Ask question

Ask question

All categories

1 year ago

12

Un Iceberg, con forma aproximada a la de un paralelepípedo (rectángulo en 3D), flota en el mar de modo que la parte fuera del ag

ua tiene 10m de altura. Cuál es la altura "h" de la parte sumergida del Iceberg. Escribir la expresión algebraica de la solución antes de sustituir los datos y después el resultado.

1 answer:

sdas [7]1 year ago

5 0

Answer:

maybe could you put it in english and i could help you

You might be interested in

A piece of copper wire with thin insulation, 200 m long and 1.00 mm in diameter, is wound onto a plastic tube to form a long sol

bearhunter [10]

Answer:

N= 3

Explanation:

For this exercise we must use Faraday's law

E = - dФ / dt

Ф = B . A = B Acos θ

tje bold indicate vectors. As it indicates that the variation of the field is linear, we can approximate the derivatives

E = - A cos θ (B - B₀) / t

The angle enters the magnetic field and the normal to the area is zero

cos 0 = 1

A = π r²

In the length of the wire there are N turns each with a length L₀ = 2π r

L = N (2π r)

r = L / 2π N

we substitute

A = L² / (4π N²)

The magnetic field produced by a solenoid is

B = μ₀ N/L I

for which

B₀ = μ₀ N/L I

The final field is zero, because the current is zero

B = 0

We substitute

E = - (L² / 4π N²) (0 - μ₀ N/L I) / t

E = μ₀ L I / (4π N t)

N = μ₀ L I / (4π t E)

The electromotive force is E = 0.80 mV = 0.8 10⁻³ V

let's calculate

N = 4π 10⁻⁷ 200 1.60 / (4π 0.120 0.8 10⁻³)]

N = 320 10⁻⁷ / 9.6 10⁻⁶

N = 33.3 10⁻¹

N= 3

7 0

2 years ago

A student pulls a cart across the floor with a force of 45 N. If the cart travels 10 m, how much work does the cart do on the st

bixtya [17]

I'm assuming we're applying the standard Integral form of the calculation of work. The solution is provided in the image.

6 0

2 years ago

A boat moves through the water of a river at 10m/s relative to the water, regardless of the boat ‘s direction . If the water in

katen-ka-za [31]

Answer:

The appropriate solution is "61.37 s".

Explanation:

The given values are:

Boat moves,

= 10 m/s

Water flowing,

= 1.50 m/s

Displacement,

d = 300 m

Now,

The boat is travelling,

= $10+1.50$

= $11.5 \ m/s$

Travelling such distance for 300 m will be:

⇒ $v = \frac{d}{t} \ sot \ t$

$=\frac{d}{v}$

On putting the values, we get

$=\frac{300}{11.5}$

$=26.08 \ s$

Throughout the opposite direction, when the boat seems to be travelling then,

= $10-1.50$

= $8.5 \ m/s$

Travelling such distance for 300 m will be:

⇒ $v=\frac{v}{t} \ sot \ t$

$=\frac{d}{v}$

On putting the values, we get

$=\frac{300}{8.5}$

$=35.29 \ s$

hence,

The time taken by the boat will be:

= $26.08+35.29$

= $61.37 \ s$

8 0

2 years ago

What is the name and symbol of the element in the second row and fourteenth column of the periodic table? Hint: Review your peri

Papessa [141]

The answer is Carbon. Beyond being the only element listed here that is located in the second row, it is also in the fourteenth column of the table if you count from left to right. Hope this helps!

4 0

3 years ago

Through what potential difference would you need to accelerate an alpha particle, starting from rest, so that it will just reach

Svetlanka [38]

Answer:

$\Delta V = 1.8 \times 10^7 V$

Explanation:

GIVEN

diameter = 15 fm = $15 \times 10^{-15}$ m

we use here energy conservation

$K_{i}+U_{i} =K_{f}+U_{f}$

there will be some initial kinetic energy but after collision kinetic energy will zero

$K_{i} + 0 = 0 + \frac{1}{4 \pi \epsilon _{0}} \frac{(2e)(92e)}{7.5 \times 10^{-15}}$

on solving these equations we get kinetic energy initial

$KE_{i} = 5.65\times 10 ^{-12} \times \frac {1 eV}{1.6 \times 10^{-19}}$

$KE_{i} = 35.33$ J ..............(i)

That is, the alpha particle must be fired with 35.33 MeV of kinetic energy. An alpha particle with charge q = 2 e

and gains kinetic energy K =e∆V ..........(ii)

by accelerating through a potential difference ∆V

Thus the alpha particle will

just reach the ${238}_U$ nucleus after being accelerated through a potential difference ∆V

equating (i) and second equation we get

e∆V = 35.33 Me V

$\Delta V = \frac{35.33}{2} MV\\\Delta V = 1.8 \times 10^7 V$

7 0

2 years ago