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olya-2409 [2.1K]

4 years ago

9

Which process is part of modern laboratory genetics? A. Natural Selection B. Evolutionary Manipulation C. Dna Engineering D. Sel

ective Breeding

2 answers:

Anit [1.1K]4 years ago

8 0

DEFINITELY C

c is the most accurate answer the others dont make sence

slamgirl [31]4 years ago

5 0

Answer:

C. Dna Engineering

Explanation:

Genetic engineering are the techniques of manipulation and recombination of genes, through a set of scientific knowledge (genetics, molecular biology, biochemistry, among others), which reformulate, reconstitute, reproduce and even create living beings. Genetic manipulation techniques were developed in the 1970s and their applications have reached several areas, such as medicine, agriculture and livestock. Genetic engineering is now part of modern laboratory genetics.

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The skateboarder weighs 75 kilogram. Calculate the potential energy of the skateboarder sliding on the track when his height abo

ArbitrLikvidat [17]

Answer:F(of gravity) = MA

F(normal force) = MA * cos(angle)

F = 72 * 9.81 * cos28

Don't have a calculator, so can't really do all the math right there. So just plug that in

Explanation:

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7 0

3 years ago

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Which of the follow are types of changes that can happen in the rock cycle

ollegr [7]

you must have a rock first of all then the cycle continues

3 0

3 years ago

The volume of an object as a function of time is calculated by v = At3+B÷t where t is time measured in seconds and v is in cubic

madam [21]

Answer:

A = m³/s³ = [L]³/[T]³ = [L³T⁻³]

B = m³s = [L³T]

Explanation:

We have the equation:

V = At³ + B/t

where, the dimensions of each variable are as follows:

V = m³ = [L]³

t = s = [T]

substituting these in equation, we get:

m³ = A(s)³ + B/s

for the homogeneity of the equation:

A(s)³ = m³

<u>A = m³/s³ = [L]³/[T]³ = [L³T⁻³]</u>

Also,

B/s = m³

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3 0

4 years ago

As sea floor spreading occurs, the oceanic plate _______.

Ira Lisetskai [31]

Becomes older

Explanation:

As sea floor spreading occurs at divergent margins, the oceanic plate becomes older. Younger plate margin are the closest to the margin whereas the older plates bushes backward away from the spreading centers.

The idea that the sea floor spreads was postulated by Harry Hess shortly after the second world war around the 1960's.
At divergent margins new crust materials from the mantle are brought to the surface.
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Learn more:

Sea floor spreading brainly.com/question/9912731

#learnwithBrainly

8 0

4 years ago

Initially, a 2.00-kg mass is whirling at the end of a string (in a circular path of radius 0.750 m) on a horizontal frictionless

drek231 [11]

Answer:

$v_f = 15 \frac{m}{s}$

Explanation:

We can solve this problem using conservation of angular momentum.

The angular momentum $\vec{L}$ is

$\vec{L} = \vec{r} \times \vec{p}$

where $\vec{r}$ is the position and $\vec{p}$ the linear momentum.

We also know that the torque is

$\vec{\tau} = \frac{d\vec{L}}{dt} = \frac{d}{dt} ( \vec{r} \times \vec{p} )$

$\vec{\tau} = \frac{d}{dt} \vec{r} \times \vec{p} + \vec{r} \times \frac{d}{dt} \vec{p}$

$\vec{\tau} = \vec{v} \times \vec{p} + \vec{r} \times \vec{F}$

but, as the linear momentum is $\vec{p} = m \vec{v}$ this means that is parallel to the velocity, and the first term must equal zero

$\vec{v} \times \vec{p}=0$

so

$\vec{\tau} = \vec{r} \times \vec{F}$

But, as the only horizontal force is the tension of the string, the force must be parallel to the vector position measured from the vertical rod, so

$\vec{\tau}_{rod} = 0$

this means, for the angular momentum measure from the rod:

$\frac{d\vec{L}_{rod}}{dt} = 0$

that means :

$\vec{L}_{rod} = constant$

So, the magnitude of initial angular momentum is :

$| \vec{L}_{rod_i} | = |\vec{r}_i||\vec{p}_i| cos(\theta)$

but the angle is 90°, so:

$| \vec{L}_{rod_i} | = |\vec{r}_i||\vec{p}_i|$

$| \vec{L}_{rod_i} | = r_i * m * v_i$

We know that the distance to the rod is 0.750 m, the mass 2.00 kg and the speed 5 m/s, so:

$| \vec{L}_{rod_i} | = 0.750 \ m \ 2.00 \ kg \ 5 \ \frac{m}{s}$

$| \vec{L}_{rod_i} | = 7.5 \frac{kg m^2}{s}$

For our final angular momentum we have:

$| \vec{L}_{rod_f} | = r_f * m * v_f$

and the radius is 0.250 m and the mass is 2.00 kg

$| \vec{L}_{rod_f} | = 0.250 m * 2.00 kg * v_f$

but, as the angular momentum is constant, this must be equal to the initial angular momentum

$7.5 \frac{kg m^2}{s} = 0.250 m * 2.00 kg * v_f$

$v_f = \frac{7.5 \frac{kg m^2}{s}}{ 0.250 m * 2.00 kg}$

$v_f = 15 \frac{m}{s}$

8 0

3 years ago