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Olin [163]
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?
Physics
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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Un proyectil es lanzado horizontalmente desde una altura de 12 metro con una velocidad de 80 m/sg. a.Calcular el tiempo de vuelo
Naily [24]

Answer:

t= 1,56 s ,   x= 124,8 m , v = (80 i^ - 15,288 j ) m/s

Explanation:

Este es un ejercicio de lanzamiento proyectiles, comencemos por encontrar el tiempo que tarda en llegar al piso

        y = y₀ + v_{oy} t – ½ g t²

en este caso la altura inicial es y₀= 12 m y llega a y=0 , como es lanzado horizontalmente la velocidad vertical es cero (v_{oy}=0)

       0 = y₀ – ½ g t²

       t= √ (2 y₀/g)

calculemos

       t= √ ( 2 12 / 9,8)

       t= 1,56 s

El alcance del proyectil es la distancia horizontal recorrida  

        x = v₀ₓ t

        x = 80 1,56

        x= 124,8 m

La velocidad de impacto cuando toca el suelo

        vx = v₀ₓ = 80 ms

        v_{y} = v_{oy} – gt

        v_{y} = - 9,8 1,56

        v_{y} = - 15,288 m/s

la velocidad es

       v = (80 i^ - 15,288 j ) m/s

Traducttion  

This is a projectile launching exercise, let's start by finding the time it takes to reach the ground

        y = y₀ + v_{oy} t - ½ g t²

in this case the initial height is i = 12 m and it reaches y = 0, as it is thrown horizontally the vertical speed is zero (v_{oy} = 0)

       0 =y₀I - ½ g t²

       t = √ (2y₀ / g)

let's calculate

       t = √ (2 12 / 9.8)

       t = 1.56 s

Projectile range is the horizontal distance traveled

        x = v₀ₓ t

        x = 80 1.56

        x = 124.8 m

Impact speed when it hits the ground

        vₓ = v₀ₓ = 80 ms

        v_{y} = v_{oy} - gt

        v_{y} = - 9.8 1.56

        v_{y} = - 15,288 m / s

the speed is

       v = (80 i ^ - 15,288 j) m / s

7 0
3 years ago
A 1.120 kg car is traveling with a speed of 40 m/s. find its energy
Aleonysh [2.5K]

Answer:

896 kJ

Explanation:

KInetic Energy = 1/2 m v^2

                         = 1/2 (1120)(40^2) = 896 000 J    or  896 kJ

4 0
2 years ago
A grandfather clock is "losing" time because its pendulum moves too slowly. Assume that the pendulum is a massive bob at the end
IgorC [24]

Answer:

d) shortening the string

Explanation:

Time period of a pendulum clock is dependent on two factors namely:length and acceleration due to gravity.

When a clock loses time, the time period of the pendulum clock increases.

This however can be corrected by decreasing the length of the pendulum.The time period of the pendulum clock is not dependent on the mass of the bob. The time period of the pendulum clock can be corrected only by changing the length of the pendulum string.

5 0
3 years ago
Read 2 more answers
17 copper wires of length l and diameter d are connected in parallel to form a single composite conductor of resistance R. What
Lubov Fominskaja [6]

Answer:

\frac{D}{d} = 4.12

Explanation:

As we know that resistance of one copper wire is given as

r = \rho \frac{L}{a}

here we know that

a = \pi (\frac{d}{2})^2

now we have

r = \rho \frac{L}{\pi (\frac{d^2}{4})}

r = \rho \frac{4L}{\pi d^2}

now we know that such 17 resistors are connected in parallel so we have

R = \frac{r}{17}

R = \rho \frac{4L}{17 \pi d^2}

Now if a single copper wire has same resistance then its diameter is D and it is given as

R = \rho \frac{4L}{\pi D^2}

now from above two equations we have

\rho \frac{4L}{\pi D^2} = \rho \frac{4L}{17 \pi d^2}

D^2 = 17 d^2

now we have

\frac{D}{d} = 4.12

3 0
2 years ago
The gravitational force between two objects depends on the masses and what factor between them
mina [271]
Mass and distance

If mass is doubled, the force of gravity between the objects is doubled
6 0
3 years ago
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