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3 years ago

What are two ways that convention currents in the ocean distribute the suns energy?

Physics

1 answer:

PtichkaEL [24]3 years ago

3 0

Answer:

The Sun dumps a lot of thermal energy by sunlight into the atmosphere, and especially into the oceans. Basically, currents arise as warm equatorial waters move to the NORTH carrying heat, and as part of this flow, the cold polar waters go south and there get heated.

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El motor de una licuadora gira a 3600 rpm, disminuye su velocidad angular hasta 2000 rpm realizando 120 vueltas. Calcular: a) La

allsm [11]

Answer:

a) α = -65,2 rad/s².

b) t = 2,57 s.

Explanation:

a) La aceleración angular se puede calcular usando la siguiente ecuación:

$\omega_{f}^{2} = \omega_{0}^{2} + 2\alpha \theta$

En donde:

$\omega_{f}$ : es la velocidad angular final = 2000 rpm = 209,4 rad/s

$\omega_{0}$ : es la velocidad angular inicial = 3600 rpm = 377,0 rad/s

α: es la aceleración angular=?

θ: es el desplazamiento o número de vueltas = 120 rev = 754,0 rad

Las conversiones de unidades se hicieron sabiendo que 1 revolución = 2π radianes y que 1 minuto = 60 segundos.

Resolviendo la ecuación (1) para α, tenemos:

$\alpha = \frac{\omega_{f}^{2} - \omega_{0}^{2}}{2\theta} = \frac{(209,4 rad/s)^{2} - (377,0 rad/s)^{2}}{2*754,0 rad} = -65,2 rad/s^{2}$

Entonces, la aceleración angular es -65,2 rad/s². El signo negativo se debe a que el motor está desacelerando.

b) El tiempo transcurrido se puede encontrar como sigue:

$\omega_{f} = \omega_{0} + \alpha t$

Resolviendo para t, tenemos:

$t = \frac{\omega_{f} - \omega_{0}}{\alpha} = \frac{209,4 rad/s - 377,0 rad/s}{-65,3 rad/s^{2}} = 2,57 s$

Por lo tanto, el tiempo transcurrido fue 2,57 s.

Espero que te sea de utilidad!

3 0

3 years ago

If the incoming photon had a frequency of 8.0 E14 Hz, would you have a photo-electron ejected?

cupoosta [38]

Yes.

In fact, from the graph we see that the threshold frequency (the minimum energy of the incoming energy needed to extract a photoelectron from the material) is $5.0 \cdot 10^{14} Hz$ (we see it because this is the frequency at which the maximum kinetic energy of the emitted electron is zero).

The incoming photon in this problem has a frequency of 8.0 E14 Hz, so above the threshold frequency, therefore it is enough to extract photoelectrons from the material.

4 0

4 years ago

Figure 23.9 shows a sliding mass on a spring. Assume there is no friction.

Tom [10]

<span>Without friction, there will be undamped simple harmonic motion. The force of the spring is proportional to the distance from the equilibrium point. The period of oscillation will be independent of the amplitude.

I hope my answer has come to your help. God bless and have a nice day ahead!</span>

4 0

3 years ago

An object's natural frequency depends partially on the material from which it is made.

laiz [17]

Yes. Partly. Also on the size and shape of the object, and also on what material it's IN at the time ... like air, water, Helium, Jello, etc.

3 0

3 years ago

Read 2 more answers

An arrow is moving at 35 m/s and travels for 5 seconds. How far did the arrow travel?

noname [10]

Answer:

175m