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2 years ago

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A wood salvage company is hoisting an old tree trunk off the bottom and out of a lake. The cable from the hoist is tied around t

he log above its center of mass. The hoist applies a force of 9,800 N to the cable to suspend the log in the lake water (FT water), and a force of 29,000 N to suspend the log above the lake surface (FT air). What are the volume and density of the log? Assume the lake water has a density of 1,007 kg/m3.

1 answer:

kupik [55]2 years ago

4 0

To solve this problem it is necessary to resort to the concepts expressed in the Buoyancy Force.

The buoyancy force is given by the equation

$F = \rho Vg$

Where,

$\rho =$ Density

V =Volume

g = Gravitational Acceleration

PART A) From the given data we can find the volume, so

$9800 = 1007*V*9.8$

$V = 0.99m^3$

PART B) The mass can be expressed from the Newton equation in which

$F = mg$

Where,

m = mass

g = Gravitational acceleration

Replacing with our values we have that

$29000 = m*9.8$

$m = 2959.18Kg$

Therefore the Density can be calculated with the ratio between the Volume and Mass

$\rho = \frac{m}{V}$

$\rho = \frac{2959.18Kg}{0.99m^3}$

$\rho = 2989.074kg/m^3$

Therefore the Density of the log is $2989.074kg/m^3$

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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!!!

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An owl has a mass of 4.00 kg. It dives to catch a mouse, losing 800.00 J of its GPE. What was the starting height of the owl, in

vesna_86 [32]

Answer:

height =20m

Explanation:

gpe=mgh

800=4×10×x

40x=800

x=20

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2 years ago

A fisherman casts his bait toward the river at an angle of 25° above the horizontal. As the line unravels, he notices that the b

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Answer:

$v_{o}=18m/s$

Explanation:

According to the exercise we know the angle which the bait was released and its maximum height

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To find the initial y-component of velocity we need to do the following steps:

$v_{y}^{2} =v_{oy}^{2}+2g(y-y_{o})$

At maximum height the y-component of velocity is 0 and from the exercise we know that the initial y position is 0

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$v_{oy}=\sqrt{2(9.8m/s^{2} )(2.9m)} =7.53m/s$

Since the bait is released at 25º

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$v_{o}=\frac{7.53m/s}{sin(25)}=18m/s$

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2 years ago

A gymnast's backflip is considered more difficult to do in the layout (straight body) position than in the tucked position. Why?

spin [16.1K]

Answer:

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Explanation:

A gymnast's backflip is considered more difficult to do in the layout (straight body) position than in the tucked position.

When the body is straight , its moment of rotational inertia is more than the case when he folds his body round. Hence rotational inertia ( moment of inertia x angular velocity ) is also greater. To achieve that inertia , there is need of greater imput of energy in the form of kinetic energy which requires greater effort.

So a gymnast's backflip is considered more difficult to do in the layout (straight body) position than in the tucked position.

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2 years ago

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Answer:

The force on second wheel is twice off the force on first wheel.

Explanation:

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Also, $\tau=Fr$

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It implies,

$Fr=mr^2\alpha \\\\F=mr\alpha$

Here, one has twice the radius of the other. Ratio of forces will be :

$\dfrac{F_1}{F_2}=\dfrac{mr_1\alpha }{mr_2\alpha }\\\\\dfrac{F_1}{F_2}=\dfrac{r_1 }{2r_1}\\\\\dfrac{F_1}{F_2}=\dfrac{1 }{2}\\\\F_2=2F_1$

So, the force on second wheel is twice off the force on first wheel.

7 0

2 years ago