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Dimas [21]
3 years ago
6

A cart moving at 2.7 m/s travels for 2 minutes, How far did it go?​

Physics
1 answer:
JulsSmile [24]3 years ago
7 0

Answer: 324m

Explanation:

Speed (v) = 2.7m/s

t= 2 min

Convert time in minutes to seconds

t = 2 x 60 = 120s

Distance (s) = ?

Speed(v)= distance(s)/time(t)

v = s / t

Distance(s) = speed x time

Distance(s) = 2.7 x 120

Distance(s) = 324m

The distance is 324m

 

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A long cylindrical insulating shell has an inner radius of a = 1.41 m and an outer radius of b = 1.67 m. The shell has a constan
Natasha2012 [34]

Answer:

a. E = 122.4 N/C

b. E = 58.2 N/C

c. E = 0

Explanation:

The electric field at an arbitrary point away from the axis of the cylinder can found by applying Gauss’ Law, which states that an electric flux through a closed surface is equal to the total charge enclosed by this surface divided by electric permittivity.

In order to apply this law, we have to draw an imaginary cylindrical surface of arbitrary height ‘h’ and radius ‘r’, which is equal to the point where the E-field is asked.

A. For the outside of the cylinder, we will draw our imaginary surface with r = 1.97.

E2\pi rh = \frac{\lambda V}{\epsilon_0} = \frac{\lambda \pi (b^2 - a^2)h}{\epsilon_0}\\E2\pi (1.97)h = \frac{(5.3\times 10^{-9})\pi(1.67^2 - 1.41^2)h}{\epsilon_0}\\E = 122.4~N/C

B. This time our imaginary surface should be inside the cylinder, therefore the enclosed charge will be less than that of part A.

E2\pi rh = \frac{\lambda V_{enc}}{\epsilon_0} = \frac{\lambda \pi (r^2 - a^2}h{\epsilon_0}\\E2\pi (1.51)h = \frac{5.3\times 10^{-9})\pi(1.51^2 - 1.41^2)h}{\epsilon_0}\\E = 58.2~N/C

C. In this case our imaginary surface will be inside the cylinder, where there is no charge at all. Therefore, the enclosed charge will be zero and the electric field will be zero.

6 0
3 years ago
Una bola de 1 kg gira alrededor de un circulovrtical en el extremo de un cuerda. El otro extremo de la cuerda esta fijo en el ce
Vladimir79 [104]

Answer:

La diferencia entre las tensiones máxima y mínima es de 19.614 newtons.

Explanation:

Puesto que la bola gira en un círculo vertical, existe claramente una diferencia entre las tensiones debido a la influencia de la gravedad y la tensión que resulta de la aceleración centrípeta experimentada por la masa. La máxima tensión ocurre cuando la bola se encuentra en el nadir (o la sima) del trayecto circular, la cual se describe por la Segunda Ley de Newton:

T_{max} - m\cdot g = m\cdot \frac{v^{2}}{L}

En cambio, la mínima tensión aparece cuando la bola se encuentra en el cénit (o la cima) del trayecto circular, descrita por la misma ley de Newton:

T_{min} + m\cdot g = m\cdot \frac{v^{2}}{L}

Donde:

T_{min}, T_{max} - Tensiones mínima y máxima, medidas en newtons.

m - Masa de la bola, medida en kilogramos.

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

L - Distancia con respecto al eje de rotación, medida en metros.

v - Rapidez tangencial, medido en metros por segundo.

Se elimina la aceleración centrípeta de ambas expresiones por igualación:

T_{min} + m\cdot g = T_{max} - m\cdot g

Ahora, la diferencia entre las tensiones máxima y mínima es:

T_{max} - T_{min} = 2\cdot m \cdot g

Si m = 1\,kg y g = 9.807\,\frac{m}{s^{2}}, entonces:

T_{max} - T_{min} = 2\cdot (1\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)

T_{max}-T_{min} = 19.614\,N

La diferencia entre las tensiones máxima y mínima es de 19.614 newtons.

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3 years ago
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