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lakkis [162]
4 years ago
7

What do viruses need in order to replicate?

Physics
2 answers:
STALIN [3.7K]4 years ago
7 0
<h2>a host cell</h2>

  • replication of viruses is the formation of biological viruses during the infection process in the target host cells

  • viruses must first get into the cell before viral replication can occur

hope that helps :))

alexandr1967 [171]4 years ago
4 0

Answer:

I believe you must have a host cell in order to replicate.

Explanation:

A virus can replicate while it is still alive; its called mutation and replication. It changes and remakes its self.

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Northern lights &amp; auroras can be seen within the what sub layer
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Auroras and northern lights can be seen within the ionosphere. This is the range where cosmic and the sun's radiation are ionized.
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3 years ago
Which of the following statements about the Coriolis effect is correct?
labwork [276]

Answer:

Option D

The Coriolis effect works at right angles to the direction of airflow

Explanation:

At the equator, Coriolis effect is negligible, basically zero while it's strongest at the poles (to imply statement B is wrong). Moreover, Coriolis effect is affected by the speed of wind and it also affects the speed of wind since when the wind speed decreases, due to friction for example, the Coriolis effect is also reduces.

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3 years ago
Please select the word from the list that best fits the definition site of sea-floor spreading
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<h3><u>Answer;</u></h3>

Mid-ocean ridges

<h3><u>Explanation</u>;</h3>
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  • Mid-ocean ridge is an undersea mountain chain where new ocean floor is produced at a divergent plate boundary.
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8 0
3 years ago
A rocket moves through outer space at 11,000 m/s. At this rate, how much time would be required to travel the distance from Eart
Charra [1.4K]
11,000 m = 11 km

11 km/s over 380,000km

380,000 / 11 = 34545.4 seconds

34545.4 / 60 = 575.7 minutes
5 0
3 years ago
A basketball star covers 2.70 m horizontally in a jump to dunk the ball (see figure). His motion through space can be modeled pr
Likurg_2 [28]

Answer:

Part a)

T = 0.81 s

Part b)

v_x = 3.33 m/s

Part c)

v_y = 3.91 m/s

Part d)

\theta = 49.55 degree

Part e)

T = 1.11 s

Explanation:

Part a)

initial vertical position = 1.02 m

maximum height = 1.80 m

\Delta y = 1.80 - 1.02

\Delta y = 0.78 m

v_f^2 - v_y^2 = 2a \Delta y

0 - v_y^2 = 2(-9.81)(0.78)

v_y = 3.91 m/s

time taken by it to reach this height

v_y = v_i + at

0 = 3.91 - 9.81 t_1

t_1 = 0.39 s

Now when it again touch the ground then its speed is given as

v_f^2 - v_y^2 = 2a \Delta y

v_f^2 - 0 = 2(9.81)(1.80 - 0.95)

v_y = 4.08 m/s

time taken by it to reach this height

4.08 = v_i + at

4.08 = 0 + 9.81 t_2

t_2 = 0.42 s

T = t_1 + t_2

T = 0.81 s

Part b)

Horizontal velocity

v_x = \frac{x}{t}

v_x = \frac{2.70}{0.81}

v_x = 3.33 m/s

Part c)

vertical velocity is the intial y direction velocity

v_y = 3.91 m/s

Part d)

Take off angle is given as

tan\theta = \frac{3.91}{3.33}

\theta = 49.55 degree

Part e)

initial vertical position = 1.20 m

maximum height = 2.50 m

\Delta y = 2.50 - 1.20

\Delta y = 1.30 m

v_f^2 - v_y^2 = 2a \Delta y

0 - v_y^2 = 2(-9.81)(1.30)

v_y = 5.05 m/s

time taken by it to reach this height

v_y = v_i + at

0 = 5.05 - 9.81 t_1

t_1 = 0.51 s

Now when it again touch the ground then its speed is given as

v_f^2 - v_y^2 = 2a \Delta y

v_f^2 - 0 = 2(9.81)(2.50 - 0.72)

v_y = 5.9 m/s

time taken by it to reach this height

5.9 = v_i + at

5.9 = 0 + 9.81 t_2

t_2 = 0.60 s

T = t_1 + t_2

T = 1.11 s

5 0
3 years ago
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