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3 years ago

Cuestionario:

Physics

1 answer:

Sati [7]3 years ago

3 0

Explanation:

1.El objetivo principal de realizar la llamada entrada en calor o calentamiento, es preparar el cuerpo para la actividad física o deportiva. Numerosas lesiones y ciertos problemas cardíacos como algunas arritmias, están asociados a la ejercitación violenta sin mediar un adecuado calentamiento.

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What is the atomic mass of oxygen?

lianna [129]

Answer:

$atomic \: mass \: of \: oxygen = 15.99aproximtly = 16 \\ thank \: you$

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2 years ago

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A 1000.kg car moving at north at 100. km/hr brakes to a stop in 50. m within a uniform acceleration. what are magnitude and dire

Marianna [84]

Answer:

Force=7.71KN in the opposite direction

Explanation:

Given the mass of the car(M)=1000Kg

The initial speed of the car(u)=100km/hr

we know that 1km/hr=5/18 m/sec

100km/hr=100*5/18 m/sec=27.77m/sec

the distance travelled before it stops (s)=50m

let the acceleration of the car be a

The final velocity of the car is 0.

we know that $v^{2}-u^{2}=2as$

$0^{2}-27.77^{2}=2(-a)(50)$

a=7.712m/ $sec^{2}$

We know that F=Ma

F=1000* 7.712=7712N=7.71KN

6 0

3 years ago

If the phase of the vibrating sources was changed so that they were vibrating completely out of phase, what effect would this ha

Over [174]

Answer:

There would be complete destructive interference.

Explanation:

This is because since the waves are completely out of phase, the phase difference is half wavelength, that is the phase angle is 180°. The vibrating sources are 180° out of phase with each other.

Since this is the case, the crest of the one source meets the trough of the other, this causes the resultant vibrational wave to cancel out, thus producing a destructive interference pattern.

Since the vibrating sources are completely out of phase, every point they meet is completely out of phase, so the resultant interference pattern would produce a complete destructive interference pattern of no wave.

4 0

3 years ago

Una placa de cobre a 20°C tiene unas dimensiones de 65cm x 78 cm. Encuentra el área de la placa a 400°C; Coeficiente de dilataci

ValentinkaMS [17]

Answer:

El área de la placa es aproximadamente 5102.752 centímetros cuadrados.

Explanation:

Asumamos que el cambio dimensional como consecuencia de la temperatura es pequeña, entonces podemos estimar el área de la placa de cobre en función de la temperatura mediante la siguiente aproximación:

$A_{f} = w\cdot l \cdot [1 + 2\cdot \alpha\cdot (T_{f}-T_{o})]$ (1)

Donde:

$w$ - Ancho de la placa, en centímetros.

$l$ - Longitud de la placa, en centímetros.

$\alpha$ - Coeficiente de dilatación, en $\frac{1}{^{\circ}C}$ .

$T_{o}$ - Temperatura inicial, en grados Celsius.

$T_{f}$ - Temperatura final, en grados Celsius.

Si sabemos que $w = 65\,cm$ , $l = 78\,cm$ , $\alpha = 17\times 10^{-6}\,\frac{1}{^{\circ}C}$ , $T_{o} = 20\,^{\circ}C$ and $T_{f} = 400\,^{\circ}C$ , entonces el área de la placa a la temperatura final: