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3 years ago

What is primary sex organs called

2 answers:

6 0

The primary sex organs for men are called testes and ovaries for the women. They produce sex cells called gametes and they secrete sex hormones.

atroni [7]3 years ago

5 0

They're called the <span>reproductive organs.

</span>For a male >> testicles or testes

For a female >> Ovaries

You might be interested in

If part of an electric circuit dissipates energy at 6 W when it draws a current of 3 A, what voltage is impressed across it?

Lena [83]

Answer:2 volts

Explanation:

Power=current x voltage

6=3 x voltage

Divide both sides by 3

6/3=(3 x voltage)/3

2=voltage

Voltage=2volts

5 0

3 years ago

Design a voltage divider to provide the following approximate voltages with respect to ground using a 30 V source: 8.18 V, 14.7

yarga [219]

Answer:

R₁ = 14.7 10³ Ω , R₂ = 8.18 10³ Ω , R₃ = 1.72 10³ Ω , R₄ = 5.4 10³ Ω 1/8 W resistor

Explanation:

For this exercise we must use a series circuit since the sum of the voltage on each resin is equal to the source voltage (V = 30 V)

Therefore we build a circuit with 4 resistors in series, in such a way that

V = i R

let the voltage

1st resistance

V = i R

R₁ = V / i

R₁ = 14.7 / 1 10⁻³

R₁ = 14.7 10³ Ω

power is

P = V i

P = 14.7 1 10⁻³

P = 14.7 10⁻³ W = 0.0147 W

a resistance of ⅛ W is indicated

2nd resistance

R₂ = 8.18 / 1 10⁻³

R₂ = 8.18 10³ Ω

Power

P = 8.18 1 10⁻³

P = 0.00818W

a 1/8 W resistor

3rd resistance

this resistance is calculated in such a way that

V₁ + V₂ + V₃ = 24.6

V₃ = 24.6 - V₁ -V₂

V₃ = 24.6 - 14.7 - 8.18

V₃ = 1.72 V

R₃ = 1.72 / 1 10⁻³

R₃ = 1.72 10³ Ω

power

P = Vi

P = 1.72 10⁻³

P = 0.00172 W

a resistance of ⅛ W

To obtain the voltage of 24.6 we use this three resistors together

4th resistance

The value of this resistance is calculated so that the sum of all the voltages reaches the source voltage

30 = V₁ + V₂ + V₃ + V₄

V₄ = 30 - V₁ -V₂ -V₃

V₄ = 30 -14.7 - 8.18 - 1.72

V₄ = 5.4 V

R₄ = 5.4 / 1 10⁻³

R₄ = 5.4 10³ Ω

Power

P = V i

P = 5.4 10⁻³

P = 0.0054 W

⅛ W resistance

The values of these resistance are commercially

Let's check the consumption of the circuit

R_total = R₁ + R₂ + R₃ + R₄

R_total = (14.7 + 8.18 + 1.72 + 5.4) 10³

R_total = 30 10³

the current circulating in the circuit is

i = V / R_total

i = 30/30 10³

i = 1 10⁻³ A

therefore it is within the order requirement.

for connections see attached diagram

8 0

3 years ago

11. A vector M is 15.0 cm long and makes an angle of 20° CCW from x axis and another vector N is 8.0 cm long and makes an angle

Alja [10]

Answer:

The magnitude of the resultant vector is 22.66 cm and it has a direction of 29.33°

Explanation:

To find the resultant vector, you first calculate x and y components of the two vectors M and N. The components of the vectors are calculated by using cos and sin function.

For M vector you obtain:

$M=M_x\hat{i}+M_y\hat{j}\\\\M=15.0cm\ cos(20\°)\hat{i}+15.0cm\ sin(20\°)\hat{j}\\\\M=14.09cm\ \hat{i}+5.13\ \hat{j}$

For N vector:

$N=N_x\hat{i}+N_y\hat{j}\\\\N=8.0cm\ cos(40\°)\hat{i}+8.0cm\ sin(40\°)\hat{j}\\\\N=6.12cm\ \hat{i}+5.142\ \hat{j}$

The resultant vector is the sum of the components of M and N:

$F=(M_x+N_x)\hat{i}+(M_y+N_y)\hat{j}\\\\F=(14.09+6.12)cm\ \hat{i}+(5.13+5.142)cm\ \hat{j}\\\\F=20.21cm\ \hat{i}+10.27cm\ \hat{j}$

The magnitude of the resultant vector is:

$|F|=\sqrt{(20.21)^2+(10.27)^2}cm=22.66cm$

And the direction of the vector is:

$\theta=tan^{-1}(\frac{10.27}{20.21})=29.93\°$

hence, the magnitude of the resultant vector is 22.66 cm and it has a direction of 29.33°

4 0

3 years ago

A one-dimensional plane wall of thickness 2l= 100 mm experiences uniform thermal energy generation of q˙= 800 w/m3 and is convec

slega [8]

Answer:

The thermal conductivity of the wall = 40W/m.C

h = 10 W/m^2.C

Explanation:

The heat conduction equation is given by:

d^2T/ dx^2 + egen/ K = 0

The thermal conductivity of the wall can be calculated using:

K = egen/ 2a = 800/2×10

K = 800/20 = 40W/m.C

Applying energy balance at the wall surface

"qL = "qconv

-K = (dT/dx)L = h (TL - Tinfinity)

The convention heat transfer coefficient will be:

h = -k × (-2aL)/ (TL - Tinfinty)

h = ( 2× 40 × 10 × 0.05) / (30-26)

h = 40/4 = 10W/m^2.C

From the given temperature distribution

t(x) = 10 (L^2-X^2) + 30 = 30°

T(L) = ( L^2- L^2) + 30 = 30°

dT/ dx = -2aL

d^2T/ dx^2 = - 2a

4 0

3 years ago

A small drop of water is suspended motionless in air by a uniform electric field that is directed upward and has a magnitude of

Gennadij [26K]

To solve this problem we will apply the concepts related to the electric field such as the smelting of the Force and the load (In this case the force is equivalent to the weight). Later we will apply the ratio of the total charge as a function of the multiplication of the number of electrons and their individual charge.

$E = \frac{mg}{q}$

Here,

m = mass

g = Acceleration due to gravity

Rearranging to find the charge,

$q = \frac{mg}{E}$

Replacing,

$q = \frac{(3.37*10^{-9})(9.8)}{11000}$

$q = 3.002*10^{-12}C$

Since the field is acting upwards the charge on the drop should be negative to balance it in air. The equation to find the number of electrons then is

$q = ne$