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Gnoma [55]
2 years ago
11

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

Physics
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
Read 2 more answers
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