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2 years ago

11

How can you decrease the amount of input force of a wheel and axle?

1 answer:

nasty-shy [4]2 years ago

7 0

Explanation:

hmm by the increasing the size of wheel and decreasing axle

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A man does 4,475 J of work in the process of pushing his 2.50 103 kg truck from rest to a speed of v, over a distance of 26.0 m.

Tcecarenko [31]

Answer:

a) 1.89 m/s b) 172.1 N

Explanation:

a)

Applying the work-energy theorem, if we can neglect the friction between truck and road, the total change in kinetic energy must be equal to the work done by the external forces.
This work, is just 4,475 J.
So we can write the following equation:

$\Delta K = \frac{1}{2} * m*v^{2} = 4,475 J$

where m= mass of the truck = 2.5*10³ kg.
So, we can find the speed v, as follows:

$v =\sqrt{\frac{2*W}{m}} =\sqrt{\frac{2*4,475J}{2.5e3kg} } = 1.89 m/s$

b)

The work done by the man, is just the horizontal force applied, times the displacement produced by the force horizontally:

$W = F*d$

We can solve for F, as follows:

$F = \frac{W}{d} = \frac{4,475 J}{26.0m} = 172.1 N$

4 0

2 years ago

Cuanto cambia la entropía de 0.50 kg de vapor de mercurio [Lv: 2.7 x 10⁵ j/kg ] al calentarse en su punto de ebullición de 357°

lord [1]

Answer:

La entropía del vapor de mercurio cambia en 214.235 joules por Kelvin.

Explanation:

Por definición de entropía ( $S$ ), medida en joules por Kelvin, tenemos la siguiente expresión:

$dS = \frac{\delta Q}{T}$ (1)

Donde:

$Q$ - Ganancia de calor, en joules.

$T$ - Temperatura del sistema, en Kelvin.

Ampliamos (1) por la definición de calor latente:

$dS = \frac{L_{v}}{T}\cdot dm$ (1b)

Donde:

$m$ - Masa del sistema, en kilogramos.

$L_{v}$ - Calor latente de vaporización, en joules

Puesto que no existe cambio en la temperatura durante el proceso de vaporización, transformamos la expresión diferencial en expresión de diferencia, es decir:

$\Delta S = \frac{\Delta m \cdot L_{v}}{T}$

Como vemos, el cambio de la entropía asociada al cambio de fase del mercurio es directamente proporcional a la masa del sistema. Si tenemos que $m = 0.50\,kg$ , $L_{v} = 2.7\times 10^{5}\,\frac{J}{kg}$ and $T = 630.15\,K$ , entonces el cambio de entropía es:

$\Delta S = \frac{(0.50\,kg)\cdot \left(2.7\times 10^{5}\,\frac{J}{kg} \right)}{630.15\,K}$

$\Delta S = 214.235 \,\frac{J}{K}$

La entropía del vapor de mercurio cambia en 214.235 joules por Kelvin.

3 0

3 years ago

A joule is an amount of energy, and a watt is a rate of using energy, defined as 1 W = 1 J / s. How many joules of energy are re

Naily [24]

How many joules of energy are required to run a 100 W light bulb for one day?

<span><span><span>A</span><span>100 </span>joules</span><span><span>B</span>100<span>W </span><span>× </span>24<span>hr </span>joules</span><span><span>C</span>100<span>W </span><span>× </span>24<span>hr </span><span>× </span>60<span>min∕hr </span>joules</span><span><span>D</span>100<span>W </span><span>× </span>24<span>hr </span><span>× </span>60<span>min∕hr </span><span>× </span>60<span>s∕min </span>joules</span></span>

4 0

3 years ago

Una onda tiene una frecuencia de 350 Hz. ¿Cuál es su período?

Alika [10]

Answer:0.00285714285 seconds

Explanation:

period=1 ➗ frequency

Period=1 ➗ 350

Period=0.00285714285 seconds

7 0

3 years ago

A spherical capacitor is formed from two concentric sphericalconducting shells separated by vacuum. The inner sphere has radius1

zubka84 [21]

Explanation:

(1). Formula to calculate the potential difference is as follows.

$\Delta V = -\int E dr$

= $-\int \frac{kq}{r^{2}} dr$

= $\frac{kq}{r_{f}} - \frac{kq}{r_{i}}$

= $\frac{kq(r_{f} - r_{i})}{r_{f}r_{i}}$

= $\frac{9 \times 10^{9} \times 3.30 \times 10^{-9}(0.1 - 0.015)}{0.1 \times 0.015}$

= 38.7 volts

Therefore, magnitude of the potential difference between the two spheres is 38.7 volts.

(2). Now, formula to calculate the energy stored in the capacitor is as follows.

E = $\frac{1}{2}QV$

= $\frac{1}{2} \times 3.30 \times 10^{-9} \times 3.87 V$

= $6.39 \times 10^{-8} J$

Thus, the electric-field energy stored in the capacitor is $6.39 \times 10^{-8} J$ .

7 0

2 years ago

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