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3 years ago

When does DNA replication occur?

Physics

2 answers:

andrew-mc [135]3 years ago

7 0

Answer: During Mitosis

Explanation: During Mitosis, DNA is replicated during the S phase (Synthesis phase) of Interphase. Interphase is basically the daily life cycle of the cell.

Digiron [165]3 years ago

5 0

Answer:

NA is replicated during the S phase (Synthesis phase) of Interphase.

Hope it helps!!!!!!!!!!

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Define average velocity <br>

maxonik [38]

Hello! mark me brainliest please

The average speed of an object is defined as the distance traveled divided by the time elapsed. Velocity is a vector quantity, and average velocity can be defined as the displacement divided by the time.

7 0

3 years ago

Read 2 more answers

Finally, you are ready to answer the main question. Cheetahs, the fastest of the great cats, can reach 50.0 miles/hour in 2.22 s

muminat

Answer:

Acceleration = 10.06 m/s²

Explanation:

1 mile = 1.6093km

1609.3m = 1 mile

1 m = $\frac{1}{1609}$ mile

50.0 miles/hour = $\frac{50 * 1609.3}{60 * 60}$ m/s

= 22.35m/s

from equation

S = Ut + 1/2 at²

v = U + at

22.35 = 0 + a * 2.22

a = 22.35 ÷ 2.22

= 10.06 m/s²

4 0

3 years ago

A hawk is flying horizontally at 18.0 m/s in a straight line, 230 m above the ground. A mouse it has been carrying struggles fre

Lisa [10]

Answer:

a) vd = 47.88 m/s

b) θ = 80.9°

c) t = 6.8 s

Explanation:

In the situation of the problem, you can assume that the trajectory of the hawk and the trajectory of the mouse form a rectangle triangle.

One side of the triangle is the horizontal trajectory of the hawk after 2.00s of flight, the other side of the triangle is the distance traveled by the mouse when it is falling down. And the hypotenuse is the trajectory of the hawk when it is trying to recover the mouse.

(a) In order to calculate the diving speed of the hawk, you first calculate the hypotenuse of the triangle.

One side of the triangle is c1 = (18.0m/s)(2.0s) = 36m

The other side of the triangle is c2 = 230m - 3m = 227 m

Then, the hypotenuse is:

$h=\sqrt{(36m)^2+(227m)^2}=229.83m$ (1)

Next, it is necessary to calculate the falling down time of the mouse, this can be done by using the following formula:

$y=y_o+v_ot+\frac{1}{2}gt^2$ (2)

yo: initial height = 230m

vo: initial vertical speed of the mouse = 0m/s

g: gravitational acceleration = -9.8m/s^2

y: final height of the mouse = 3 m

You replace the values of the parameters in (2) and solve for t:

$3=230-4.9t^2\\\\t=\sqrt{\frac{227}{4.9}}=6.8s$

The hawk traveled during 2.00 second in the horizontal trajectory, hence, the hawk needed 6.8s - 2.0s = 4.8 s to travel the distance equivalent to the hypotenuse to catch the mouse.

You use the value of h and 4.8s to find the diving speed of the hawk:

$v_d=\frac{229.83m}{4.8s}=47.88\frac{m}{s}$