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Rufina [12.5K]
4 years ago
5

A watch manufacturer claims that its watches gain or lose no more than 8 seconds in a year. How accurate are these watches, expr

essed as a percentage?
Physics
1 answer:
olchik [2.2K]4 years ago
7 0

Answer:

Accurate at 99.9999746%

Explanation:

First of all, let's calculate how many seconds we have in a year which has 365 days.

1day = 24 × 60 × 60 seconds = 86400 s

Thus, for 365 days, we have;

86400 × 365 = 31536000 seconds

Now, since 1 year has 31536000 seconds, let's find the ratio of efficiency.

We are told that the watch manufacturer claims that its watches gain or lose no more than 8 seconds in a year. Thus, efficiency is given by;

Efficiency = (8/31536000) × 100% = 0.00002536783 %

Thus, it will be be accurate at;

100% - 0.00002536783% ≈ 99.9999746%

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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

i_{2}=4/65\: A

i_{3}=11/65\: A

i_{4}=1/65\: A

i_{5}=3/65\: A

     

Finalmente:

  • La corriente correspondiente a R8 y R9 es 0.23 A
  • La corriente correspondiente a R7 es 0.17 A.

Pudes aprender mas de mallas aquí:

https://brainly.lat/tarea/11593276

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3 years ago
Which process is represented by the PV diagram?
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Hi!


The answer would be A. Isobaric Process


<h3>Explanation:</h3>

Isobaric process is a process where the pressure inside a system remains unchanged. In the Pressure Volume graph given, you can see that the pressure (y axis) remains constant with an increasing volume ( x axis). An example of this would be heating a container with a movable piston. Now, the degree of pressure is dependent on the frequency of collisions of particles inside a system on the walls. If this frequency changes, the pressure changes (proportionally). In our example, heating a container with a movable piston results in the particles inside the container to gain kinetic energy and move faster, meaning an increased frequency of collisions (higher pressure), but at the system time the increase in pressure results in the piston being pushed outwards, causing the volume of the container to increase. This results in decreased frequency of collision of the particles with the walls of the container (lesser pressure). This results in the a zero net effect on the pressure.


Hope this helps!

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Answer:

1) They are electrically neutral

2) They have slightly more weight than protons

Explanation:

The given atomic particle found in the nucleus has the following characteristics;

The location of the particle = The nucleus

The (numbers of the) particle does not contribute to (change) the atomic number of the element

The particles found within the nucleus of an atom are; Neutrons and protons

The particle within the nucleus that determines the atomic number = The number of protons

Therefore, the particle referenced in the question is the neutrons

The two characteristics of the neutron are;

1) The neutrons are neutral, electrically

2) Neutrons have slightly more weight than protons

3) Neutrons are magnetic

4) Neutrons are very small

5) Neutrons consist of three quarks; One 'Up', and two 'Down' quarks

Therefore, two characteristics of the particle are;

1) They are electrically neutral and 2) They are slightly heavier than protons.

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