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nirvana33 [79]
3 years ago
5

what is the purpose of a control sample that is known to be from bighead carp in edna surveillance? a. to produce multiple bandi

ng patterns b. to amplify the dna in small samples c. to see if samples contain the target dna d. to identify each species of carp found
Physics
2 answers:
garri49 [273]3 years ago
8 0

Correct answer choice is :


C) To see if samples contain the target dna


Explanation:


In many North American rivers, populations of various kinds of non-native cyprinid fishes are present, including black carp, grass carp, bighead carp, silver carp, common carp, and goldfish. All six of these classes are found in the Mississippi River basin and following their aggression has shown complex, especially where wealth is low. The bighead carp is a species of freshwater fish, one of several Asian carps. It is one of the most intensively exploited fishes in aquaculture, with a yearly global product of over three million tonnes in 2013, mostly from China.

professor190 [17]3 years ago
4 0
I think the correct answer from the choices listed above is option C. The purpose of a control sample that is known to be from bighead carp in edna surveillance is to see if samples contain the target dna. A control sample is a set of sample that <span>does not receive treatment by the researchers.</span>
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What other uses does petroleum have besides being used for fuel and lubricants
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Notice that all the initial spring potential energy was transformed into gravitational potential energy. If you compressed the s
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<u><em>The  question doesn't provide enough data to be solved, but I'm assuming some magnitudes to help you to solve your own problem</em></u>

Answer:

<em>The maximum height is 0.10 meters</em>

Explanation:

<u>Energy Transformation</u>

It's referred to as the change of one energy from one form to another or others. If we compress a spring and then release it with an object being launched on top of it, all the spring (elastic) potential energy is transformed into kinetic and gravitational energies. When the object stops in the air, all the initial energy is now gravitational potential energy.

If a spring of constant K is compressed a distance x, its potential energy is

\displaystyle P_E=\frac{Kx^2}{2}

When the launched object (mass m) reaches its max height h, all that energy is now gravitational, which is computed as

U=mgh

We have then,

U=P_E

\displaystyle mgh=\frac{Kx^2}{2}

Solving for h

\displaystyle h=\frac{Kx^2}{2mg}

We have little data to work on the problem, so we'll assume some values to answer the question and help to solve the problem at hand

Let's say: x=0.2 m (given), K=100 N/m, m=2 kg

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\displaystyle h=\frac{(100)0.2^2}{2(2)(9.8)}

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8 0
3 years ago
A ball is dropped from rest from the top of a building of height h. At the same instant, a second ball is projected vertically u
uranmaximum [27]

Answer:

a) t = \sqrt{\frac{h}{2g}}

b) Ball 1 has a greater speed than ball 2 when they are passing.

c) The height is the same for both balls = 3h/4.

Explanation:

a) We can find the time when the two balls meet by equating the distances as follows:

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

Where:

y_{0_{1}}: is the initial height = h

v_{0_{1}}: is the initial speed of ball 1 = 0 (it is dropped from rest)

y = h - \frac{1}{2}gt^{2}     (1)

Now, for ball 2 we have:

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

Where:

y_{0_{2}}: is the initial height of ball 2 = 0

y = v_{0_{2}}t - \frac{1}{2}gt^{2}    (2)

By equating equation (1) and (2) we have:

h - \frac{1}{2}gt^{2} = v_{0_{2}}t - \frac{1}{2}gt^{2}

t=\frac{h}{v_{0_{2}}}

Where the initial velocity of the ball 2 is:

v_{f_{2}}^{2} = v_{0_{2}}^{2} - 2gh

Since v_{f_{2}}^{2} = 0 at the maximum height (h):

v_{0_{2}} = \sqrt{2gh}

Hence, the time when they pass each other is:

t = \frac{h}{\sqrt{2gh}} = \sqrt{\frac{h}{2g}}

b) When they are passing the speed of each one is:

For ball 1:

v_{f_{1}} = - gt = -g*\sqrt{\frac{h}{2g}} = - 0.71\sqrt{gh}

The minus sign is because ball 1 is going down.

For ball 2:

v_{f_{2}} = v_{0_{2}} - gt = \sqrt{2gh} - g*\sqrt{\frac{h}{2g}} = (\sqrt{1} - \frac{1}{\sqrt{2}})*\sqrt{gh} = 0.41\sqrt{gh}

Therefore, taking the magnitude of ball 1 we can see that it has a greater speed than ball 2 when they are passing.

c) The height of the ball is:

For ball 1:

y_{1} = h - \frac{1}{2}gt^{2} = h - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h

For ball 2:

y_{2} = v_{0_{2}}t - \frac{1}{2}gt^{2} = \sqrt{2gh}*\sqrt{\frac{h}{2g}} - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h

Then, when they are passing the height is the same for both = 3h/4.

I hope it helps you!                  

7 0
3 years ago
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