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3 years ago

10

What is the correct path of sperm cells through the male reproductive system?

1 answer:

irakobra [83]3 years ago

8 0

Answer:

What is the correct path of sperm cells through the male reproductive system?

Epididymis, seminiferous tubules, urethra, vas deferens

<u>Seminiferous tubules, epididymis, vas deferens, urethra </u>

Urethra, seminiferous tubules, epididymis, vas deferens

Seminiferous tubules, vas deferens, epididymis, urethra

Hope this helps :)

Have a great day !

5INGH

Explanation:

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How many earths could fit inside jupiter

Snezhnost [94]

1,300 or more.

hope this helped :)

8 0

3 years ago

A 0.10 g honeybee acquires a charge of +23 pC while flying.

kari74 [83]

Answer:

a) $\frac{F}{w} =2.347\times 10^{-6}\ N$

b) $E=4.2609\times 10^7\ N.C^{-1}$ parallel to the earth surface.

In this case according to the Fleming's left hand rule the direction of movement of bee must be in a direction parallel to the earth surface and perpendicular to the electric field at the same time.

Explanation:

Given:

mass of the bee, $m=10^{-4}\ kg$

charge acquired by the bee, $q_2=23\times 10^{-12}\ C$

a.

Electrical field near the earth surface, $E=100\ N.C^{-1}$

Now the electric force on the bee:

we know:

$F=\frac{1}{4\pi.\epsilon_0} \times \frac{q_1.q_2}{r^2}$

$F=E.q_2$

$F=100\times 23\times 10^{-12}$

$F=23\times 10^{-10}\ N$

The weight of the bee:

$w=m.g$

$w=10^{-4}\times 9.8$

$w=9.8\times10^{-4}\ N$

Therefore the ratio :

$\frac{F}{w} =\frac{23\times 10^{-10}}{9.8\times10^{-4}}$

$\frac{F}{w} =2.347\times 10^{-6}\ N$

b.

The condition for the bee to hang is its weight must get balanced by the electric force acing equally in the opposite direction.

So,

$F=9.8\times10^{-4}\ N$

$E.q_2=9.8\times10^{-4}\ N$

$E\times 23\times 10^{-12}=9.8\times10^{-4}\ N$

$E=4.2609\times 10^7\ N.C^{-1}$ parallel to the earth surface.

In this case according to the Fleming's left hand rule the direction of movement of bee must be in a direction parallel to the earth surface and perpendicular to the electric field at the same time.

3 0

3 years ago

If a body having mass 40kg started moving initially with rest and it takes a velocity of 20m/sec in time 4 seconds. Find the val

baherus [9]

${\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{Mass \ of \ the \ body \ (m) = 40 \ kg}$

$\:\:\:\:\bullet\:\:\:\sf{Final \ velocity \ of \ the \ body \ (v) = 20 \ m/s}$

$\:\:\:\:\bullet\:\:\:\sf{Initial \ velocity \ of \ the \ body \ (u) = 0}$

$\\$

${\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{Force \ exerted \ by \ the \ body \ ( F)}$

$\\$

${\mathfrak{\underline{\purple{\:\:\: Solution:-\:\:\:}}}} \\ \\$

☯ <u>Using 1st equation of motion </u>

$\\$

$\dashrightarrow\:\: \sf{v = u + at}$

$\\$

$\dashrightarrow\:\: \sf{20 = 0 + a(4)}$

$\\$

$\dashrightarrow\:\: \sf{20 = 4a}$

$\\$

$\dashrightarrow\:\: \sf{\dfrac{\cancel{20}}{\cancel{4}} = a}$

$\\$

$\dashrightarrow\:\: \sf{a = 5}$

$\\$

☯ <u>Now, Finding the force exerted </u>

$\\$

$\dashrightarrow\:\: \sf{F = ma}$

$\\$

$\dashrightarrow\:\: \sf{F = 40 \times 5}$

$\\$

$\dashrightarrow\:\: \sf{F = 200 \ N}$

$\\$

☯ <u>Hence</u>, $\\$

$\:\:\:\:\star\:\:\:\sf{The \ force \ exerted \ by \ the \ body \ is \ 200N}$

8 0

3 years ago

A boy jumps from rest, straight down from the top of a cliff. He falls halfway down to the water below in 0.866 s. How much time

Iteru [2.4K]

<h2>Entire trip takes 1.22 seconds.</h2>

Explanation:

We have equation of motion s = ut + 0.5 at²

Initial velocity, u = 0 m/s

Acceleration, a = 9.81 m/s²

Time, t = 0.866 s

Substituting

s = ut + 0.5 at²

s = 0 x 0.866 + 0.5 x 9.81 x 0.866²

s = 3.68 m

Halfway is 3.68 m

Total height = 2 x 3.68 = 7.36 m

We have equation of motion s = ut + 0.5 at²

Initial velocity, u = 0 m/s

Acceleration, a = 9.81 m/s²

Time, t = ?

Displacement, s = 7.36 m

Substituting

s = ut + 0.5 at²

7.36 = 0 x t + 0.5 x 9.81 x t²

t = 1.22 s

Entire trip takes 1.22 seconds.

7 0

3 years ago

While Bob is demonstrating the gravitational force on falling objects to his class, he drops an 1.0 lb bag of feathers from the

BartSMP [9]

In the experiment of free fall bob released a bag of mass 1 lb

so here we can say that initial speed of the bag is Zero

time taken by the bag to free fall is given as

t = 1.5 s

also the acceleration of free fall is given as

a = 9.8 m/s^2

now we will use kinematics equation here for finding the distance of free fall

$d = v_i * t + \frac{1}{2} at^2$

$d = 0 + \frac{1}{2}*9.8* 1.5^2$

$d = 4.9 * 2.25$

$d = 11.025 m$

so the bag will fall down by total distance of 11.025 m from its initial released position.

3 0

3 years ago