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3 years ago

Write the function of gonads

Physics

1 answer:

navik [9.2K]3 years ago

7 0

Answer:

The gonads, the primary reproductive organs, are the testes in the male and the ovaries in the female. These organs are responsible for producing the sperm and ova, but they also secrete hormones and are considered to be endocrine glands.

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Fnet = fa + ff or fnet = fa - ff

skelet666 [1.2K]

Answer:first of all what is your question and i can give and example which is Use them when you have 2 forces named Fa & FF or Fg & Ff acting in opposite directions on an object and you need to know the resultant of your 2 forces.

Explanation:

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3 years ago

A racquetball strikes a wall with a speed of 30 m/s and rebounds in the opposite direction with a speed of 26 m/s. The collision

Fudgin [204]

Answer:

The average acceleration of the ball during the collision with the wall is $a=2,800m/s^{2}$

Explanation:

Known Data

We will asume initial speed has a negative direction, $v_{i}=-30m/s$ , final speed has a positive direction, $v_{f}=26m/s$ , $\Delta t=20ms=0.020s$ and mass $m_{b}$ .

Initial momentum

$p_{i}=mv_{i}=(-30m/s)(m_{b})=-30m_{b}\ m/s$

final momentum

$p_{f}=mv_{f}=(26m/s)(m_{b})=26m_{b}\ m/s$

Impulse

$I=\Delta p=p_{f}-p_{i}=26m_{b}\ m/s-(-30m_{b}\ m/s)=56m_{b}\ m/s$

Average Force

$F=\frac{\Delta p}{\Delta t} =\frac{56m_{b}\ m/s}{0.020s} =2800m_{b} \ m/s^{2}$

Average acceleration

$F=ma$ , so $a=\frac{F}{m_{b}}$ .

Therefore, $a=\frac{2800m_{b} \ m/s^{2}}{m_{b}} =2800m/s^{2}$

8 0

3 years ago

You go to the hardware store to buy a new 50 ft garden hose. You find you can choose between hoses of ½ inch and 5/8 inch inner

omeli [17]

To solve this problem it is necessary to consider two concepts. The first of these is the flow rate that can be defined as the volumetric quantity that a channel travels in a given time. The flow rate can also be calculated from the Area and speed, that is,

Q = V*A

Where,

A= Cross-sectional Area

V = Velocity

The second concept related to the calculation of this problem is continuity, which is defined as the proportion that exists between the input channel and the output channel. It is understood as well as the geometric section of entry and exit, defined as,

$Q_1 = Q_2$

$V_1A_1=V_2A_2$

Our values are given as,

$A_1=\frac{1}{2}^2*\pi=0.785 in^2$

$A_2=\frac{5}{8}^2*\pi=1.227 in^2$

Re-arrange the equation to find the first ratio of rates we have:

$\frac{V_1}{V_2}=\frac{A_2}{A_1}$

$\frac{V_1}{V_2}=\frac{1.227}{0.785}$

$\frac{V_1}{V_2}=1.56$

The second ratio of rates is

$\frac{V2}{V1}=\frac{A_1}{A2}$

$\frac{V2}{V1}=\frac{0.785}{1.227}$

$\frac{V2}{V1}=0.640$

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3 years ago

If you pull a wool sweater over your head, you will often end up with a wild hairdo. Your hair is hard to comb and wants to stan

Sergio [31]

Becasue when you rubbed your hair while you were putting on your sweater it caused it to rub against together causing electricity thingy lol. and thats why your hair goes straigh tup.

5 0

3 years ago

In a football game, the running back takes a handoff and begins running toward midfield at 3.21 yards/s . As he moves through hi

Inessa [10]

Given Information:

Initial speed = u = 3.21 yards/s

Acceleration = α = 1.71 yards/s²

Final speed = v = 7.54 yards/s

Required Information:

Distance = s = ?

Answer:

Distance = s = 13.61

Explanation:

We are given the speeds and acceleration of the runner and we want to find out how much distance he covered before being tackled.

We know from the equations of motion,

v² = u² + 2αs

Where u is the initial speed of the runner, v is the final speed of the runner, α is the acceleration of the runner and s is the distance traveled by the runner.

Re-arranging the above equation for distance yields,

2αs = v² - u²

s = (v² - u²)/2α

s = (7.54² - 3.21²)/2×1.71

s = 46.55/3.42

s = 13.61 yards

Therefore, the runner traveled a distance of 13.61 yards before being tackled.

8 0

3 years ago

Write the function of gonads​

Write the function of gonads