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Bess [88]
3 years ago
15

Does someone know how to do math with that equation

Physics
1 answer:
mestny [16]3 years ago
8 0

Answer:

f = 5 cm

Explanation:

using the thin lens equation, given as follows:

\frac{1}{f} = \frac{1}{d_{o}}+\frac{1}{d_{i}}\\

where,

f = focal length = ?

do = the distance of object from lens = 20 cm

di = the distance of image from lens = 6.6667 cm

Therefore,

\frac{1}{f} = \frac{1}{20\ cm}+\frac{1}{6.6667\ cm}\\\\\frac{1}{f} =  0.199999\ cm^{-1}\\\\f = \frac{1}{0.199999\ cm^{-1}}\\\\

<u>f = 5 cm</u>

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A race car moves along a circular track at a speed of 0.512 m/s. If the car's centripetal acceleration is 15.4 m/s2, what is the
tiny-mole [99]

Answer:

The radius is  r = 0.0170 \ m  

Explanation:

From the question we are told that

      The speed at which the race car moves is v  =  0.512 \ m/s

       The centripetal acceleration is  a _r  =  15.4 \ m/s

Generally the centripetal acceleration is mathematically represented as

           a_r =  \frac{v^2 }{r}

=>        15.4  =  \frac{0.512^2 }{ r}

=>       r = 0.0170 \ m  

 

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3 years ago
An object with 5 protons, 5 electrons and 6 neutrons would have what charge
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Answer:

0

Explanation:

The object would be neutral. There are equal numbers of protons and electrons, so the positive and negative charges cancel one another.

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4 years ago
Most ___________________ are located on the ocean floor.
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Answer:

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Explanation:

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4 0
3 years ago
The process of evaluating evidence to form an attitude is called
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5 0
3 years ago
AYUDA!!!!!
ZanzabumX [31]

Answer:

a) El valor de la densidad es 0.79 \frac{g}{cm^{3} } o 790 \frac{g}{cm^{3} }

b) El peso especifico es 7749.9\frac{N}{m^{3} }

Explanation:

a) La densidad se define como la propiedad que tiene la materia, ya sean sólidos, líquidos o gases, para comprimirse en un espacio determinado. En otras palabras, la densidad es una magnitud que permite medir la cantidad de masa que hay en determinado volumen de una sustancia. Entonces, la expresión para el cálculo de la densidad es el cociente entre la masa de un cuerpo y el volumen que ocupa:

d=\frac{m}{V}

En este caso:

  • masa= 237 g= 0,237 kg (siendo 1000 g= 1 kg)
  • volumen= 300 cm³= 0,0003 m³ (siendo 1 cm³= 0,000001 m³)

Reemplazando:

d=\frac{237 g}{300cm^{3} } →  d=0.79 \frac{g}{cm^{3} }

d=\frac{0,237 kg}{0,0003m^{3} }→  d=790 \frac{g}{cm^{3} }

<u><em>El valor de la densidad es 0.79 </em></u>\frac{g}{cm^{3} }<u><em> o 790 </em></u>\frac{g}{cm^{3} }<u><em></em></u>

b) El peso específico es la relación existente entre el peso y el volumen que ocupa una sustancia en el espacio.

Entonces, en este caso, siendo el peso:

P= m*g= 0,237 kg* 9,81 \frac{m}{s^{2} }= 2,32497 N

el peso especifico es calculado como:

Pe=\frac{Peso}{Volumen}= \frac{2,32497N}{0,0003 m^{3} }

Pe= 7749.9\frac{N}{m^{3} }

<u><em>El peso especifico es 7749.9</em></u>\frac{N}{m^{3} }<u><em></em></u>

8 0
3 years ago
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