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2 years ago

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The eiffel tower has a mass of 7.3 million kilograms and a height of 324 meters. its base is square with a side length of 125 me

ters. the steel used to make the tower occupies a volume of 930 cubic meters. air has a density of 1.225 kg per cubic meter. suppose the tower was contained in a cylinder. find the mass of the air in the cylinder. is this more or less than the mass of the tower?

1 answer:

uranmaximum [27]2 years ago

4 0

Since the tower base is square with a side length of 125 m,

Therefore,

$(125\ m)^2+ (125\ m)^2=31250 m^2$

Square root of 31250 = 176.776953 (Diameter) , so this is the diameter of the cylinder to enclose it, and radius, r = 88.38834765 m and height, h = 324 m.

The volume of cylinder,

$=\pi r^2h=3.14(88.38834765 m)^2\times 324 m =7948168.803\ m^3$

Thus, the mass of the air in the cylinder,

$=1.225\ kg/m^3 \times 7948168.803\ m^3=9736506.78\ kg$

Hence, the mass of the air in the cylinder is this more than the mass of the tower.

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To solve this problem we will apply the concepts related to energy conservation. Here we will use the conservation between the potential gravitational energy and the kinetic energy to determine the velocity of this escape. The gravitational potential energy can be expressed as,

$PE= \frac{GMm}{d}$

The kinetic energy can be written as,

$KE= \frac{1}{2} mv^2$

Where,

$G = 6.67*10^{-11}m^3/kg\cdot s^2$ Gravitational Universal Constant

$m = 5.972*10^{24}kg$ Mass of Earth

$h = 56*10^6m$ Height

$r = 6.378*10^6m$ Radius of Earth

From the conservation of energy:

$\frac{1}{2} mv^2 = \frac{GMm}{d}$

Rearranging to find the velocity,

$v = \sqrt{\frac{2Gm}{d}} \rightarrow$ Escape velocity at a certain height from the earth

If the height of the satellite from the earth is h, then the total distance would be the radius of the earth and the eight,

$d = r+h$

$v = \sqrt{\frac{2Gm}{r+h}}$

Replacing the values we have that

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Therefore the escape velocity is 3.6km/s

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2 years ago

Essam is abseiling down a steep cliff. How much gravitational potential energy does he lose for every metre he descends? His mas

Dafna11 [192]

Answer:

720 J

Explanation:

The gravitational potential energy that Essam loses for every metre is given by:

$\Delta U=mg \Delta h$

where

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By substituting the numbers into the formula, we find

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2 years ago

Read 2 more answers

A square loop of wire has a perimeter of 4.00 mm and is oriented such that two of its parallel sides form a 13.0 ∘∘ angle with t

vivado [14]

Answer:

a) (2.436 × 10⁻⁷) Wb

b) (7.308 × 10⁻⁷) Wb

Explanation:

Magnetic flux is the dot product of the magnetic field vector and the Area vector.

Mathematically, it is given as

Φ = BA cos θ

where B = magnetic field strength

A = Cross sectional Area of the loop enclosed

θ = angle between the magnetic field and the plane of the area.

a) B = 0.250 T

To find A, the perimeter of the loop is given as 4.00 mm.

Perimeter of a square = 4L

4L = 4.00

L = 1.00 mm = 0.001 m

The area is given as L²

A = (0.001)² = 0.000001 m²

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Φ = BA cos θ

Φ = 0.25 × 0.000001 × cos 13°

Φ = 0.0000002436 Wb = (2.436 × 10⁻⁷) Wb

b) Another loop lies in the same plane but has an irregular shape, resembling a starfish. Its area is three times greater than that of the square loop

Φ = BA cos θ

B = 0.25 T

A = 3 × 0.000001 = 0.000003 m²

θ = 13°

Φ = 0.25 × 0.000003 × cos 13°

Φ = 0.0000007308 Wb = (7.308 × 10⁻⁷) Wb

Hope this Helps!!!

