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3 years ago

Pea plants have 2 advantages as genetic specimens:

Physics

1 answer:

jeka943 years ago

3 0

Answer:

it can be cross pollinated as well as self pollinated

it has short life style

pea plant has many contrasting character in pair example tall, short.

large number of offspring are produced from hybrid plants

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Tutorial Exercise An unstable atomic nucleus of mass 1.83 10-26 kg initially at rest disintegrates into three particles. One of

kogti [31]

Answer:

A) v3 = -[6.29 × 10^(6)]j^ - [7.06 × 10^(6)]i^

B) K_total = 373.08 × 10^(-15) J

Explanation:

We are given;

Mass of unstable atomic nucleus; M = 1.83 × 10^(-26) kg

Mass of first particle; m1 = 5.03 × 10^(-27) kg

Speed of first particle in y-direction; v1 = (6 × 10^(6) m/s) j^

Mass of second particle; m2 = 8.47 × 10^(-27) kg

Speed of second particle in x - direction; v2 = (4 × 10^(6) m/s) i^

Now, we don't have the mass of the third particle but since we are told the unstable atomic nucleus disintegrates into 3 particles, thus;

M = m1 + m2 + m3

1.83 × 10^(-26) = (5.03 × 10^(-27)) + (8.47 × 10^(-27)) + m3

m3 = (1.83 × 10^(-26)) - (13.5 × 10^(-27))

m3 = 4.8 × 10^(-27) kg

A) Applying law of conservation of momentum, we have;

MV = (m1 × v1) + (m2 × v2) + (m3 × v3)

Now, the unstable atomic nucleus was at rest before disintegration, thus V = 0 m/s.

Thus, we now have;

0 = (m1 × v1) + (m2 × v2) + (m3 × v3)

We want to find the velocity of the third particle v3. Let's make it the subject of the formula;

v3 = [(m1 × v1) + (m2 × v2)]/(-m3)

Plugging in the relevant values, we have;

v3 = [(5.03 × 10^(-27) × 6 × 10^(6))j^ + (8.47 × 10^(-27) × 4 × 10^(6))i^]/(-4.8 × 10^(-27))

v3 = [(30.18 × 10^(-21))j^ + (33.88 × 10^(-21))i^]/(-4.8 × 10^(-27))

v3 = -[6.29 × 10^(6)]j^ - [7.06 × 10^(6)]i^

B) Formula for kinetic energy is;

K = ½mv²

Now,total kinetic energy is;

K_total = K1 + K2 + K3

K1 = ½ × 5.03 × 10^(-27) × (6 × 10^(6))²

K1 = 90.54 × 10^(-15) J

K2 = ½ × 8.47 × 10^(-27) × (4 × 10^(6))²

K2 = 67.76 × 10^(-15)

To find K3, let's first find the magnitude of v3 because it's still in vector form.

Thus;

v3 = √[(-6.29 × 10^(6))² + (-7.06 × 10^(6))²]

v3 = 9.46 × 10^(6) m/s

K3 = ½ × 4.8 × 10^(-27) × (9.46 × 10^(6))²

K3 = 214.78 × 10^(-15) J

K_total = (90.54 × 10^(-15)) + (67.76 × 10^(-15)) + (214.78 × 10^(-15))

K_total = 373.08 × 10^(-15) J

7 0

3 years ago

How much heat is needed to change 1.25 kg of steak at 100°C to water at 100°C?

cricket20 [7]

The heat required to change 1.25 kg of steak is 2825 kJ /kg.

<u>Explanation</u>:

Given, mass m = 1.25 kg, Temperature t = 100 degree celsius

To calculate the heat required,

Q = m $\times$ L

where m represents the mass in kg,

L represents the heat of vaporization.

When a material in the liquid state is given energy, it changes its phase from liquid to vapor and the energy absorbed in this process is called heat of the vaporization. The heat of vaporization of the water is about 2260 kJ/kg.

Q = 1.25 $\times$ 2260

Q = 2825 kJ /kg.

