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3 years ago

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Un montañero de 65kg de masa ha ascendido a la cima del Everest, la montaña más alta del mundo de 8848m de altura sobre el nivel

del mar. Calcula el trabajo que ha realizado para subir los últimos 500m

1 answer:

andrew-mc [135]3 years ago

7 0

Answer:

El trabajo realizado para subir los últimos 500 metros es 318727,5 joules.

Explanation:

Por la definición de trabajo sabemos que el montañero debió contrarrestar trabajo causado por la gravedad terrestre. Si asumimos que el cambio de la altura es muy pequeño en comparación con el radio del planeta (6371 kilómetros vs. 0,5 kilómetros), entonces podemos considerar que la aceleración gravitacional es constante y la ecuación de trabajo ( $\Delta W$ ), medido en joules, que reducida a:

$\Delta W = m\cdot g\cdot \Delta z$ (1)

Donde:

$m$ - Masa del montañero, medido en kilogramos.

$g$ - Aceleración gravitacional, medida en metros por segundo al cuadrado.

$\Delta z$ - Distancia vertical de ascenso del montañero, medida en metros.

Si tenemos que $m = 65\,kg$ , $g = 9,807\,\frac{m}{s^{2}}$ y $\Delta z = 500\,m$ , entonces el trabajo realizado por el montañero para subir es:

$\Delta W = (65\,kg)\cdot \left(9,807\,\frac{m}{s^{2}} \right)\cdot (500\,m)$

$\Delta W = 318727,5\,J$

El trabajo realizado para subir los últimos 500 metros es 318727,5 joules.

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A 60 kg acrobat is in the middle of a 10 m long tightrope. The center of the rope dropped 30 cm in relation to the ends that are

Zigmanuir [339]

Answer:

The tension in each half of the rope, is approximately 4,908.8 N

Explanation:

The mass of the acrobat, m = 60 kg

The length of the rope, l = 10 m

The extent by which the center dropped = 30 cm = 0.3 m

Let, 'T' represent the tension in each half of the rope

Weight, W = Mass, m × The acceleration due to gravity, g

∴ W = m × g

The acceleration due to gravity, g ≈ 9.8 m/s²

∴ The weight of the acrobat, W = 60 kg × 9.8 m/s² ≈ 588 N

The angle the dropped rope makes with the horizontal, θ is given as follows;

θ = arctan((0.3 m)/(5 m)) = arctan(0.06) ≈ 3.434°

At equilibrium, the sum of vertical forces, $\Sigma F_y$ = 0

The vertical component of the tension, $T_y$ , in each half of the rope is given as follows;

$T_y$ = T × sin(θ)

∴ $\Sigma F_y$ = W + T × sin(θ) + T × sin(θ) = W + 2 × T × sin(θ)

Plugging in the values, with θ = arctan(0.06) for accuracy, we get;

588 N + 2 × T × sin(arctan(0.06) = 0

∴ 2 × -T × sin(arctan(0.06) = 588 N

-T= 588 N/(2 × sin(arctan(0.06)) = 4,908.81208 N ≈ 4,908.8 N

The tension in each half of the rope, T ≈ 4,908.8 N.

4 0

3 years ago

BRAINLIESTTT!! HELP ME!!!

likoan [24]

The efficiency of the scissor is 200%.

<u>Explanation:</u>

Efficiency is defined as the ratio of output of any instrument or device or machine to the input supplied to it. So the greater the output the greater will be the efficiency of the device.

As here the work done by us on the system is said to be 10 J so this will be equal to the input work done on the system. And the work done by the system i.e., the scissor is 20 J, so this will be the output work.

So, the efficiency is the ratio of output to input as shown below.

Efficiency = $(\frac {20} {10} ) \times 100$ = 200

So, the efficiency of the scissor is 200%.

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3 years ago

A completely inelastic collision occurs between two balls of wet putty that move directly toward each other along a vertical axi

Marrrta [24]

Answer:

h = 2.64 meters

Explanation:

It is given that,

Mass of one ball, $m_1=3\ kg$

Speed of the first ball, $v_1=20\ m/s$ (upward)

Mass of the other ball, $m_2=2\ kg$

Speed of the other ball, $v_2=-12\ m/s$ (downward)

We know that in an inelastic collision, after the collision, both objects move with one common speed. Let it is given by V. Using the conservation of momentum to find it as :

$V=\dfrac{m_1v_1+m_2v_2}{m_1+m_2}$

$V=\dfrac{3\times 20+2\times (-12)}{3+2}$

V = 7.2 m/s

Let h is the height reached by the combined balls of putty rise above the collision point. Using the conservation of energy as :

$mgh=\dfrac{1}{2}mV^2$

$h=\dfrac{V^2}{2g}$

$h=\dfrac{7.2^2}{2\times 9.8}$

h = 2.64 meters

So, the height reached by the combined mass is 2.64 meters. Hence, this is the required solution.

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3 years ago

An infinitely long straight wire has a uniform linear charge density of Derive the 4. equation for the electric field a distance

marshall27 [118]

Answer:

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Explanation:

Let the linear charge density of the charged wire is given as

$\frac{q}{L} = \lambda$

here we can use Gauss law to find the electric field at a distance r from wire

so here we will assume a Gaussian surface of cylinder shape around the wire

so we have

$\int E. dA = \frac{q}{\epsilon_0}$

here we have

$E \int dA = \frac{\lambda L}{\epsilon_0}$

$E. 2\pi r L = \frac{\lambda L}{\epsilon_0}$

so we have

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

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AleksAgata [21]

Answer:

B.C. D. G.

Explanation:

A vector quantity, has both magnitude and direction. A tip to remember is if you can add a direction to it! You wouldnt say 30 pounds north, but you would say 30 mph north.

<em>I hope this helped! Comment if you have any questions! :)</em>

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3 years ago