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2 years ago

.Golden Rice is an example of genetically modified organism. true or false

Physics

2 answers:

Rashid [163]2 years ago

8 0

Answer:

true it would be because rice is a different breed no but ye it's true I don't have an explanation

MissTica2 years ago

7 0

Answer:

True

Explanation:

Golden Rice is an example of genetically modified organism.

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A 20 KeV electron emits two bremsstrahlung photons as it is being brought to rest in two successive decelerations. The wavelengt

Degger [83]

Answer:

λ₁ = 87.5 10⁻¹² m , λ₂ = 2.175 10⁻¹⁰ m, E₂ = 5.8 10³ eV

Explanation:

In this case you can use the law of conservation of energy, all the energy of the electron is converted into energized emitted photons

Let's reduce to the SI system

E₀ = 20 10³ eV (1.6 10⁻¹⁹ J / 1eV) = 3.2 10⁻¹⁵ J

Δλ = 1.30 A = 0.13 nm = 0.13 10⁻⁹ m

Ef = E₁ + E₂

E₀ = Ef

E₀ = E₁ + E₂

The energy can be found with the Planck equation

E = h f

c = λ f

f = c / λ

E = hc / λ

They indicate that the wavelength of the second photon is

λ₂ = λ₁ +0.130 10⁻⁹

We replace

E₀ = hv / λ₁ + hc / ( λ₁ + 0.130 10⁺⁹)

E₀ / hv = 1 / λ₁ + 1 / ( λ₁ + 0.13 10⁻⁹)

3.2 10⁻¹⁵ / (6.63 10⁻³⁴ 3 10⁸) = ( λ₁ + 0.13 10⁻⁹ + λ₁) / λ₁ ( λ₁ + 0.13 10⁻⁹)

1.6 10¹⁰ ( λ₁² +0.13 10⁻⁹ λ₁) = 2 λ₁ + 0.13 10⁻⁹

λ₁² + 0.13 10⁻⁹ λ₁ = 1.25 10⁻¹⁰ λ₁ + 8.125 10⁻²¹

λ₁² + 0.005 10⁻⁹ λ₁ = 8.125 10⁻²¹

λ₁² + 5 10⁻¹² λ₁ - 8.125 10⁻²¹ = 0

Let's solve the second degree equation

λ₁ = [-5 10⁻¹² ±√((5 10⁻¹²)² + 4 8.125 10⁻²¹)] / 2

λ₁ = [-5 10⁻¹² ±√(25 10⁻²⁴ +32.5 10⁻²¹)] / 2 = [-5 10⁻¹² ±√ (32525 10⁻²⁴)] / 2

λ₁ = [-5 10⁻¹² ± 180 10⁻¹²] / 2

λ₁ = 87.5 10⁻¹² m

λ₂ = -92.5 10⁻¹² m

We take the positive wavelength

The wavelength of the photons is

λ₁ = 87.5 10⁻¹² m

λ₂ = λ₁ + 0.13 10⁻⁹

λ₂ = 87.5 10⁻¹² + 0.13 10⁻⁹

λ₂ = 0.2175 10⁻⁹ m = 2.175 10⁻¹⁰ m

The energy after the first deceleration is

E₂ = E₀ –E₁

E₂ = E₀ –hc / λ₁

E₂ = 3.2 10⁻¹⁵ - 6.63 10⁺³⁴ 3 10⁸ / 87.5 10⁻¹²

E₂ = 3.2 10⁻¹⁵ - 2.27 10⁻¹⁵

E₂ = 0.93 10⁻¹⁵ J

E₂ = 0.93 10⁻¹⁵ J (1 eV / 1.6 10⁻¹⁹ J)

E₂ = 5.8 10³ eV

7 0

3 years ago

What investigations are best for demonstrating cause and effect relationship

poizon [28]

The cause would be how the idea started or developed and the effect would be how they both effected you together

8 0

3 years ago

Read 2 more answers

The weights in atwoods machine, starting at rest, attain a velocity of 2ft/sec in one sec. Find the ratio of the masses

Orlov [11]

Refer to the figure shown below.
Let m₁ and m₂ e the two masses.
Let a = the acceleration.
Let T = tension over the frictionless pulley.

