Ask question

Ask question

All categories

3 years ago

5

A steel aircraft carrier is 403 m long when moving through the icy North Atlantic at a temperature of 1.22 °C. By how much does

the carrier lengthen when it is traveling in the warm Mediterranean Sea at a temperature of 24.0 °C?

1 answer:

Lorico [155]3 years ago

8 0

Answer:

it expands by 11 cm.

Explanation:

First we need the coefficient of linear thermal expansion of steel

a = 12e-6/K

Now we use the equation for linear thermal expansion

L1 = L0 * (1 + a * (t1 - t0))

Where

L0: length before expansion

L1: expanded length

t0: temperature before expansion

t1: temperature expanded

L1 = 403 * (1 + 12e-6 * (24 - 1.22)) = 403.11 m

It expanded 11 cm.

You might be interested in

Two kids are playing on a newly installed slide, which is 3 m long. John, whose mass is 30 kg, slides down into William (20 kg),

yuradex [85]

Answer:

$v=3.564\ m.s^{-1}$

$\Delta v =2.16\ m.s^{-1}$

Explanation:

Given:

mass of John, $m_J=30\ kg$

mass of William, $m_W=30\ kg$

length of slide, $l=3\ m$

(A)

height between John and William, $h=1.8\ m$

<u>Using the equation of motion:</u>

$v_J^2=u_J^2+2 (g.sin\theta).l$

where:

v_J = final velocity of John at the end of the slide

u_J = initial velocity of John at the top of the slide = 0

Now putting respective :

$v_J^2=0^2+2\times (9.8\times \frac{1.8}{3})\times 3$

$v_J=5.94\ m.s^{-1}$

<u>Now using the law of conservation of momentum at the bottom of the slide:</u>

<em>Sum of initial momentum of kids before & after collision must be equal.</em>

$m_J.v_J+m_w.v_w=(m_J+m_w).v$

where: v = velocity with which they move together after collision

$30\times 5.94+0=(30+20)v$

$v=3.564\ m.s^{-1}$ is the velocity with which they leave the slide.

(B)

frictional force due to mud, $f=105\ N$

<u>Now we find the force along the slide due to the body weight:</u>

$F=m_J.g.sin\theta$

$F=30\times 9.8\times \frac{1.8}{3}$

$F=176.4\ N$

<em><u>Hence the net force along the slide:</u></em>

$F_R=71.4\ N$

<em>Now the acceleration of John:</em>

$a_j=\frac{F_R}{m_J}$

$a_j=\frac{71.4}{30}$

$a_j=2.38\ m.s^{-2}$

<u>Now the new velocity:</u>

$v_J_n^2=u_J^2+2.(a_j).l$

$v_J_n^2=0^2+2\times 2.38\times 3$

$v_J_n=3.78\ m.s^{-1}$

Hence the new velocity is slower by

$\Delta v =(v_J-v_J_n)$

$\Delta v =5.94-3.78= 2.16\ m.s^{-1}$

8 0

3 years ago

Inductors in parallel. Two inductors L1 = 1.31 H and L2 = 2.24 H are connected in parallel and separated by a large distance so

mote1985 [20]

Answer:

a) 0.83H

b) 3.22/N Henry

Explanation:

Given two inductors L1 = 1.31 H and L2 = 2.24 H connected in parallel, their equivalent inductance derivative is similar to that of resistance in parallel.

Derivation:

If the voltage across an inductor

VL = IXL

I is the current

XL is the inductive reactance

XL = 2πfL

VL = I(2πfL)

L is the inductance.

From the formula, I = V/2πfL

Given two inductors in parallel, different current will flow through them.

I1 = V/2πfL1 (current in L1)

I2 = V/2πfL2 (current in L2)

Total current I = I1+I2

I = V/2πfL1 + V/2πfL2

I = V/2πf{1/L1+1/L2}

V/2πfL = V/2πf{1/L1+1/L2}

1/L = 1/L1+1/L2 (equivalent inductance in parallel)

Given L1 = 1.31 H and L2 = 2.24

1/L = 1/1.31 + 1/2.24

1/L = 0.763 + 0.446

1/L = 1.209

L = 1/1.209

L = 0.83H

The equivalent inductance is 0.83H

b) Given similar inductors L = 3.22H in parallel, the equivalent inductance will be:

1/L = 1/3.22+1/3.22+1/3.22+1/3.22+1/3.22

+1/3.22+1/3.22+1/3.22+1/3.22+1/3.22

1/L = 10/3.22 (since that all have the same denominator)

L = 3.22/10

If N = 10, the generalization of 10 similar inductors in parallel will be:

L = 3.22/N Henry

4 0

3 years ago

How to find the time with only the distance or height from the ground

Harman [31]

You use acceleration due to gravity
and 1/2 atsqr=d
therefore 1/2 * 9.8 * tsqr= d

5 0

4 years ago

Why is a protective apron or lab coat important to use when working with acids?

Kipish [7]

Answer:

Acids break down fabrics and can cause burns if the acids are strong.

Explanation:

A protective apron or lab coat is important when working with acids because acids break down fabrics and can cause burns if the acids are strong.

An acid is a substance that interacts with water to produce excess hydroxonium ions in an aqueous solution.
A strong acid ionizes completely in solution.
When they come in contact with a fabric, they break them down violently.
So, if they come in contact with the skin, it causes a violent break down of body tissues.
The apron acts a protective layer.

4 0

3 years ago

Read 2 more answers

PLEASE HELP ASAP!! What are 3 types of forces acting on your marshmallow (or other flying object) when it impacts it’s target? C

Alekssandra [29.7K]

•Every action has an equal and opposite reaction (the object is putting force on the target, and the target is putting an equal amount of force back)
•Am object in motion (the object) will stay in motion unless an outside force acts upon it (the Target)

And as for the third one I really don’t know, those seem to be the only two, I’m sorry. I did what a could, Hope it kinda helps :)

8 0

4 years ago