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3 years ago

Si un ciclista se mueve a una velocidad de 5 m/s y acelera 1 m/s2, a los 10 segundos su velocidad será

Physics

1 answer:

MrRa [10]3 years ago

5 0

Answer:

A los 10 segundos su velocidad será 15 $\frac{m}{s}$

Explanation:

La aceleración de un objeto es una magnitud que indica cómo cambia la velocidad del objeto en una unidad de tiempo.

En otras palabras, la aceleración relaciona los cambios de la velocidad con el tiempo en el que se producen, es decir que mide cómo de rápidos son los cambios de velocidad:

Una aceleración grande significa que la velocidad cambia rápidamente.
Una aceleración pequeña significa que la velocidad cambia lentamente.
Una aceleración cero significa que la velocidad no cambia.

La aceleración "a" puede ser calculada mediante la expresión:

$a=\frac{vfinal - vinicial}{tiempo}$

En este caso:

a= 1 $\frac{m}{s^{2} }$

vfinal= ?
vinicial= 5 $\frac{m}{s}$
tiempo= 10 s

Reemplazando:

$1\frac{m}{s^{2} }=\frac{vfinal - 5\frac{m}{s} }{10 s}$

Resolviendo se obtiene:

1 $\frac{m}{s^{2} }$ *10 s= vfinal - 5 $\frac{m}{s}$

10 $\frac{m}{s}$ = vfinal - 5 $\frac{m}{s}$

10 $\frac{m}{s}$ + 5 $\frac{m}{s}$ = vfinal

15 $\frac{m}{s}$ = vfinal

A los 10 segundos su velocidad será 15 $\frac{m}{s}$

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In a machining operation that approximates orthogonal cutting, the cutting tool has a rake angle =100 . The chip thickness ratio

Rudik [331]

Answer:

The percentage of the total energy dissipated into shear plane is 89.46%.

Explanation:

Given that,

Rake angle = 10°

Thickness ratio= 0.5

Cutting Force = 400 N

Thrust force = 200 N

Speed =3 m/s

Suppose the shear force is 345.21 N.

We need to calculate the shear plane angle

Using formula shear angle

$\tan\phi=\dfrac{r\cos\alpha}{1-r\sin\alpha}$

Put the value in to the formula

$\tan\phi=\dfrac{0.5\cos10}{1-0.5\sin10}$

$\tan\phi=0.539$

$\phi=\tan^{-1}(0.539)$

$\phi=28.32^{\circ}$

We need to calculate the shear velocity

Using formula of shear velocity

$v_{2}=\dfrac{v\cos\alpha}{\cos(\phi-\alpha)}$

Put the value into the formula

$v_{2}=\dfrac{3\times\cos10}{\cos(28.32-10)}$

$v_{2}=3.11\ m/s$

We need to calculate the percentage of the total energy dissipated into shear plane

Using formula of energy dissipated

$\%d=\dfrac{P_{s}}{P}\times100$

$\%d=\dfrac{F_{s}\times v_{c}}{F_{c}\times v}\times100$

Put the value into the formula

$\%d=\dfrac{345.21\times3.11}{400\times3}\times100$

$d=89.46\%$

Hence, The percentage of the total energy dissipated into shear plane is 89.46%.

6 0

3 years ago

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SIZIF [17.4K]

The correct answer is the Sun.i hope that helped! if you have any questions or concerns about the answer i gave you please let me know!!

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4 years ago

Limestone cave can develop when Limestone rock is weathered. The weathering of the rock leaves an empty space that forms the cav

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Answer:

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Explanation:

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que 2. Why do we keep frequency constant instead of keeping vibrating length constam second law of vibrating string?

ella [17]

Answer:

The second law of a vibrating string states that for a transverse vibration in a stretched string, the frequency is directly proportional to the square root of the string's tension, when the vibrating string's mass per unit length and the vibrating length are kept constant

The law can be expressed mathematically as follows;

$f = \dfrac{1}{2\cdot l} \cdot \sqrt{\dfrac{T}{m} }$

The second law of the vibrating string can be verified directly, however, the third law of the vibrating string states that frequency is inversely proportional to the square root of the mass per unit length cannot be directly verified due to the lack of continuous variation in both the frequency, 'f', and the mass, 'm', simultaneously

Therefore, the law is verified indirectly, by rearranging the above equation as follows;

$m = \dfrac{1}{ l^2} \cdot \dfrac{T}{4\cdot f^2} }$

From which it can be shown that the following relation holds with the limits of error in the experiment

m₁·l₁² = m₂·l₂² = m₃·l₃² = m₄·l₄² = m₅·l₅²

Explanation:

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What does the slope of a velocity versus time graph represent?

liberstina [14]

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