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kkurt [141]
4 years ago
8

A pendulum makes 60 vibrations in 15 secs what’s its frequency

Physics
1 answer:
o-na [289]4 years ago
8 0

Answer:

4 hertz

Explanation:

The defination of freqyency = the total no of cycle made by a wave in one second .

so,

cycle or vibrations=60

tame taken = 15

now,

frequency = no. of cycle / time taken

= 60/15

=4 hertz

hence, the its frequency = 4hertz

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Una placa de cobre a 20°C tiene unas dimensiones de 65cm x 78 cm. Encuentra el área de la placa a 400°C; Coeficiente de dilataci
ValentinkaMS [17]

Answer:

El área de la placa es aproximadamente 5102.752 centímetros cuadrados.

Explanation:

Asumamos que el cambio dimensional como consecuencia de la temperatura es pequeña, entonces podemos estimar el área de la placa de cobre en función de la temperatura mediante la siguiente aproximación:

A_{f} = w\cdot l \cdot [1 + 2\cdot \alpha\cdot (T_{f}-T_{o})] (1)

Donde:

w - Ancho de la placa, en centímetros.

l - Longitud de la placa, en centímetros.

\alpha - Coeficiente de dilatación, en \frac{1}{^{\circ}C}.

T_{o} - Temperatura inicial, en grados Celsius.

T_{f} - Temperatura final, en grados Celsius.

Si sabemos que w = 65\,cm, l = 78\,cm, \alpha = 17\times 10^{-6}\,\frac{1}{^{\circ}C}, T_{o} = 20\,^{\circ}C and T_{f} = 400\,^{\circ}C, entonces el área de la placa a la temperatura final:

A_{f} = (65\,cm)\cdot (78\,cm)\cdot \left[1+\left(17\times 10^{-6}\,\frac{1}{^{\circ}C} \right)\cdot (400\,^{\circ}C-20\,^{\circ}C)\right]

A_{f} = 5102.752\,cm^{2}

El área de la placa es aproximadamente 5102.752 centímetros cuadrados.

4 0
3 years ago
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miss Akunina [59]

Answer:

Option 2

Explanation:

Weight of the box is being acted downwards due to gravity

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3 years ago
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A loop of wire is placed in a magnetic field such that it has a flux, LaTeX: \phi ϕ, through it. The loop is compressed so that
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Answer:

B'= 3.333 B

Explanation:

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