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2 years ago

A block of mass m1 = 3.5 kg moves with velocity v1 = 6.3 m/s on a frictionless surface. it collides with block of mass m2 = 1.7

maxonik [38]

First, let's find the speed $v_i$ of the two blocks m1 and m2 sticked together after the collision.
We can use the conservation of momentum to solve this part. Initially, block 2 is stationary, so only block 1 has momentum different from zero, and it is:
$p_i = m_1 v_1$
After the collision, the two blocks stick together and so now they have mass $m_1 +m_2$ and they are moving with speed $v_i$ :
$p_f = (m_1 + m_2)v_i$
For conservation of momentum
$p_i=p_f$
So we can write
$m_1 v_1 = (m_1 +m_2)v_i$
From which we find
$v_i = \frac{m_1 v_1}{m_1+m_2}= \frac{(3.5 kg)(6.3 m/s)}{3.5 kg+1.7 kg}=4.2 m/s$

The two blocks enter the rough path with this velocity, then they are decelerated because of the frictional force $\mu (m_1+m_2)g$ . The work done by the frictional force to stop the two blocks is
$\mu (m_1+m_2)g d$
where d is the distance covered by the two blocks before stopping.
The initial kinetic energy of the two blocks together, just before entering the rough path, is
$\frac{1}{2} (m_1+m_2)v_i^2$
When the two blocks stop, all this kinetic energy is lost, because their velocity becomes zero; for the work-energy theorem, the loss in kinetic energy must be equal to the work done by the frictional force:
$\frac{1}{2} (m_1+m_2)v_i^2 =\mu (m_1+m_2)g d$
From which we can find the value of the coefficient of kinetic friction:
$\mu = \frac{v_i^2}{2gd}= \frac{(4.2 m/s)^2}{2(9.81 m/s^2)(1.85 m)}=0.49$

3 0

2 years ago

La luz pasa del medio A al medio B formando un ángulo de 35° con la frontera horizontal entre ambos. Si el ángulo de refracción

zaharov [31]

Answer:

Índice de refracción entre los dos medios = 1,43

Refractive index between the two media = 1.43

Explanation:

El índice de refracción entre dos medios se explica mejor entendiendo primero la refracción.

Cuando las olas se mueven de un medio a otro, a menudo experimentan un cambio de dirección con respecto al medio en el que viajan.

Por lo tanto, el índice de refracción se expresa como el seno del ángulo de incidencia dividido por el seno del ángulo de refracción.

El seno del ángulo de incidencia y la refracción utilizados en esta fórmula de índice de refracción se miden respectivamente con respecto a la vertical.

En esta pregunta Ángulo de incidencia = 35° a la horizontal = (90° - 35°) a la vertical = 55° a la vertical.

Ángulo de refracción = 35°

Índice de refracción entre los dos medios

= (Sin 55°) ÷ (Sin 35°)

= 0.8192 ÷ 0.5736

= 1.428 = 1.43 a 2 d.p.

¡¡¡Espero que esto ayude!!!

English Translation

The light passes from medium A to medium B at an angle of 35 ° with the horizontal border between the two. If the angle of refraction is also 35 °, what is the relative refractive index between the two media?

Solution

The refractive index between two media is best explained by first understanding refraction.

When waves move from one medium to another, they often experience a change in direction with respect to the medium in which they are travelling.

Hence, refractive index is expressed as the sine of angle of incidence dibided by the sine of angle of refraction.

The sine of angle of incidence and refraction used in this refractive index formula are both respectively measured with respect to the vertical.

In this question,

Angle of incidence = 35° to the horizontal = (90° - 35°) to the vertical = 55° to the vertical.

Angle of refraction = 35°

Refractive index between the two media

= (Sin 55°) ÷ (Sin 35°)

= 0.8192 ÷ 0.5736

= 1.428 = 1.43 to 2 d.p.

Hope this Helps!!!

3 0

2 years ago