Write the equations of motion.
m₂g - T = m₂a (1)
T - m₁g = m₁a (2)

Add equations (1) and (2).
m₂g - T + T - m₁g = (m₁ + m₂)a
(m₂ - m₁)g = (m₁ + m₂)a

Divide through by m₁.
(m₂/m₁ - 1)g = (1 + m₂/m₁)a

Define r = m₂/m₁ as the ratio of the two masses. Then
(r - 1)g = (1 +r)a
r(g-a) = a + g
r = (g - a)/(g + a)

With = 2 ft/s from rest, the acceleration is
a = 2/32.2 = 0.062 ft/s²
Therefore
r = (32.2 - 0.062)/(32.2 + 0.062) = 0.9962

Answer:
The ratio of masses is 0.9962 (heavier mass divided by the lighter mass).

8 0

3 years ago

What is the magnitude of the magnetic field at a point midway between them if the top one carries a current of 19.5 A and the bo

Phantasy [73]

Answer:

The magnetic field will be $\large{\dfrac{1.4 \times 10^{-4}}{d}} T$ , '2d' being the distance the wires.

Explanation:

From Biot-Savart's law, the magnetic field ( $\large{\overrightarrow{B}}$ ) at a distance ' $r$ ' due to a current carrying conductor carrying current ' $I$ ' is given by

$\large{\overrightarrow{B} = \dfrac{\mu_{0}I}{4 \pi}} \int \dfrac{\overrightarrow{dl} \times \hat{r}}{r^{2}}}$

where ' $\overrightarrow{dl}$ ' is an elemental length along the direction of the current flow through the conductor.

Using this law, the magnetic field due to straight current carrying conductor having current ' $I$ ', at a distance ' $d$ ' is given by

$\large{\overrightarrow{B}} = \dfrac{\mu_{0}I}{2 \pi d}$

According to the figure if ' $I_{t}$ ' be the current carried by the top wire, ' $I_{b}$ ' be the current carried by the bottom wire and ' $2d$ ' be the distance between them, then the direction of the magnetic field at 'P', which is midway between them, will be perpendicular towards the plane of the screen, shown by the $\bigotimes$ symbol and that due to the bottom wire at 'P' will be perpendicular away from the plane of the screen, shown by $\bigodot$ symbol.

Given $\large{I_{t} = 19.5 A}$ and $\large{I_{B} = 12.5 A}$

Therefore, the magnetic field ( $\large{B_{t}}$ ) at 'P' due to the top wire

$B_{t} = \dfrac{\mu_{0}I_{t}}{2 \pi d}$

and the magnetic field ( $\large{B_{b}}$ ) at 'P' due to the bottom wire

$B_{b} = \dfrac{\mu_{0}I_{b}}{2 \pi d}$

Therefore taking the value of $\mu_{0} = 4\pi \times 10^{-7}$ the net magnetic field ( $\large{B_{M}}$ ) at the midway between the wires will be

$\large{B_{M} = \dfrac{4 \pi \times 10^{-7}}{2 \pi d} (I_{t} - I_{b}) = \dfrac{2 \times 10^{-7}}{d} = \dfrac{41.4 \times 10 ^{-4}}{d}} T$

5 0

3 years ago

A jet plane is cruising at 310 m/s when suddenly the pilot turns the engines up to full throttle. after traveling 4.0 km , the j

Inessa [10]

For this problem, we would be using the formula: Vf^2 = Vi^2 + 2ad
where:
Vf = 400m/s
Vi = 300m/s
a = ?
d = 4.0km
= 4000m

400^2 = 300^2 + 2a4000
a = [ 160000 - 90000 ] / 8000
a = 8.75m/s^2
rounding it off to 2 significant figures, will give us 8.8 m/s^2.

4 0

3 years ago