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Oduvanchick [21]
3 years ago
14

Arden is making a pros and cons list about genetic engineering for a debate at school. Which part of her list is incorrect?

Physics
2 answers:
romanna [79]3 years ago
4 0

Answer:

Con 2

Explanation:

<em>The part of Arden's list that is incorrect would be con 2, </em><em>that edited gene is more likely to mutate and cause disease.</em>

<u>Genetic engineering is the science of modifying the gene of organisms through the process of biotechnology in order to serve specific needs.</u>

Genetic engineering has a wide variety of applications, including the production of vaccines and drugs in medicine and improved breeds of organisms in agriculture.

However, genetically modified organisms sometimes present ecological and or genetic problems. They can alter the dynamics of population in the environments where they are introduced, affect gene flow, and affect biogeochemistry of the environment.

Edited genes in genetically modified organisms have the same chances of mutation just like every other gene. That a gene is edited does not make it more likely to mutate and cause disease.

Tju [1.3M]3 years ago
4 0

Answer:

Con 2

Explanation:

I took the test and got it correct

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(Superposition, quadratic formuła) The Earth (mass M) is at origin, and the Moon (mass m ) is located at a distance d. (a) Using
jekas [21]

Answer:

a)    Fa = G m2 [M / r² - m / (d-r)²]  

b) r2 = 31 10⁶ m

Explanation:

The equation of the law of universal gravitation is

            F = G m1 m1/ r²

The value of the gravitation constant is 6.67 10-11 N m²/kg². This force is always attractive.

Let's calculate the value of that force on the spacecraft, add the strength

Earth's force to the ship

            F1 = G M m2 / r²

The moon force ship

            F2 = G m m2 / (d-r)²

Total force is

            Fa = F1 — F2

            Fa = G M M2 / r² - G m m2 / (d-r)²

            Fa = G m2 [M / r² - m / (d-r)²]

This is the force on the spaceship

b)  Let's look for the point where the force is zero, for this we can see that the value of the bracket must be zero

           Fa = 0

           [M / r² - m / (d-r)²] = 0

            M / r² = m / (d-r)²

           (d-r)² = m/M   r²

           d² -2rd + r² - m/M   r² = 0

           r² [m/M - 1] + r 2d - d² = 0

This is a second degree equation for r, we solve the to find the results.

           r = {-2d ±√[4d² - 4 [m/M -1] (-d²)]} / (2 [m/M-1])

           r = {-2d ± √ [4d² (1 + (m/M-1)]} / 2(m/M-1)

           r = {-2d ± 2d √(m/M)}  / (2(m/M-1))

           r = 2d {-1 ± √(m/M)} / 2(m/M-1)

           r = d [-1 ± √(m/M)] /  (m/M-1)

To find the explicit value we substitute the values ​​that we can find in tables

          m = 7.36 1022 kg

          M = 5.98 1024 kg

          d = 380000 km (1000m / 1 km) = 380 10⁶ m

          r = 380 10⁶ [-1 ±√(7.36 10²² / 5.98 10²⁴)] /(7.36 10²² / 5.98 / 10²⁴ -1)

          r = 380 10⁶ [-1 - √ (1.23 10²)] / (123-1)

          r = 380 106 [-1 ± 11] / 122

 

          r1 = 380 10⁶ 10/122 = 380 10⁶ 0.08197

          r1 =  31 10⁶ m

          r2 = 380 106 [-12/122] = 380 10 6 0.09836

          r2 = -37 10⁶ m

The correct distance is the positive r2 = 31 10⁶ m

c) let's use Newton's second law, to find the acceleration in the spacecraft

 

          F = m a

          a = Fa / m2 = G m2 [M / r² - m / (d-r)²] / m2

          a = G [M/r² - m/(d-r)²]

Since we have acceleration, we can use the definition of kinematics

           a = dv / dt = dv / dr dr / dt = dv / dr v

           v dv = a dr

           v dV = G [M /r² - m /(d-r)²] dr

We integrate

            ½ (V² - Vo²) = G [M (-1 /r) -m (1 / (d-r)

We evaluate between the initial point where we can assume that the initial velocity is zero for the Xo position and the final point with velocity v at the points

            V² = 2G [M (1 / Xo - 1 /X) - m (1 / (d-X) - 1 /(d-xo)]

            V² = [-M /X -m /(d-X)] 2G + constant

We now use the definition of speed

            v = dx / dt

            dx = V dt

We substitute, perform the integral and simplify, if we can make the constant zero

             dx = √([-M / X -m / (d-X)] 2G) dt

             dx / √([-M / X -m / (d-X)] 2G = dt

8 0
3 years ago
A stationary block resting on the ground is pulled up with a tension force of 100N, but does not leave the ground.
tekilochka [14]

Answer:

The normal force the ground exerts on the block, F = -300 N

Explanation:

Given data,

The block pulled up with a tension force, T = 100 N

The weight of the block, W = 300 N

The weight of the block is due to the force of attraction of gravitation.

The surface exerts a force that is equal and opposite to the force acting on the block due to gravitation.

The weight of the block,

                                  W = mg

                                          300 N

The normal force the ground exerts on the block,

                                  F = - mg

                                    = - 300 N

Hence, the normal force the ground exerts on the block, F = -300 N

6 0
3 years ago
Which object has the most thermal energy?
den301095 [7]

Answer:

D is the answer

Explanation:

D is the most highest one so

the answer is D

3 0
2 years ago
Read 2 more answers
Is hydrogen good conductor or poor conductor
m_a_m_a [10]

Answer:

Good one

Explanation:

Hydrogen has the highest thermal conductivity of any gas.

7 0
2 years ago
Read 2 more answers
The work function for magnesium is 3.70 ev. what is its cutoff frequency?
alexandr402 [8]

The cutoff frequency for magnesium is 8.93 x 10¹⁴ Hz.

<h3>What is cutoff frequency?</h3>

The work function is related to the frequency as

W0 = h x fo

where, fo = cutoff frequency and h is the Planck's constant

Given is the work function for magnesium is  3.70 eV.

fo = 3.7 x 1.6 x 10⁻¹⁹ / 6.626 x 10⁻³⁴

fo = 8.93 x 10¹⁴ Hz.

Thus, the cut off frequency is 8.93 x 10¹⁴ Hz.

Learn more about cutoff frequency.

brainly.com/question/14378802

#SPJ1

7 0
2 years ago
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