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4 years ago

A wave with a period of 1⁄3 second has a frequency of A. 67 Hz. B. 33 Hz. C. 300 Hz. D. 3 Hz.

1 answer:

3 0

A wave with a period of 1⁄3 second has a frequency of D. 3 Hz. To calculate this we will use the formula that represents the correlation between a frequency (f) and a time period (T): T = 1/f. Or: f = 1/T. The unit for the time period is second "s" while the unit for frequency is Hertz "Hz" (=1/s). We know that T = 1/3 s. That means that f = 1/(1/3s) = 3 1/s = 3 Hz.

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How do gases respond to changes in pressure and temperature?

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They compress or expand depending on amount of pressure or depending on the temperature

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3 years ago

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A 1kg ball is dropped (from rest) 100m onto a spring with spring constant 125N/m. How much does the spring compress?

Anastasy [175]

Answer:

3.95 m

Explanation:

m = 1 kg, h = 100 m, k = 125 N/m

Let the spring is compressed by y.

Use the conservation of energy

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m x g x h = 1/2 x ky^2

1 x 9.8 x 100 = 0.5 x 125 x y^2

y^2 = 15.68

y = 3.95 m

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3 years ago

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3 years ago

En un experimento de calorimetría, 0.50 kg de un metal a 100°C se añaden a 0.50 kg de agua a 20°C en un vaso de calorímetro de a

Maru [420]

Answer:

c=0.14J/gC

Explanation:

2) The specific heat will be the same because it is a property of the substance and does not depend on the medium.

We can use the expression for heat transmission

$Q=mc(T_2-T_1)$

In this case the heat given by the metal (which is at a higher temperature) is equal to that gained by the water, that is to say

$Q_1=-Q_2$

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c = 4.18J / g ° C

replacing we have

$c_{metal}*(500g)(100\°C-25\°C)=-(250g)(4.18\frac{J}{g\°C})(20\°C-25\°C)\\c_{metal}=0.14\frac{J}{g\°C}$

I hope this is useful for you

2) El calor específico será igual porque es una propiedad de la sustancia y no depende del medio.

Podemos usar la expresión para la transmisión de calor

$Q=mc(T_2-T_1)$

En este caso el calor cedido por el metal (que está a mayor temperatura) es igual al ganado por el agua, es decir

$Q_1=-Q_2$

para el agua tenemos que

c=4.18J/g°C

reemplazando tenemos

$c_{metal}*(500g)(100\°C-25\°C)=-(250g)(4.18\frac{J}{g\°C})(20\°C-25\°C)\\c_{metal}=0.14\frac{J}{g\°C}$

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3 years ago