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3 years ago

A wave x meters long has a speed of y meters per second. The frequency of the wave is

1 answer:

6 0

The answer is B. y/x hertz.

Frequency of a wave is defined as the ratio between the speed and wavelength of the wave. The unit of frequency is hertz (Hz) which is also equivalent to s^-1 or "per second". Following the units of wavelength (m) and speed (m/s), it can also be inferred that frequency is equal to (m/s)/m or y/x.

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What is the specific heat of the masses in this experiment? Infer the substance the masses are made of and explain your inferenc

Liono4ka [1.6K]

The metal whose specific heat capacity is close to the obtained value is aluminum.

<h3>What is specific heat capacity</h3>

The specific heat capacity of an object is the heat required to raise a unit mass of the substance by 1 kelvin.

$Q= mc\Delta \theta$

where;

c is the specific heat capacity
Δθ is change in temperature

Let the mass of the water = 50 g

mass of the metal for this first trial = 50 g

The heat gained by the water is calculated as follows

$Q = 50 \times 4.184 \times 8.4\\\\Q = 1757.28 \ J$

Specific heat capacity of the metal for the first trial is calculated as follows;

Heat gained by water = Heat lost by metal

$C = \frac{Q}{m\Delta T} = \frac{1757.28}{50\times 8.4} = 4.184 \ J/g^oC$

Specific heat capacity of the metal for the second trial;

mass of metal = 200 - mass of water = 150 g

$C_2 = \frac{1757.28}{150 \times 15.2} = 0.77 \ J/g^oC$

Specific heat capacity of the metal for the third trial;

$C_3 = \frac{1757.28}{250 \times 20.8} = 0.34\ J/g^oC$

Specific heat capacity of the metal for the fourth trial;

$C_4 = \frac{1757.28}{350 \times 25.4} = 0.19\ J/g^oC$

Specific heat capacity of the metal for the fifth trial;

$C_5 = \frac{1757.28}{450 \times 29.6} = 0.13\ J/g^oC$

Average specific heat capacity

$C = \frac{4.184 + 0.77 + 0.34+ 0.19 + 0.13 }{5} = 1.12 \ J/g^oC = 1120 J/kg^oC$

The metal whose specific heat capacity is close to the above value is aluminum.

Learn more about specific heat capacity here: brainly.com/question/16559442

8 0

2 years ago

A single-turn circular loop of wire that has a radius of 3.5 cm lies in the plane perpendicular to a spatially uniform magnetic

Lena [83]

Explanation:

Given that,

Radius of the circular loop, r = 3.5 cm = 0.035 m

(a) During a 0.12-s time interval, the magnitude of the field increases uniformly from 0.2 T to 0.5 T. Due to the change in the magnetic field, an emf will induced in it. The magnitude of induced emf is given by :

$\epsilon=-\dfrac{d(BA)}{dt}\\\\\epsilon=-\pi r^2\dfrac{B_f-B_i}{dt}\\\\\epsilon=-\pi (0.035)^2\dfrac{0.5-0.2}{0.12}\\\\\epsilon=-9.62\times 10^{-3}\ volts$

So, the magnitude of the emf induced in the loop during the time interval is $9.62\times 10^{-3}\ V$ .

(b) The negative sign shows that the direction of induced emf in the loop is in anitclockwise direction.

5 0

3 years ago

A child looks at his reflection in a spherical Christmas tree ornament 8.0 cm in diameter in season that the image of his face i

malfutka [58]

From the information given,

diameter of ornament = 8

radius = diameter/2 = 8/2

radius of curvature, r = 4

Recall,

focal length, f = radius of curvature/2 = 4/2

f = 2

Recall,

magnification = image d

8 0

2 years ago

Human activities can have an impact on Earth's systems. Which Earth system is most affected when factories burn harmful chemical

IgorC [24]

Answer:

Hey

Your answer would be Earth's atmospheric system

Over time of cousre it will effect Earth's intire surface.

DON'T DRIVE MORE THAT YOU HAVE TO, SAVE THE EARTH!

7 0

3 years ago

Determine la resistencia equivalente de la "escalera" de

defon

Las respuestas a cada inciso son:

a) La resistencia equivalente de la "escalera" de resistores iguales de 125 Ω que se muestra en la figura adjunta es:

$R_{t} = 341.7 \: \Omega$

b) La corriente a través de cada uno de los tres resistores de la izquierda si se conecta una batería de 50.0 V entre los puntos A y B es:

Correspondiente a R8 y R9 es 0.23 A
Correspondiente a R7 es 0.17 A.

a) En la imagen adjuntada correspondiente a la Figura 26-40, podemos observar que las resistencias 1, 2 y 3 están en serie, por lo tanto la ressitencia equivalente entre estas 3 es:

$R' = R_{1} + R_{2} + R_{3}$

De aquí en adelante tendremos presente que las todas las resistencias son iguales entre sí y por ende igual a 125 Ω. Las notaciones del 1 al 9 son para poder mostrar la resolución del problema.

Entonces:

$R' = 3R$

Ahora, esta resistencia está en paralelo con la resistencia R₄, por lo tanto la resistencia equivalente entre estas dos es:

$\frac{1}{R''} = \frac{1}{R'} + \frac{1}{R_{4}} = \frac{1}{3R} + \frac{1}{R} = \frac{4R}{3R^{2}}$

$R'' = \frac{3}{4}R$

Luego, esta resistencia está en serie con las resistencias R₅ y R₆, por lo tanto:

$R''' = R'' + R_{5} + R_{6} = \frac{3}{4}R + 2R = \frac{11}{4}R$

Esta resistencia está ahora en paralelo con R₇, entonces:

$\frac{1}{R''''} = \frac{1}{R'''} + \frac{1}{R_{7}} = \frac{4}{11R} + \frac{1}{R} = \frac{15R}{11R^{2}}$

$R'''' = \frac{11}{15}R$

Finalmente, esta resistencis está en serie con las resistencias R₈ y R₉, por lo tanto la resistencia total es:

$R_{t} = R'''' + R_{8} + R_{9} = \frac{11}{15}R + 2R = \frac{41}{15}R = \frac{41}{15}*125 \: \Omega = 341.7 \: \Omega$

b) Para este inciso debemos usar la Ley de Kirchhoff, pues tenemos tres mallas. Supondremos que las corriente de cada malla fluiran en sentido horario, por lo tanto las ecuaciones de para cada malla serán:

Malla 1

Segun la ley de Ohm tenemos:

$V-i_{1}R-i_{3}R=0$ (1)

Malla 2

$-i_{2}R-i_{5}R-i_{2}R+i_{3}R=0$ (2)

Malla 3

$-i_{4}R-i_{4}R-i_{4}R+i_{5}R=0$ (3)

Recordemos tambien que:

$i_{1}=i_{2}+i_{3}$ (4)

$i_{2}=i_{4}+i_{5}$ (5)

Lo que debemos hacer ahora es resolver el sistema de ecuaciones y encontrar los valore de las corrientes. Por lo tanto, los valores de las corrientes serán:

$i_{1}=3/13\: A$