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geniusboy [140]

3 years ago

9

Experiments using "optical tweezers" measure the elasticity of individual DNA molecules. For small enough changes in length, the

elasticity has the same form as that of a spring. A DNA molecule is anchored at one end, then a force of 1.5nN (1.5×10−9N) pulls on the other end, causing the molecule to stretch by 5.0nm (5.0×10−9m). What is the spring constant of that DNA molecule?

1 answer:

GalinKa [24]3 years ago

8 0

Answer:

Spring constant, k = 0.3 N/m

Explanation:

It is given that,

Force acting on DNA molecule, $F=1.5\ nN=1.5\times 10^{-9}\ N$

The molecule got stretched by 5 nm, $x=5\times 10^{-9}\ m$

Let k is the spring constant of that DNA molecule. It can be calculated using the Hooke's law. It says that the force acting on the spring is directly proportional to the distance as :

$F=-kx$

$k=\dfrac{F}{x}$

$k=\dfrac{1.5\times 10^{-9}\ N}{5\times 10^{-9}\ m}$

k = 0.3 N/m

So, the spring constant of the DNA molecule is 0.3 N/m. Hence, this is the required solution.

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Answer:

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A rocket at rest with a mass of 942 kg is acted on by an average net force of 6,731 N upwards for 21 s. What is the final veloci

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Answer:

Explanation:

Givens

m = 942

F = 6731

t = 21 seconds

vi = 0

vf = ?

Formula

F = m * (vf - vi ) / t

Solution

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So far in your life, you may have assumed that as you are sitting in your chair right now, you are not accelerating. However, th

tia_tia [17]

Answer:

a) $a=33.73mm/s^{2}$

b) mg>N

c) $\%_{change}=0.343\%$

d) $a=24.07mm/s^{2}$

Explanation:

In order to solve part a) of the problem, we can start by drawing a free body diagram of the presented situation. (see attached picture).

In this case, we know the centripetal acceleration is given by the following formula:

$a_{c}=\omega ^{2}r$

where:

$\omega=\frac{2\pi}{T}$

we know the period of rotation of the earth is about 24 hours, so:

$T=24hr*\frac{3600s}{1hr}=86400s$

so we can now find the angular speed:

$\omega=\frac{2\pi}{86400s}$

$\omega=72.72x10^{-6} rad/s^{2}$

So the centripetal acceleration will be:

$a_{c} =(72.72x10^{-6} rad/s^{2})^{2}(6478x10^{3}m)$

which yields:

$a_{c}=33.73mm/s^{2}$

b)

In order to answer part b, we must draw a free body diagram of us sitting on a chair. (See attached picture.)

So we can do a sum of forces in equilibrium:

$\sum F=0$

so we get that:

$N-mg+ma_{c} = 0$

and solve for the normal force:

$N=mg-ma_{c}$

In this case, we can clearly see that:

$mg>mg-ma_{c}$

therefore mg>N

This is because the centripetal acceleration is pulling us upwards, that will make the magnitude of the normal force smaller than the product of the mass times the acceleration of gravity.

c)

So let's calculate our weight and normal force:

Let's say we weight a total of 60kg, so:

$mg=(60kg)(9.81m/s^{2})=588.6N$

and let's calculate the normal force:

$N=m(g-a_{c})$

$N=(60kg)(9.81m/s^{2}-33.73x10^{-3}m/s^{2})$

N=586.58N

so now we can calculate the percentage change:

$\%_{change} = \frac{mg-N}{mg}x100\%$

so we get:

$\%_{change} = \frac{588.6N-586.58N}{588.6N} x 100\%$

$\%_{change}=0.343\%$

which is a really small change.

d) In order to find this acceleration, we need to start by calculating the radius of rotation at that point of earth. (See attached picture).

There, we can see that the radius can be found by using the cos function:

$cos \theta = \frac{AS}{h}$

In this case:

$cos \theta = \frac{r}{R_{E}}$

so we can solve for r, so we get:

$r= R_{E}cos \theta$

in this case we'll use the average radius of earch which is 6,371 km, so we get:

$r = (6371x10^{3}m)cos (44.4^{o})$

which yields:

r=4,551.91 km

and now we can calculate the acceleration at that point:

$a=\omega ^{2}r$

$a=(72.72x10^{-6} rad/s)^{2}(4,551.91x10^{3}m$

$a=24.07 mm/s^{2}$

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Answer:

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