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3 years ago

What is the changing relationship between volume and surface area as an object get bigger?

1 answer:

soldier1979 [14.2K]3 years ago

8 0

As the cube size increases or the cell gets bigger , then the surface area to volume ratio - SA:V ratio decreases. When an object/cell is very small, it has a large surface area to volume ratio, while a large object/ cell has a small surface area to volume ratio.

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Given a constant acceleration and assuming linear motion, derive equations for velocity and position of a body with respect to t

uysha [10]

Answer:

v = at + u

$x = ut+\frac{1}{2}at^{2}+x_{0}$

Explanation:

acceleration, a = constant

As we know that acceleration is the rate of change of velocity

$a=\frac{dv}{dt}$

$dv=adt$

integrate on both sides

$\int dv=\int adt$

v = at + u

Where, u is the integrating constant and here it is equal to the initial velocity

Now we know that the rate of change of displacement is called velocity

$v = \frac{dx}{dt}$

$dx=vdt=(u+at) dt$

Integrate on both sides

$\int dx=\int (u+at) dt$

$x = ut+\frac{1}{2}at^{2}+x_{0}$

where, xo is the integrating constant which is initial position of the particle.

8 0

3 years ago

HELPPP I NEED HELP ASAP NOW

Zarrin [17]

Answer:

Your answers would be

4. A. sperm and testosterone.

7. C. prostate, penis, Testes

uterus, vagina, fallopian tubes

10. B. Protein

11. A. carbohydrate

12. B. amino acids (I'm not positive on this i haven't taken bio in years

27. D. Respiratory system

Explanation:

yeah

8 0

3 years ago

A pendulum with a length of 1 meter is released from an initial angle of 15.0° after 1000s its amplitude has been reduced by fri

Setler [38]

Answer:

0.366×10^{-3} / s

Explanation:

θ = θmax e^{-bt/2m}

Given that

θ = 5.50°

θmax = 15.0°

So that we have

ln (θ / θmax) = -bt /2m

= - ln(5.50°/ 15.0°) / 1000s = b /2m

= b / 2m = 0.366×10^{-3} / s

3 0

3 years ago

a stone is thrown horizonttaly from a cliff of a hill with an initial velocity of 30m/s it hits the ground at a horizontal dista

ELEN [110]

Answer:

a) Time = 2.67 s

b) Height = 35.0 m

Explanation:

a) The time of flight can be found using the following equation:

$x_{f} = x_{0} + v_{0_{x}}t + \frac{1}{2}at^{2}$ (1)

Where:

$x_{f}$ : is the final position in the horizontal direction = 80 m

$x_{0}$ : is the initial position in the horizontal direction = 0

$v_{0_{x}}$ : is the initial velocity in the horizontal direction = 30 m/s

a: is the acceleration in the horizontal direction = 0 (the stone is only accelerated by gravity)

t: is the time =?

By entering the above values into equation (1) and solving for "t", we can find the time of flight of the stone:

$t = \frac{x_{f}}{v_{0}} = \frac{80 m}{30 m/s} = 2.67 s$

b) The height of the hill is given by:

$y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}$

Where:

$y_{f}$ : is the final position in the vertical direction = 0

$y_{0}$ : is the initial position in the vertical direction =?

$v_{0_{y}}$ : is the initial velocity in the vertical direction =0 (the stone is thrown horizontally)

g: is the acceleration due to gravity = 9.81 m/s²

Hence, the height of the hill is:

$y_{0} = \frac{1}{2}gt^{2} = \frac{1}{2}9.81 m/s^{2}*(2.67 s)^{2} = 35.0 m$

I hope it helps you!

5 0

3 years ago

Traveler A starts from rest at a constant acceleration of 6 m/s^2. Two seconds later, traveler B starts with an initial velocity

Troyanec [42]

Answer:

3. 3.5 s

Explanation:

The position of traveller A is given by the equation:

$x_A(t) = \frac{1}{2}a t^2$

where

$a = 6 m/s^2$ is the acceleration of A

t is the time measured from when A started the motion

The position of traveller B instead is given by

$x_B(t) = u_B (t-2) + \frac{1}{2}a(t-2)^2$

where a (acceleration) is the same as traveller A, and

$u_B = 20 m/s$

is B's initial velocity. We can verify that the formula is correct by substituting t=2, and we get $x_B=0$ , which means that B starts its motion 2 seconds later.

Traveller B overtakes traveller A when the two positions are the same, so:

$x_A = x_B\\\frac{1}{2}at^2 = u_B (t-2) + \frac{1}{2}a(t-2)^2\\\frac{1}{2}at^2 = u_B t - 2u_B +\frac{1}{2}at^2 +2a-2at\\u_Bt-2at = 2u_B-2a\\t=\frac{2u_B-2a}{u_B-2a}=\frac{2(20)-2(6)}{20-2(6)}=3.5 s$

4 0

3 years ago