All categories

3 years ago

How does a man find the g-spot of a woman?

1 answer:

4 0

The Grafënburg spot is an erogenous region which is located inside the vagina. It is located on the upper wall of the vagina and is usually located 2-3 cm deep. However, there appears to be variation in the perception of Grafënburg spot stimulation amount females. Some women may have trouble finding their g spot due to lack of perception during stimulation, while other women may experience pain and discomfort upon stimulation. This is why communication between sexual partners. By communicating with your partner you can work together in finding the Grafënburg spot, and discuss if stimulation of the area provides satisfaction, or if it is just causing discomfort.

As far as clinical significance, I can not think of any valid reason for a physician to examine the Grafënburg spot, so I suggest that you direct your academic interests to more relevant health related topics.

You might be interested in

Wheat production W in a given year depends on the average temperature T and the annual rainfall R. Scientists estimate that the

Nimfa-mama [501]

Answer:

-1.1

Explanation:

Data provided in the question:

Average temperature is rising at a rate, $\frac{dT}{dt}$ = 0.15°C/year

Rate of change rainfall, $\frac{dR}{dt}$ = - 0.1 cm/year

$\frac{\delta W}{\delta T}$ = -2

$\frac{\delta W}{\delta R}$ = 8

Now,

we need to calculate $\frac{dW}{dt}$

since,

The wheat production (W) is dependent on the rainfall (R) and the Temperature (T)

thus, Using the chain rule , we have

$\frac{dW}{dt}$ = $\frac{\delta W}{dT}\times\frac{dT}{dt}$ + $\frac{\delta W}{dR}\times\frac{dR}{dt}$

on substituting the respective values, we get

$\frac{dW}{dt}$ = -2 × 0.15 + 8 × (-0.1)

$\frac{dW}{dt}$ = -0.3 - 0.8

$\frac{dW}{dt}$ = -1.1

3 0

2 years ago

When a plant sweats or emits water is called

Alla [95]

Answer:

Overall, this uptake of water at the roots, transport of water through plant tissues, and release of vapor by leaves is known as transpiration

Explanation:

4 0

2 years ago

Read 2 more answers

27. What is the major reactant needed for the Citric Acid Cycle (aka Krebs Cycle)?

sp2606 [1]

Answer: b. Acetyl Co-A

Explanation:

Acetyl CoA produced through pyruvate, amino acids, and fatty acids are oxidized in the Krebs cycle in CO2, obtaining as products NADH, FADH2 and GTP (ATP). Parallel to this oxidation, the Krebs cycle produces compounds used as precursors for biosynthesis. As it is a cycle, an oxaloacetate molecule could, in principle, oxidize an amount indefinite of acetyl CoA. Acetyl-CoA is formed from the oxidative decarboxylation of pyruvate, sequentially performed by pyruvate dehydrogenase -PDH (complex multi enzymatic of 3 enzymes), in the mitochondrial matrix.

5 0

2 years ago

I need help

Mnenie [13.5K]

Answer:

A. Sediment

Explanation:

As water flows, it both breaks down rocks and picks up soil. When the pieces of rock and soil are travelling in water, they are called sediment.

3 0

3 years ago

Read 2 more answers

Reticulate and elementary bodies are two forms of A. Mycoplasma. B. Caulobacter. C. Chlamydia. D. Myxobacteria.

Aleksandr [31]

Answer:

C. Chlamydia. is the correct answer.

Explanation:

Chlamydia is a bacterial sexually transmitted disease, caused by obligate intracellular Chlamydia trachomatis.

In the life of chlamydia, there are two stages of developmental(Reticulate and elementary bodies are two forms of Chlamydia.)

The elementary body is the infectious substance, they are non replicating and have a rigid outer membrane that binds to the receptors present on the host cell and it initiates the infection and due to the rigid outer membrane there, no fusion between the phagosome and the lysosome hence oppose the intracellular killing.

Reticulate bodies are the metabolically active form of a chlamydia and non-infectious.

6 0

3 years ago