\,\mathrm s)=(64.0\,\mathrm m)\,\mathbf i+(64.0\,\mathrm m)\,\mathbf j

so the particle is located at (<em>x</em>, <em>y</em>) = (64.0, 64.0).

Get the speed at <em>t</em> = 8.00 s by evaluating \mathbf v(t) at the same time:

\mathbf v(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)(8.00\,\mathrm s)\,\mathbf j

\mathbf v(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(16.0\dfrac{\rm m}{\rm s}\right)\,\mathbf j

This is the <em>velocity</em> at <em>t</em> = 8.00 s. Get the <em>speed</em> by computing the magnitude of this vector:

\|\mathbf v(8.00\,\mathrm s)\|=\sqrt{\left(8.00\dfrac{\rm m}{\rm s}\right)^2+\left(16.0\dfrac{\rm m}{\rm s}\right)^2}=8\sqrt5\dfrac{\rm m}{\rm s}\approx17.9\dfrac{\rm m}{\rm s}

5 0
3 years ago
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