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3 years ago

7

A 5 kg bucket is lifted from the ground into the air by pulling in 10 meters of rope with linear density of 2 kg/m at a constant

speed. The bucket starts with 100 cm^3 of water and leaks at a constant rate. It finishes draining just as it reaches the 10 meter mark. How much work was spent lifting the bucket, rope, and water?

1 answer:

iren [92.7K]3 years ago

7 0

Answer:

Workdone W = 1465.1 J

Explanation:

The weight of the water = density × volume

weight of the water = 1000 kg/m³ × 100 cm³

weight of the water = 1000 kg/m³ × 0.0001 m³

weight of the water = 0.1 kg

weight of the bucket = 5 kg

weight of the rope = $2*10- 2x\\\\$

= $20- 2x$

Leakage = $\frac{0.1}{10} * x$

= $0.01 x$

Total weight = $5 + 20 - 2x - 0.01 x$

= $25 - 2.01 x$

Force = wg

Force = ( $25 - 2.01 x$ )g

Force = 9.8 ( $25 - 2.01 x$ )

Finally; the amount of work spent in lifting the bucket, rope, and water is calculated as follows:

$W = \int\limits^{10}_0 {(25-2.01 x)} \, 9.8 dx \\\\W = 9.8 (25x - \frac{2.01x^2}{2}})^{10}__0\\\\$

$W = 9.8 (25*10-2.01*50)\\\\W = 1465.1 \ \ J$

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NASA is giving serious consideration to the concept of solar sailing. A solar sailcraft uses a large, low mass sail and the ener

Stella [2.4K]

Answer:

Reflective

Explanation:

The radiation pressure of the wave that totally absorbed is given by;

$P_{abs}= \frac{I}{C}$

and While the radiation pressure of the wave totally reflected is given by;

$P_{ref}= \frac{2I}{C}$

Now compare the two-equation you can clearly see that the pressure due to reflection is larger than absorption therefore the sail should be reflective.

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3 years ago

What is the velocity of an object that has been in free fall for 1.5s?

Crank

Answer:

D. 15 m/s downward

Explanation:

v = at + v₀

v = (-9.8 m/s²) (1.5 s) + (0 m/s)

v = -14.7 m/s

Rounded to two significant figures, the answer is D, 15 m/s downward.

8 0

3 years ago

An inflatable raft (unoccupied) floats down a river at an approximately constant speed of 5.6 m/s. A child on a bridge, 71 m abo

saveliy_v [14]

Answer:

21.28 m

Explanation:

height, h = 71 m

velocity of raft, v = 5.6 m/s

let the time taken by the stone to reach to raft is t.

use second equation of motion for stone

$h = ut + \frac{1}{2}at^{2}$

u = 0 m/s, h = 71 m, g = 9.8 m/s^2

71 = 0 + 0.5 x 9.8 x t^2

t = 3.8 s

Horizontal distance traveled by the raft in time t

d = v x t = 5.6 x 3.8 = 21.28 m

3 0

3 years ago

Two ships of equal mass are 110 m apart. What is the acceleration of either ship due to the gravitational attraction of the othe

dusya [7]

Answer:

Acceleration of the ship, $a=2.14\times 10^{-7}\ m/s^2$

Explanation:

It is given that,

Mass of both ships, $m=39000\ metric\ tons=39\times 10^6\ kg$

Distance between two ships, d = 110 m

The gravitational force between two ships is given by :

$F=G\dfrac{m^2}{d^2}$

$F=6.67\times 10^{-11}\ Nm^2/kg^2\times \dfrac{(39\times 10^6\ kg)^2}{(110\ m)^2}$

F = 8.38 N

Let a is the acceleration. Now, using second law of motion as :

$a=\dfrac{F}{m}$

$a=\dfrac{8.38\ N}{39\times 10^6\ kg}$

$a=2.14\times 10^{-7}\ m/s^2$

So, the acceleration of either ship due to the gravitational attraction of the other is $2.14\times 10^{-7}\ m/s^2$ . Hence, this is the required solution.

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3 years ago

A polarized light that has an intensity I0 = 60.0 W/m² is incident on three polarizing disks whose planes are parallel and cente

nikitadnepr [17]

Answer:

The transmitted intensity through all polarizers is $I_3 =41.31 W/m^2$

Explanation:

According to Malu's law the intensity of a polarized light having an initial intensity $I_0$ is mathematically represented as

$I = I_0cos^2 \theta$

Now considering the polarizer(The polarizing disk) the equation above becomes

$I = I_0 (cos^2 \theta)^n$

Where n is the number of polarizers

Substituting $60.0W/m^2$ for the initial intensity 3 for the n and 20° for the angle of rotation

$I_3 = 60 (cos^220)^3$

$=41.31 W/m^2$

6 0

3 years ago