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kolbaska11 [484]
3 years ago
6

Thoughts about Genetically Modified Crops

Physics
2 answers:
Alex787 [66]3 years ago
4 0
I think that GMOs, also known as genetically modified crops or organisms, can be used in good and bad ways. They can provide crops to survive longer, and produce a bigger portion. However, on the other side of that, it is claimed that it creates a large amount of greenhouse gas emissions, which can turn into climate change.
vazorg [7]3 years ago
3 0
There r many pros and cons, also using a special type of bacteria is really helpful in genetic modification
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they can blend in with their suroundings

Explanation:

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Which atmospheric gas is used by plants and given off by animals? A. Carbon dioxide. B. Nitrogen C. Oxygen D. Argon​
Alexxx [7]

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A

Explanation:

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3 years ago
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A store owner has 11 baskets of mangoes for sale at k8.26.How much will he receive in total when he sells them​
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11*2.86 which is 31.46
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4 years ago
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A beam of light with a frequency range from 3.01 × 10 14 Hz to 6.10 × 10 14 Hz is incident on a metal surface. If the work funct
Sever21 [200]

Answer:

The maximum kinetic energy of the photoelectrons ejected from the surface is 5.22×10^-20 J.

Explanation:

let h = 6.626×10^-34 J×s be the planck constant.

let f be the frequency of light.

let Ф = 2.20×1.60×10^-19 = 3.52×10^-19 J be the work function.

then, the relationship  between the kinetic energy of photoelectrons K, the energy provided by the light E and the work function of the material is given by:

K = h×f - Ф

  = (6.626×10^-34)×(6.10×10^14) - 3.52×10^-19

  = 5.22×10^-20 J

Therefore, the maximum kinetic energy of the photoelectrons ejected from the surface is 5.22×10^-20 J.

5 0
3 years ago
horizontal circular platform rotates counterclockwise about its axis at the rate of 0.919 rad/s. You, with a mass of 73.5 kg, wa
VARVARA [1.3K]

Answer:

The total angular momentum is 292.59 kg.m/s

Explanation:

Given that :

Rotation of the horizontal circular platform \omega = 0.919 rad/s

mass of the platform (m) = 90.7 kg

radius (R) = 1.91 m

mass of the poodle m_p = 20.5 kg

Your mass  m' = 73.5 kg

speed v = 1.05 m/s with respect to the platform

V_p = \frac{1.05}{2} \ m /s \\ \\ = 0.525  \ m /s  \\  \\r = \frac{R}{2}

r = 0.955

Mass of the mutt m_m = 18.5 kg

r' = \frac{3}{4} \ R

Your angular momentum is calculated as:

Your angular velocity relative to the platform is \omega' = \frac{v}{R}  = \frac{1.05}{1.91 } = 0.5497 \ rad/s

Actual \ \omega_y = \omega - \omega ' = (0.919 - 0.5497) \ rad/s = 0.3693 \ rad/s

I_y = m'R^2 = 73.5 *1.91^2= 268.14 \ kgm^2

L_Y = I_y*\omega_y= 268.14*0.3683= 98.76 \ kg.m/s

For poodle :

Relative \ \ \omega' = \frac{V_p}{R/2} = 0.550 \  rad/s

Actual \omega_p = \omega - \omega' =  0.919 -0.550 = 0.369 \ rad/s

I_p = m_p(\frac{R}{2} )^2  = 20.5(\frac{1.91}{2} )^2 = 18.70 \ kgm^2

L_p = I_p *\omega_p = 18.70*0.369 = 6.9003 \  kgm/s

I_M = m_m(\frac{3}{y4} R)^2 = 18.5 (\frac{3}{4} 1.91)^2 = 37.96 \  kgm^2

L_M = I_M \omega = 37.96 * 0.919= 34.89 \ kg.m/s

Disk I = \frac{mr^2}{2} =  \frac{90.7*1.91^2}{2}= 165.44 \ kgm^2

L_D = I \omega = 165.44*0.919 =152.04 \ kg.m/s

Total angular momentum of system is:

L = L_D +L_Y+L_P+L_M

= (152.04 + 98.76 + 6.9003 + 34.89) kg.m/s

= 292.59 kg.m/s

6 0
4 years ago
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