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Mashutka [201]
2 years ago
11

A 40.0 N crate is pulled up a 5.0 m inclined plane. The angle of inclination of the plane to horizontal is 37o. If there is a co

nstant force of friction of 10.0 N between the crate and the surface, what is the net gain in potential energy by the crate? Group of answer choices 120 J -120 J 210 J -210 J
Physics
1 answer:
Digiron [165]2 years ago
4 0

The net gain in potential energy by the crate is 120 J.

<h3>What is potential energy?</h3>

Potential energy is the energy that an object has stored in it as a result of its position or condition. Potential energy exists in a stretched spring, a book held above your head, and a bicycle on top of a hill. The joule, which is represented by the letter "J," is the standard unit for calculating potential energy.

<h3 /><h3>Calculation:</h3>

We use the idea of potential energy, which is given by the equation,

PE = mgh

where,

m = mass of the crate

g = gravitational acceleration

h = height

Given,

mg = 40 N

L = 5 m

angle of inclination, ∅ = 37°

h = L sin∅

h = 5 (sin 37°)

h = 5 (3/5)

h = 3 m

Put the values in the above formula,

PE = mgh

PE = 40 (3)

PE = 120 J

Therefore, the net gain in potential energy by the crate is 120 J.

<h3 />

Learn more about the calculation of work and force here:

brainly.com/question/28043403

#SPJ4

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Answer:

Período del tambor: T = 0.25\,s, fuerza sobre la prenda: F \approx 80.852\,N, velocidad lineal del tambor: v \approx 10.053\,\frac{m}{s}, velocidad angular del tambor: \omega \approx 25.133\,\frac{rad}{s}.

Explanation:

La expresión tiene un error por omisión, su forma correcta queda descrita a continuación:

<em>"Una prenda de 320 gramos de ropa gira en el interior de una lavadora si dicha lavadora tiene un radio de 40 centímetros y gira con una frecuencia de 4 hertz. Halle </em><em>a)</em><em> el período, </em><em>b) </em><em>la velocidad angular, </em><em>c) </em><em>la fuerza con la que gira la prenda y </em><em>d) </em><em>la velocidad lineal de la lavadora."</em>

El tambor gira a velocidad angular constante (\omega), en radianes por segundo, lo cual significa que la prenda experimenta una aceleración centrífuga (a), en metros por segundo al cuadrado. En primer lugar, calculamos el período de rotación del tambor (T), en segundos:

T = \frac{1}{f} (1)

Donde f es la frecuencia, en hertz.

(f = 4\,hz)

T = \frac{1}{4\,hz}

T = 0.25\,s

Ahora determinamos la fuerza aplicada sobre la prenda (F), en newtons:

F = m\cdot a (2)

F = \frac{4\pi^{2}\cdot m \cdot r}{T^{2}} (2b)

Donde:

m - Masa de la prenda, en kilogramos.

r - Radio interior del tambor, en metros.

(m = 0.32\,kg, r = 0.4\,m, T = 0.25\,s)

F = \frac{4\pi^{2}\cdot (0.32\,kg)\cdot (0.4\,m)}{(0.25\,s)^{2}}

F \approx 80.852\,N

La velocidad lineal de la lavadora es:

v = \frac{2\pi\cdot r}{T} (3)

(r = 0.4\,m, T = 0.25\,s)

v = \frac{2\pi\cdot (0.4\,m)}{0.25\,s}

v \approx 10.053\,\frac{m}{s}

Y la velocidad angular del tambor de la lavadora:

\omega = \frac{2\pi}{T}

(T = 0.25\,s)

\omega = \frac{2\pi}{0.25\,s}

\omega \approx 25.133\,\frac{rad}{s}

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The mass of a mass-and-spring system is displaced 10 cm from its equilibrium position and released. A frequency of 4.0 Hz is obs
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Answer:

It will remain same i.e. 4.0 Hz

Explanation:

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Since, the frequency does not depend on the initial displacement of the mass, the frequency would remain the same i.e. 4.0 Hz.

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