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3 years ago

Who performed classic experiments that supported the semiconservative model of dna replication?

Physics

1 answer:

Salsk061 [2.6K]3 years ago

6 0

Meselson and Stahl

<u>Explanation:</u>

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The classic experiment that supported the semiconservative model of dna replication was performed by Matthew Meselson and Franklin W. Stahl. In this model, the two strands of DNA unwind from each other, and each acts as a template for synthesis of a new, complementary strand. This results in two DNA molecules with one original strand and one new strand. They used E. coli bacteria as a model system.

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A 50.-kilogram rock rolls off the edge of a cliff. if it is traveling at a speed of 24.2 m/s when it hits the ground, what is th

ElenaW [278]

The correct answer to the question is : 29.88 m.

EXPLANATION :

As per the question, the mass of the rock m = 50 Kg.

The rock is rolling off the edges of the cliff.

The final velocity of the rock when it hits the ground v = 24 .2 m/s.

Let the height of the cliff is h.

The potential energy gained by the rock at the top of the cliff = mgh.

Here, g is known as acceleration due to gravity, and g = $9.8\ m/s^2$

When the rock rolls off the edge of the cliff, the potential energy is converted into kinetic energy.

When the rock hits the ground, whole of its potential energy is converted into its kinetic energy.

The kinetic energy of the rock when it touches the ground is given as -

Kinetic energy K.E = $\frac{1}{2}mv^2$ .

From above we know that -

Kinetic energy at the bottom of the cliff = potential energy at a height h

$\frac{1}{2}mv^2=\ mgh$

⇒ $v^2=\ 2gh$

⇒ $h=\ \frac{v^2}{2g}$

⇒ $h=\ \frac{(24.2)^2}{2\times 9.8}$

⇒ $h=\ 29.88\ m$

Hence, the height of the cliff is 29.88 m

5 0

3 years ago

What are two different examples of positive exeleration

sdas [7]

Answer:Accelerating your car to get up to speed on a freeway

An airplane accelerating to take off

Accelerating out of the starting blocks at a track meet

Explanation:

5 0

3 years ago

PLEASE HELP FOR PHYSICS!

Stolb23 [73]

Hi there!

Recall Newton's Law of Universal Gravitation:

$\large\boxed{F_g = G\frac{m_1m_2}{r^2}}$

Where:

Fg = Force of gravity (N)

G = Gravitational Constant

m1, m2 = masses of objects (kg)

r = distance between objects (m)

Plug in the given values stated in the problem:

$F_g = (6.673*10^{-11})\frac{50 * 50}{25^2} = \boxed{2.669 * 10^{-10} N}$

8 0

2 years ago

eimsori [14]

Explanation:

$d = \frac{1}{2} a {t}^{2} + vt \\ d = 75 \\ t = 5 \\ v = 0 \\ \\ 75 = \frac{1}{2} \times 25a + 0 \\ a = 6 \frac{m}{ {s}^{2} } \\$

7 0

3 years ago

CHEGG You stretch a spring with spring constant k = 1.2x104 N/m to extend 6.0 cm away from its equilibrium position. How much do

lubasha [3.4K]

Answer:

The elastic potential energy of the spring change during this process is 21.6 J.

Explanation:

Given that,

Spring constant of the spring, $k=1.2\times 10^4\ N/m$

It extends 6 cm away from its equilibrium position.

We need to find the elastic potential energy of the spring change during this process. The elastic potential energy of the spring is given by the formula as follows :

$E=\dfrac{1}{2}kx^2\\\\E=\dfrac{1}{2}\times 1.2\times 10^4\times (0.06)^2\\\\E=21.6\ J$

So, the elastic potential energy of the spring change during this process is 21.6 J.

4 0

3 years ago