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3 years ago

What were the two different alleles for height in the pea plants that Mendel studied?

1 answer:

8 0

You might want to go to this website, http://www.indiana.edu/~p1013447/dictionary/mendel.htm
Welcome, And i hope this helps :P

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A 55-kg woman is wearing high heels.

Grace [21]

Answer:

Pressure, $P=1.90\times 10^7\ Pa$

Explanation:

It is given that,

Mass of the woman, m = 55 kg

Diameter of the circular cross section, d = 6 mm

Radius, r = 3 mm = 0.003 m

Let P is the pressure exerted on the floor. It is equal to the force acting on woman per unit area. It is given by :

$P=\dfrac{F}{A}$

$P=\dfrac{mg}{\pi r^2}$

$P=\dfrac{55\times 9.8}{\pi (0.003)^2}$

$P=1.90\times 10^7\ Pa$

So, the pressure exerted on the floor is $1.90\times 10^7\ Pa$ . Hence, this is the required solution.

7 0

3 years ago

True or False: An example of an Isometric movement is a plank. please help 2 mins left! PE

Anon25 [30]

Answer:

Yes

Explanation:

The plank (also called a front hold, hover, or abdominal bridge) is an isometric core strength exercise that involves maintaining a position similar to a push-up for the maximum possible time

5 0

2 years ago

Read 2 more answers

What happens to a planes wing when air moves taster over the top of?

Anon25 [30]

Answer:

it begins to decrease it's altitude

Explanation:

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3 years ago

What is a strain streched to 5 mm? in a wire of length 0.5m

Pie

5mm ................

7 0

3 years ago

In the Hunger Games movie, Katniss Everdeen fires a 0.0200-kg arrow from ground level to pierce an apple up on a stage. The spri

egoroff_w [7]

Answer:

a) $v=99.8584\ m.s^{-1}$

b) $v'=99.366\ m.s^{-1}$

Explanation:

Given:

mass of the arrow, $m=0.02\ kg$

stiffness constant of the bow, $k=330\ N,m^{-1}$

distance of pulling back the arrow on the bow from its mean position, $\Delta x=0.55\ m$

height of the apple targeted, $h=5\ m$

Force on the arrow due to the stiffness of the bow:

$F=k.\Delta x$

$F=330\times 0.55$

$F=181.5\ N$

Now the acceleration of the arrow upwards:

$a=\frac{F}{m}$

$a=\frac{181.5}{0.02}$

$a=9075\ m.s^{-2}$

a) For the course of motion when the arrow leaves the bow after the stretch is relaxed we consider that the arrow left the bow after its string goes to the mean position. During this phase the arrow also faces gravity in the downward direction.

Using the equation of motion:

$v^2=u^2+2(a-g).\Delta x$