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3 years ago

Which occurs during transcription?

Physics

2 answers:

ludmilkaskok [199]3 years ago

5 0

D. mRNA is synthesized from a strand of DNA.

Transcription, hence the root script, means to write. This is the creation of the mRNA "messenger" information.

Sati [7]3 years ago

4 0

It uses DNA as a template to make an RNA molecule. RNA then leaves the nucleus and goes to a ribosome in the cytoplasm, where translation occurs. It is the transfer of genetic instructions in DNA to messenger RNA (mRNA). During transcription, a strand of mRNA is made that is complementary to a strand of DNA.

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Jessica stretches her arms out 0.60 m from the center of her body while holding a 2.0 kg mass in each hand. She then spins aroun

Juliette [100K]

Answer:

a.) L = 2.64 kgm^2/s

b.) V = 4.4 m/s

Explanation: Jessica stretches her arms out 0.60 m from the center of her body. This will be considered as radius.

So,

Radius r = 0.6 m

Mass M = 2 kg

Velocity V = 1.1 m/s

Angular momentum L can be expressed as;

L = MVr

Substitute all the parameters into the formula

L = 2 × 1.1 × 0.6 = 1.32kgm^2s^-1

the combined angular momentum of the masses will be 2 × 1.32 = 2.64 kgm^2s-1

b. If she pulls her arms into 0.15 m,

New radius = 0.15 m

Using the same formula again

L = 2( MVr)

2.64 = 2( 2 × V × 0.15 )

1.32 = 0.3 V

V = 1.32/0.3

V = 4.4 m/s

Her new linear speed will be 4.4 m/s

4 0

3 years ago

In mechanics in what do we apply v=u+at

Trava [24]

Answer:

v=u+at is the first equation of motion. In this v=u+at equation, u is initial velocity.

3 0

2 years ago

What is the frequency of a wave having a period equal to 18 seconds

Ivanshal [37]

Explanation:

The time taken by a wave crest to travel a distance equal to the length of wave is known as wave period.

The relation between wave period and frequency is as follows.

T = \frac{1}{f}T=

where, T = time period

f = frequency

It is given that wave period is 18 seconds. Therefore, calculate the wave period as follows.

T = \frac{1}{f}T=

or, f = \frac{1}{T}f=

= \frac{1}{18 sec}

18sec

= 0.055 per second (1cycle per second = 1 Hertz)

or, f = 5.5 \times 10^{-2} hertz5.5×10 −2 hertz

<h3>Thus, we can conclude that the frequency of the wave is 5.5 \times 10^{-2} hertz5.5×10 −2 hertz .</h3>

3 0

3 years ago

A plane electromagnetic wave, with wavelength 5 m, travels in vacuum in the positive x direction with its electric vector E, of

Dafna1 [17]

Answer:

Explanation:

E₀ = 229.1 V/m

E = E₀ / √2 = 229.1 / 1.414 = 162 V/m

B = E / c ( c is velocity of em waves )

= 162 / (3 x 10⁸) = 54 x 10⁻⁸ T

rate of energy flow = ( E x B ) / μ₀

= 162 x 54 x 10⁻⁸ / 4π x 10⁻⁷

= 69.65 W per m².

5 0

3 years ago

A ball is dropped from rest at a point 12 m above the ground into a smooth, frictionless chute. The ball exits the chute 2 m abo

Nonamiya [84]

Answer:

29,7 m

Explanation:

We need to devide the problem in two parts:

A) Energy

B) MRUV

Energy:

Since no friction between pint (1) and (2), then the energy conservatets:

Energy = constant ----> Ek(cinética) + Ep(potencial) = constant

Ek1 + Ep1 = Ek2 + Ep2

Ek1 = 0 ; because V1 is zero (the ball is "dropped")

Ep1 = m*g*H1

Ep2= m*g*H2

Then:

Ek2 = m*g*(H1-H2)

By definition of cinetic energy:

m*(V2)²/2 = m*g*(H1-H2) ---> $V2 = \sqrt{(2*g*(H1-H2)}$

Replaced values: V2 = 14,0 m/s

MRUV:

The decomposition of the velocity (V2), gives a for the horizontal component:

V2x = V2*cos(α)

Then the traveled distance is:

X = V2*cos(α)*t.... but what time?

The time what takes the ball hit the ground.

Since: Y3 - Y2 = V2*t + (1/2)*(-g)*t²

In the vertical axis:

Y3 = 0 ; Y2 = H2 = 2 m

Reeplacing:

-2 = 14*t + (1/2)*(-9,81)*t²

solving the ecuation, the only positive solution is:

t = 2,99 sec ≈ 3 sec

Then, for the distance:

X = V2*cos(α)*t = (14 m/s)*(cos45°)*(3sec) ≈ 29,7 m

6 0

3 years ago

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