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3 years ago

If the designers of submarines do not appropriately account for hydrostatic pressure,

Physics

1 answer:

andre [41]3 years ago

7 0

Answer:

A) the pressure could crush the submarine

Explanation:

The submarines are mainly used for defense purposes.
In world war 1 and 2 the submarines played a vital role in the war.
The submarines work according to the principle of buoyancy.
According to this principle, when seawater is filled inside the tank in a required amount, the submarines sink.
When the seawater is expelled, the submarine rises to above the sea.
Most of the time, it dives deep into the sea, the hydrostatic pressure should be taken into account.
If the hydrostatic pressure is not considered, the pressure could crush the submarine.

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A series circuit has a capacitor of 10−5 F, a resistor of 3 × 102 Ω, and an inductor of 0.2 H. The initial charge on the capacit

Alexus [3.1K]

Given the values to proceed to solve the exercise, we resort to the solution of the exercise through differential equations.

The problem can be modeled through a linear equation, in the form:

$10^5 Q +300Q'+0.2Q''=0$

With the initial conditions as,

$Q(0) = 10^{-6}$

$Q'(0)= 0$

Where Q(t) is the charge.

<em>The general solution of a linear equation is given as:</em>

Applying this definiton in our differential equation we have that

$Q(t) = C_1e^{at}+C_2e^{bt}$

To find b and a we use the first equation and find the roots:

$r_{a,b} = \frac{-300 \pm \sqrt{(300)^3-4(0.2)*10^5}}{0.4}$

$r_{a,b} = {-1000,-500}$

Then we have

$Q(t) = C_1e^{-1000t}+C_2e^{-500t}$

To find the values of the Constant we apply the initial conditions, then

$Q(0)= 10^{-6} = C_1+C_2$

And for the derivate:

$Q'(t) = -1000C_1e^{-1000t}-500C_2e^{-500t}$

$0 = -1000C_1e^{-1000(0)}-500C_2e^{-500(0)}$

$0 = -1000C_1-500C_2$

We have a system of 2x2:

$(1) 10^{-6} = C_1+C_2$

$(2) 0 = -1000C_1-500C_2$

Solving we have:

$C_1 = -10^{-6}$

$C_2 = 2*10^{-6}$

The we can replace at the equation and we have that the Charge at any moment is given by,

$Q(t) = (-10^{-6})e^{-1000t}+( 2*10^{-6})e^{-500t}$

If we obtain the derivate we find also the Current, then

$I(t)= 10^{-3}e^{-1000t}-10^{-3}e^{-500t}$

7 0

3 years ago

Talking to adults can be intimidating and uncomfortable if you do not yet feel like one. However, learning how to effectivelycom

Korvikt [17]

Answer: i didnt look at your question its to long

Explanation:

8 0

4 years ago

A 290-turn solenoid having a length of 32 cm and a diameter of 11 cm carries a current of 0.30 A. Calculate the magnitude of the

iris [78.8K]

The magnitude of the magnetic field inside the solenoid is 3.4×10^(-4) T.

To find the answer, we need to know about the magnetic field inside the solenoid.

<h3>What's the expression of magnetic field inside a solenoid?</h3>

Mathematically, the expression of magnetic field inside the solenoid= μ₀×n×I

n = no. of turns per unit length and I = current through the solenoid

<h3>What's is the magnetic field inside the solenoid here?</h3>

Here, n = 290/32cm or 290/0.32 = 906

I= 0.3 A

So, Magnetic field= 4π×10^(-7)×906×0.3 = 3.4×10^(-4) T.

Thus, we can conclude that the magnitude of the magnetic field inside the solenoid is 3.4×10^(-4) T.

Learn more about the magnetic field inside the solenoid here:

brainly.com/question/22814970

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2 years ago

It is said that a gas fills all the space available to it. Why then doesn't the atmosphere go off into space?

Alex777 [14]

Earth has its own atmosphere. That is one reason all the water that has been on Earth has been recycled through the water cycle. It never leaves Earth’s atmosphere.

7 0

3 years ago

Rewrite the below linear equation in slope-intercept form:

Bezzdna [24]

Answer:

I don't think any options are correct.

Explanation:

When solved for y the process goes as follows:

5x+ 8y = 16

5x + 8y - 5x = 16 - 5x

8y = 16 - 5x

8y/8 = (16-5x)/8

y = 2- (5/8)x

5 0

4 years ago