7 0

3 years ago

Which statement about activation energy is true? A. It is not affected by catalysts. B. It has no effect on the rate of the reac

HACTEHA [7]

C it is the energy required to break existing chemical bonds, it is the amount of energy that a reaction requires in order for the reactants to successfully collide and react

6 0

3 years ago

Read 2 more answers

A 2 eV electron encounters a barrier 5.0 eV high and width a. What is the probability b) 0.5 that it will tunnel through the bar

denis23 [38]

Answer:

The tunnel probability for 0.5 nm and 1.00 nm are $5.45\times10^{-4}$ and $7.74\times10^{-8}$ respectively.

Explanation:

Given that,

Energy E = 2 eV

Barrier V₀= 5.0 eV

Width = 1.00 nm

We need to calculate the value of $\beta$

Using formula of $\beta$

$\beta=\sqrt{\dfrac{2m}{\dfrac{h}{2\pi}}(v_{0}-E)}$

Put the value into the formula

$\beta = \sqrt{\dfrac{2\times9.1\times10^{-31}}{(1.055\times10^{-34})^2}(5.0-2)\times1.6\times10^{-19}}$

$\beta=8.86\times10^{9}$

(a). We need to calculate the tunnel probability for width 0.5 nm

Using formula of tunnel barrier

$T=\dfrac{16E(V_{0}-E)}{V_{0}^2}e^{-2\beta a}$

Put the value into the formula

$T=\dfrac{16\times 2(5.0-2.0)}{5.0^2}e^{-2\times8.86\times10^{9}\times0.5\times10^{-9}}$

$T=5.45\times10^{-4}$

(b). We need to calculate the tunnel probability for width 1.00 nm

$T=\dfrac{16\times 2(5.0-2.0)}{5.0^2}e^{-2\times8.86\times10^{9}\times1.00\times10^{-9}}$

$T=7.74\times10^{-8}$

Hence, The tunnel probability for 0.5 nm and 1.00 nm are $5.45\times10^{-4}$ and $7.74\times10^{-8}$ respectively.

6 0

3 years ago

In an engine, a piston oscillates with simple harmonic motion so that its position varies according to the following expression,

eduard

(a) 4.06 cm

In a simple harmonic motion, the displacement is written as

$x(t) = A cos (\omega t + \phi)$ (1)

where

A is the amplitude

$\omega$ is the angular frequency

$\phi$ is the phase

t is the time

The displacement of the piston in the problem is given by

$x(t) = (5.00 cm) cos (5t+\frac{\pi}{5})$ (2)

By putting t=0 in the formula, we find the position of the piston at t=0:

$x(0) = (5.00 cm) cos (0+\frac{\pi}{5})=4.06 cm$

(b) -14.69 cm/s

In a simple harmonic motion, the velocity is equal to the derivative of the displacement. Therefore:

$v(t) = x'(t) = -\omega A sin (\omega t + \phi)$ (3)

Differentiating eq.(2), we find

$v(t) = x'(t) = -(5 rad/s)(5.00 cm) sin (5t+\frac{\pi}{5})=-(25.0 cm/s) sin (5t+\frac{\pi}{5})$

And substituting t=0, we find the velocity at time t=0:

$v(0)=-(25.00 cm/s) sin (0+\frac{\pi}{5})=-14.69 cm/s$

(c) -101.13 cm/s^2

In a simple harmonic motion, the acceleration is equal to the derivative of the velocity. Therefore:

$a(t) = v'(t) = -\omega^2 A cos (\omega t + \phi)$

Differentiating eq.(3), we find

$a(t) = v'(t) = -(5 rad/s)(25.00 cm/s) cos (5t+\frac{\pi}{5})=-(125.0 cm/s^2) cos (5t+\frac{\pi}{5})$

And substituting t=0, we find the acceleration at time t=0:

$a(0)=-(125.00 cm/s) cos (0+\frac{\pi}{5})=-101.13 cm/s^2$

(d) 5.00 cm, 1.26 s

By comparing eq.(1) and (2), we notice immediately that the amplitude is

A = 5.00 cm

For the period, we have to start from the relationship between angular frequency and period T:

$\omega=\frac{2\pi}{T}$

Using $\omega = 5.0 rad/s$ and solving for T, we find

$T=\frac{2\pi}{5 rad/s}=1.26 s$

4 0

3 years ago

Pea plants have 2 advantages as genetic specimens:​

Pea plants have 2 advantages as genetic specimens: