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3 years ago

1. List one way mitosis and meiosis are the same and one way they are different.

1 answer:

6 0

1. They both are same in the process, I mean they both are types of cell division they both form and give rise to new cells from the existing ones
difference would be in their division a mitotic cell gives rise to 2 cells where as a meiotic cell gives rise to 4.
2. the advantage of sexual reproduction is that , it produces genetic variation, disadvantage is it's a very slow process and takes lot of time
3. the advantage is that ,it's a very rapid process as it doesn't involve complex interactions. of genes and chromosomal recombination, the disadvantage would be is that they don't produce variation which can have adverse affect on their survival, even due to slightest change in the environment
4. Undergoing both the type of cell division is good as in human mitotic division is meant for growth and development whereas meiotic divisions are necessary and important for gametes formation and for raising new generation

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One of your delivery trucks traveled 1,200 miles on 55 gallons of gas. How many miles per gallon did the truck get? (Round off y

Aleksandr-060686 [28]

The word "Per" means divide

"miles per gallon" is the same as "miles / gallon"

The truck went 1,200 miles
on 55 gallons

1,200 ÷ 55 = 21.81

7 0

3 years ago

Read 2 more answers

In the 1920s what did Edmund hubble notice about the galaxies

zlopas [31]

Hubble noticed that the galaxies were moving away from us, which meant the universe was expanding.

This is why constellations change over time. In some years, the Big Dipper won't actually look like a dipper anymore.

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3 years ago

Three identical charges q form an equilateral triangle of side a with two charges on the x-axis and one on the positive y-axis.

shusha [124]

Answer:

$F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2} )$

Explanation:

Given:

- Three identical charges q.

- Two charges on x - axis separated by distance a about origin

- One on y-axis

- All three charges are vertices

Find:

- Find an expression for the electric field at points on the y-axis above the uppermost charge.

- Show that the working reduces to point charge when y >> a.

Solution

- Take a variable distance y above the top most charge.

- Then compute the distance from charges on the axis to the variable distance y:

$r = \sqrt{(\frac{\sqrt{3}*a }{2} + y)^2 + (a/2)^2 }$

- Then compute the angle that Force makes with the y axis:

cos(Q) = sqrt(3)*a / 2*r

- The net force due to two charges on x-axis, the vertical components from these two charges are same and directed above:

F_1,2 = 2*F_x*cos(Q)

- The total net force would be:

F_net = F_1,2 + kq / y^2

- Hence,

$F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2} )$

- Now for the limit y >>a:

$F_n = k*q*(\frac{2*y(1 + \frac{\sqrt{3}*a }{2*y}) }{y^3((1+ \frac{\sqrt{3}*a }{2*y})^2 + (a/y*2)^2)^1.5 }) +\frac{1}{y^2} )$

- Insert limit i.e a/y = 0

$F_n = k*q*(\frac{2}{y^2} +\frac{1}{y^2}) \\\\F_n = 3*k*q/y^2$

Hence the Electric Field is off a point charge of magnitude 3q.

8 0

3 years ago

How far from the base of the platform does she land?

seropon [69]

When Janet leaves the platform, she's moving horizontally at 1.92 m/s. We assume that there's no air resistance, and gravity has no effect on horizontal motion. There's no horizontal force acting on Janet to make her move horizontally any faster or slower than 1.92 m/s.

She's in the air for 1.1 second before she hits the water.

Moving horizontally at 1.92 m/s for 1.1 second, she sails out away from the platform

(1.92 m/s) x (1.1 sec) = <em>2.112 meters</em>

3 0

3 years ago

Across:

vladimir2022 [97]

Answer:

it would be c i just had it

Explanation:

welcome................

6 0

2 years ago