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2 years ago

Some steps in mitosis are shown below in the incorrect order:

Physics

2 answers:

AysviL [449]2 years ago

6 0

Answer:

Step C, Step D, Step A, Step B i sure

DENIUS [597]2 years ago

3 0

Mark the other person as brainest I was gonna answer but no need

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Raindrops fall vertically at 7.5 m/s relative to the Earth. What does an observer in a car moving at 20.2 m/s in a straight line

Vilka [71]

Answer:

vDP = 21.7454 m/s

θ = 200.3693°

Explanation:

Given

vDE = 7.5 m/s

vPE = 20.2 m/s

Required: vDP

Assume that

vDE to be in direction of - j

vPE to be in direction of i

According to relative motion concept the velocity vDP is given by

vDP = vDE - vPE (I)

Substitute in (I) to get that

vDP = - 7.5 j - 20.2 i

The magnitude of vDP is given by

vDP = √((- 7.5)²+(- 20.2)²) m/s = 21.7454 m/s

θ = Arctan (- 7.5/- 20.2) = 20.3693°

θ is in 3rd quadrant so add 180°

θ = 20.3693° + 180° = 200.3693°

4 0

3 years ago

How do heat insulators work

a_sh-v [17]

The heat is transferred to one material to another, however insulators minimize that transfer, keeping it in the area, warming it.

4 0

3 years ago

Read 2 more answers

A convex security mirror has a radius of curvature of 12.0 cm. What is the magnification of a pare 3.0 m from the mirror?

Makovka662 [10]

Answer:

magnification will be -0.025

Explanation:

We have given the radius of curvature = 12 cm

And object distance = 3 m

So focal length $f=\frac{R}{2}=\frac{12}{2}=6cm$

Now for mirror we know that $\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$

So $\frac{1}{0.06}=\frac{1}{3}+\frac{1}{v}$

$16.66-0.333=\frac{1}{v}$

v = 0.750 m

Now magnification of the mirror is $m=\frac{-v}{u}=\frac{-0.750}{3}=-0.025$

5 0

3 years ago

a container of water is knocked off a 10.0 meter high ledge with a horizontal velocity of 1.00 meters/second. calculate the time

Evgen [1.6K]

Answer:

1.43 s

Explanation:

The time it takes for the container to reach the ground is determined only by the vertical motion of the container, which is a free-fall motion, so a uniformly accelerated motion with a constant acceleration of g=9.8 m/s^2 towards the ground.

The vertical distance covered by an object in free fall is given by

$S=ut + \frac{1}{2}at^2$

where

u = 0 is the initial vertical speed

t is the time

a= g = 9.8 m/s^2 is the acceleration

since u=0, it can be rewritten as

$S=\frac{1}{2}gt^2$

And substituting S=10.0 m, we can solve for t, to find the duration of the fall:

$t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2(10.0 m)}{9.8 m/s^2}}=1.43 s$

3 0

3 years ago

Where are alkaline earth metals found on the periodic table?

mojhsa [17]

Answer:

Explanation:

8 0

2 years ago