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babymother [125]
3 years ago
9

What is the result of a mutation during replication

Physics
2 answers:
Sonbull [250]3 years ago
8 0

The correct answer is A. The strands are different. Hope this helps! :)

ss7ja [257]3 years ago
6 0

Answer:

Its A

Explanation: I took the quiz ewe

You might be interested in
Example 4.6 provides a nice example of the overlap between kinematics and dynamics. It is known that the plane accelerates from
kodGreya [7K]

Answer:

ax = 2.60m/s^{2}, t = 26.92s

Explanation:

The acceleration of the plane can be determined by means of the kinematic equation that correspond to a Uniformly Accelerated Rectilinear Motion.

(vx)f^{2} = (vx)i^{2} + 2ax \Lambda x (1)

Where (vx)f^{2} is the final velocity, (vx)i^{2} is the initial velocity, ax is the acceleration and  \Lambda x is the distance traveled.

Equation (1) can be rewritten in terms of ax:

(vx)f^{2} - (vx)i^{2} = 2ax \Lambda x

2ax \Lambda x = (vx)f^{2} - (vx)i^{2}

ax = \frac{(vx)f^{2} - (vx)i^{2}}{2 \Lambda x}  (2)

Since the plane starts from rest, its initial velocity will be zero ((vx) = 0):

Replacing the values given in equation 2, it is gotten:

ax = \frac{(70m/s)^{2} - (0m/s)^{2}}{2(940m)}

ax = \frac{4900m/s}{2(940m)}

ax = \frac{4900m/s}{1880m}

ax = 2.60m/s^{2}

So, The acceleration of the plane is 2.60m/s^{2}    

Now that the acceleration is known, the next equation can be used to find out the time:

(vx)f = (vx)i + axt (3)

Rewritten equation (3) in terms of t:

t = \frac{(vx)f - (vx)i}{ax}

t = \frac{70m/s - 0m/s}{2.60m/s^{2}}

t = 26.92s

<u>Hence, the plane takes 26.92 seconds to reach its take-off speed.</u>

5 0
3 years ago
Which statement is true about the formation of bonds?
Ivenika [448]
There are no true statements on the list you attached.
7 0
3 years ago
Is the facceleration due to gravity the same on Earth and Earth's moon?
Alchen [17]

Answer:

Earth's average surface gravity is about 9.8 meters per second per second. ... The Moon's surface gravity is weaker because it is far less massive than Earth. A body's surface gravity is proportional to its mass but inversely proportional to the square of its radius.

Explanation:

3 0
3 years ago
If it takes 0.5 seconds to travel 5 Millimeters? What's the mph?
Angelina_Jolie [31]
I think it would me 5/0.5 = 10mph
3 0
3 years ago
A small propeller airplane can comfortably achieve a high enough speed to take off on a runway that is 1/4 mile long. A large, f
QveST [7]

Answer:

1 mile

Explanation:

We can use the following equation of motion to solve for this problem:

v^2 - v_0^2 = 2a\Delta s

where v m/s is the final take-off velocity of the airplane, v_0 = 0 initial velocity of the can when it starts from rest, a is the acceleration of the airplanes, which are the same, and \Delta s is the distance traveled before takeoff, which is minimum runway length:

v^2 - 0^2 = 2a\Delta s

\Delta s = \frac{v^2}{2a}

From here we can calculate the distance ratio

\frac{\Delta s_1}{\Delta s_2} = \frac{v_1^2/2a_1}{v_2^2/2a_2}

\frac{\Delta s_1}{\Delta s_2} = \left(\frac{v_1}{v_2}\right)^2\frac{a_2}{a_1}

Since the 2nd airplane has the same acceleration but twice the velocity

\frac{\Delta s_1}{\Delta s_2} = 0.5^2* 1

\Delta s_2 = 4 \Delta s_1 = 4*(1/4) = 1 mile

So the minimum runway length is 1 mile

6 0
3 years ago
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