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3 years ago

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When resting, a person has a metabolic rate of about 3.0X10^5 joules per hour. The person is submered neck deep into a tub conta

ining 1.2X10^3 kg of water at 21.00 degrees C. If the heat from the person only goes into the water, find the water temperature after half an hour.
Given : Specific heat of water at 15 degrees C. 4186 j/kgc

1 answer:

geniusboy [140]3 years ago

6 0

Answer:

21.02986 °C

Explanation:

c = Specific heat of water at 15 °C = 4186 J/Kg°C

$\Delta T$ = Change in temperature

$T_1$ = Initial temperature = 21 °C

$T_2$ = Final temperature

m = Mass of person = $1.2\times 10^3\ kg$

t = Time taken = 30 minutes

Heat is given by

$Qt=mc\Delta T\\\Rightarrow Qt=mc(T_2-T_1)\\\Rightarrow \frac{3\times 10^5}{60}\times 30=1.2\times 10^3\times 4186(T_2-21)\\\Rightarrow T_2-21=\frac{3\times 10^5\times 30}{60\times 1.2\times 10^3\times 4186}\\\Rightarrow T_2-21=0.02986\\\Rightarrow T_2=0.02986+21\\\Rightarrow T_2=21.02986\ ^{\circ}C$

The water temperature after half an hour is 21.02986 °C

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The ray diagram for the given object consists of 2 cm height of object, 4 cm object distance and 3 cm focal length.

<h3>Image formed by a diverging lens</h3>

Diverging lens is called a concave lens. The working of the lens is dependent on the refraction of the light rays as they pass through the lens.

Image formed by a diverging lens is always virtual, erect and diminished; smaller than the object and is located on the same side of the lens as the object.

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2 years ago

In lab, your instructor generates a standing wave using a thin string of length L = 1.65 m fixed at both ends. You are told that

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Answer:

On the standing waves on a string, the first antinode is one-fourth of a wavelength away from the end. This means

$\frac{\lambda}{4} = 0.275~m\\\lambda = 1.1~m$

This means that the relation between the wavelength and the length of the string is

$3\lambda/2 = L$

By definition, this standing wave is at the third harmonic, n = 3.

Furthermore, the standing wave equation is as follows:

$y(x,t) = (A\sin(kx))\sin(\omega t) = A\sin(\frac{\omega}{v}x)\sin(\omega t) = A\sin(\frac{2\pi f}{v}x)\sin(2\pi ft) = A\sin(\frac{2\pi}{\lambda}x)\sin(\frac{2\pi v}{\lambda}t) = (2.45\times 10^{-3})\sin(5.7x)\sin(59.94t)$

The bead is placed on x = 0.138 m. The maximum velocity is where the derivative of the velocity function equals to zero.

$v_y(x,t) = \frac{dy(x,t)}{dt} = \omega A\sin(kx)\cos(\omega t)\\a_y(x,t) = \frac{dv(x,t)}{dt} = -\omega^2A\sin(kx)\sin(\omega t)$

$a_y(x,t) = -(59.94)^2(2.45\times 10^{-3})\sin((5.7)(0.138))\sin(59.94t) = 0$

For this equation to be equal to zero, sin(59.94t) = 0. So,

$59.94t = \pi\\t = \pi/59.94 = 0.0524~s$

This is the time when the velocity is maximum. So, the maximum velocity can be found by plugging this time into the velocity function:

$v_y(x=0.138,t=0.0524) = (59.94)(2.45\times 10^{-3})\sin((5.7)(0.138))\cos((59.94)(0.0524)) = 0.002~m/s$

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3 years ago

Calcula el valor de la velocidad de las ondas sonoras en el agua sabiendo que su

dybincka [34]

La velocidad de las ondas sonoras es aproximadamente 1469,694 metros por segundo.
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1) Inicialmente, debemos determinar la velocidad de las ondas sonoras a través del agua ( $v$ ), en metros por segundo:

$v = \sqrt{\frac{K}{\rho} }$ (1)

Donde:

$K$ - Módulo de compresibilidad, en newtons por metro cuadrado.
$\rho$ - Densidad del agua, en kilogramos por metro cúbico.

Si sabemos que $\rho = 1\times 10^{3}\,\frac{kg}{m^{3}}$ y $K = 2,16\times 10^{9}\,\frac{N}{m^{2}}$ , entonces la velocidad de las ondas sonoras es:

$v = \sqrt{\frac{2,16\times 10^{9}\,\frac{N}{m^{2}}}{1\times 10^{3}\,\frac{kg}{m^{3}} } }$

$v\approx 1469,694\,\frac{m}{s}$

La velocidad de las ondas sonoras es aproximadamente 1469,694 metros por segundo.

2) Luego, determinamos la longitud de onda ( $\lambda$ ), en metros, mediante la siguiente fórmula:

$\lambda = \frac{v}{f}$ (2)

Donde $f$ es la frecuencia de las ondas sonoras, en hertz.

Si sabemos que $v\approx 1469,694\,\frac{m}{s}$ y $f = 1000\,hz$ , entonces la longitud de onda de las ondas sonoras es:

$\lambda = \frac{1469,694\,\frac{m}{s} }{1000\,hz}$

$\lambda = 1,470\,m$

La longitud de onda de las ondas sonoras es 1,470 metros.

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2 years ago

A) an electron has an initial speed of 226000 m/s. if it undergoes an acceleration of 4.0 x 1014 m/s2, how long will it take to

KIM [24]

initial speed of 226000 m/s

acceleration of 4.0 x 1014 m/s2,

speed of 781000 m/s

What is Acceleration?

Acceleration is a rate of change of velocity with respect to time with respect to direction and speed.

A point or an object moving in a straight line is accelerated if it speeds up or slows down.

Acceleration formula can be written as,

a = (v - u ) / t m/s²

As we have to find the time taken, the formula can be altered as,

$t = \frac{v-u}{a}$

where, t - time taken to reach a final speed

v - final velocity

u - initial velocity

a - acceleration.

Substituting all the given values,

$t =\frac{781000 - 226000} {4* 1014}$

= 1.3875 × 10⁻⁹ seconds.

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1 year ago

1.Predict the frequency of a tuning fork that emits a sound with a wavelength of 0.385 m.

lina2011 [118]

Answer:

1.) Frequency F = 890.9 Hz

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Explanation:

1.) Given that the wavelength = 0.385m

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To predict the frequency, let us use the formula V = F λ

Where (λ) = wavelength = 0.385m

343 = F × 0.385

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F = 890.9 Hz

2.) Given that the frequency = 384Hz

Using the formula again

V = F λ

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3 years ago