Ask question

Ask question

All categories

2 years ago

12

Of the following, which is the best reason for why a woman's risk for cardiovascular disease increases after menopause?

1 answer:

MatroZZZ [7]2 years ago

6 0

Answer:

The best reason for the increased risk of cardiovascular diseases in women after menopause is the significant decrease in the levels of Estrogen in the body.

Detailed answer:

Due to the approach of menopause the estrogen levels in the blood decrease drastically.
Estrogen helps in relaxing the vessels and keeping them in their natural diameter.
But due to a drastic decrease in its level due to menopause, there is a high chance of cholesterol building up on the walls of the artery in the heart, which can cause many cardiovascular diseases.

You might be interested in

What are the two factors that affect the frictional force between two surfaces

Talja [164]

Static friction and normal force? I would Google to double check if I'm right.

6 0

3 years ago

If you see fire or smoke coming from your vehicle's hood, you should pull over and get away from the vehicle as soon as possible

hammer [34]

The given statement "If you see fire or smoke coming from your vehicle's hood, you should pull over and get away from the vehicle as soon as possible." is True.

What to do when there is smoking engine in your car?

You should take rapid action if the smoke seems to be coming from a fire under your bonnet.

shut off your vehicle

Don't try to prop the bonnet open; instead, pull the bonnet release lever.

Remove everyone from the vehicle and maintain your distance.

Ask for the fire service by dialing 999.

If it is safe to do so, warn oncoming traffic.

Learn more about the safety precaution with the help of the given link:

brainly.com/question/2347255

#SPJ4

8 0

2 years ago

What is the energy stored, in units of nanoJoules, on a 14.8 nF capacitor when the voltage applied across the capacitor is 7.2 V

AVprozaik [17]

Answer:

383.6 nJ

Explanation:

The energy stored in a capacitor is given by the formula:

$E=\frac{1}{2}CV^2$

where

C is the capacitance

V is the voltage applied

In this problem, we have

C = 14.8 nF is the capacitance of the capacitor

V = 7.2 V is the voltage

Substituting into the equation, we find:

$E=\frac{1}{2}(14.8 nF)(7.2 V)^2=383.6 nJ$

5 0

3 years ago

Read 2 more answers

How many joules are in 10,000 Electron Volts

Semenov [28]

1.602177e-15 joules are in 10,000 electron volts

5 0

3 years ago

Read 2 more answers

Two parallel-plate capacitors, 6.0 mF each, are connected in parallel to a 10 V battery. One of the capacitors is then squeezed

Lady_Fox [76]

The additional charge transferred to the capacitors by the battery is 60 μC.

The increase in the total charge stored on the capacitors is 60 μC.

<u>Explanation</u>:

C = εоA / d

If the separation is halved, then the capacitance will be doubled and according to the equation q = CV, the charge will be doubled too.

Initial charge of the capacitor is:

q = CV

= (6e - 6) $\times$ (10)

= 60 μC

Final charge of the capacitor:

q = (2C)V

= (2 $\times$ 6e - 6) $\times$ (10)

= 120 μC

additional charge transmitted is:

q' = 120 - 60

= 60 μC

initial total charge:

$q_{i}$ = (C1 + C2) V

= (6 + 6) $\times$ (10)

= 120 μF

final total charge:

$q_{f}$ = (C1 + C2) V

= (2 $\times$ 6 + 6) $\times$ (10)

= 180 μF

Increase in the charge:

q' = 180 - 120

= 60 μC

6 0

3 years ago