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Sergeeva-Olga [200]

3 years ago

10

How long human can live inside the water without oxygen?

2 answers:

kirill115 [55]3 years ago

7 0

After five to ten minutes of not breathing, you are likely to develop serious and possibly irreversible brain damage.

viva [34]3 years ago

5 0

It really depends on how active they are. If they don't train for their lung compassion they could probably last for about 10-15 seconds. But if they exercise and do their lung compassion exerciser they can hold there breath for about 20-40 seconds and minutes. But you cannot live inside water or else you'll die.

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A long coaxial cable consists of an inner cylindrical conductor with radius a and an outer coaxial cylinder with inner radius b

Natasha_Volkova [10]

Answer:

Part a)

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Part b)

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Part d)

As we know that due to induction of charge there will be same charge appear on the inner and outer surface of the cylinder but the sign of the charge must be different

On the inner side of the cylinder there will be negative charge induce on the inner surface and on the outer surface of the cylinder there will be same magnitude charge with positive sign.

Explanation:

Part a)

By Guass law we know that

$\int E. dA = \frac{q}{\epsilon_0}$

$E. 2\pi rL = \frac{\lambda L}{\epsilon_0}$

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Part b)

Outside the outer cylinder we will again use Guass law

$\int E. dA = \frac{q}{\epsilon_0}$

$E. 2\pi rL = \frac{\lambda L}{\epsilon_0}$

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Part d)

As we know that due to induction of charge there will be same charge appear on the inner and outer surface of the cylinder but the sign of the charge must be different

On the inner side of the cylinder there will be negative charge induce on the inner surface and on the outer surface of the cylinder there will be same magnitude charge with positive sign.

4 0

3 years ago

edward travels 150 kilometers due west and then 200 kilometers in a direction 60 degrees north of west.what is his displacement

Naily [24]

The displacement of Edward in the westerly direction is determined as 338.32 km.

<h3>What is displacement of Edward?</h3>

The displacement of Edward can be determined from different methods of vector addition. The method applied here is triangular method.

The angle between the 200 km north west and 150 km west = 60 + 90 = 150⁰

The displacement is the side of the triangle facing 150⁰ = R

R² = a² + b² - 2abcosR

R² = 150² + 200² - (2x 150 x 200)xcos(150)

R² = 62,500 - (-51,961.52)

R² = 114,461.52

R = 338.32 km

Learn more about displacement here: brainly.com/question/321442

#SPJ1

3 0

1 year ago

A car has an acceleration of 4m/s^2 to the east for 5 seconds. What is the change in velocity?

Natali5045456 [20]

Answer:

a= 4m/s^2

t = 5 secs

v = at

v = 4 × 5

v =20 m/s

3 0

2 years ago

The radius of the aorta is about 1 cm and the blood flowing through it has a speed of about 30 cm/s. Calculate the average speed

puteri [66]

Answer:

The average speed of the blood in the capillaries is 0.047 cm/s.

Explanation:

Given;

radius of the aorta, r₁ = 1 cm

speed of blood, v₁ = 30 cm/s

Area of the aorta, A₁ = πr₁² = π(1)² = 3.142 cm²

Area of the capillaries, A₂ = 2000 cm²

let the average speed of the blood in the capillaries = v₂

Apply continuity equation to determine the average speed of the blood in the capillaries.

A₁v₁ = A₂v₂

v₂ = (A₁v₁) / (A₂)

v₂ = (3.142 x 30) / (2000)

v₂ = 0.047 cm/s

Therefore, the average speed of the blood in the capillaries is 0.047 cm/s.

4 0

2 years ago

Two parallel plates of area 2.34*10-3 M2 have 7.07*10-7C of charge placed on them. A6.62*10-5C charge q1 is placed between the p

ira [324]

Answer:

The force exerted on the $q_1$ is $F = 2.25*10^{3} \ N$

Explanation:

From the question we are told that

The area is $A = 2.34*10^{-3} \ m^2$

The magnitude of charge placed on them is $q = 7.07 * 10^{-7} C$

The charge placed between the plate is $q_1 = 6.62 *10^{-5} C$

The electric field generated around the plate is mathematically represented as

$E = \frac{q}{A \epsilon_o}$

Substituting values

$E = \frac{7.07*10^{-7}}{2.34*10^{-3} * 8.85 *10^{-12}}$

$E = 34*10^{6} \ V/m$

The force exerted the charge $q_1$ is mathematically represented as

$F = q_1 * E$

Substituting values

$F = 6.62 *10^{-5} * 34*10^{6}$

$F = 2.25*10^{3} \ N$

8 0

3 years ago