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slega [8]
2 years ago
8

Air escapes from a balloon at a rate of 2 60 ( ) 1 R t t   3 ft / min , where t is measured in minutes. How much air (in 3 ft

) escapes during the first minute?
Physics
1 answer:
kotykmax [81]2 years ago
4 0

Answer:

Given R (t) = 60/(1+t²), ft /min

To find the amount of air in ft³, during the first min,

R (t) = 60/(1+t²

at t=1min, R is the air amount in ft³/min

Take the integral, and evaluate over [0,1]

Integral Of R (t) = 60/(1+t²) = 60 tan⁻¹(t),

60(Tan⁻¹(1) - Tan⁻¹(0)) = 60(pi/2) = 30 π

Therefore, it means that in the first minute, 30π ft³ of air escaped.

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A 20~\mu F20 μF capacitor has previously charged up to contain a total charge of Q = 100~\mu CQ=100 μC on it. The capacitor is t
sertanlavr [38]

Explanation:

The given data is as follows.

       C = 20 \times 10^{-6} F

        R = 100 \times 10^{3} ohm

        Q_{o} = 100 \times 10^{-6} C

          Q = 13.5 \times 10^{-6} C

Formula to calculate the time is as follows.

          Q_{t}  = Q_{o} [e^{\frac{-t}{\tau}]

       13.5 \times 10^{-6} = 100 \times 10^{-6} [e^{\frac{-t}{2}}]

               0.135 = e^{\frac{-t}{2}}

         e^{\frac{t}{2}} = \frac{1}{0.135}

                         = 7.407

           \frac{t}{2} = ln (7.407)

                      t = 4.00 s

Therefore, we can conclude that time after the resistor is connected will the capacitor is 4.0 sec.

4 0
2 years ago
A bee wants to fly to a flower located due North of the hive on a windy day. The wind blows from East to West at speed 6.68 m/s.
Aleonysh [2.5K]

Answer:  53.31\° East of North

Explanation:

We have the following data:

Speed of the wind from East to West: 6.68 m/s

Speed of the bee relative to the air:  8.33 m/s

If we graph these speeds (which in fact are velocities because are vectors) in a vector diagram, we will have a right triangle in which the airspeed of the bee (its speed relative to te air) is the hypotense and the two sides of the triangle will be the <u>Speed of the wind from East to West</u> (in the horintal part) and the <u>speed due North relative to the ground</u> (in the vertical part).

Now, we need to find the direction the bee should fly directly to the flower (due North):

sin \theta=\frac{Windspeed-from-East-to-West}{Speed-bee-relative-to-air}

sin \theta=\frac{6.68 m/s}{8.33 m/s}

Clearing \theta:

\theta=sin^{-1} (\frac{6.68 m/s}{8.33 m/s})

\theta=53.31\°

6 0
2 years ago
4.A photon of green light strikes an unknown metal and an electron is emitted. The voltage is set to 2 volts. The electron canno
Anarel [89]
4) The correct answer is:
<span>B. An electron will be emitted in the second experiment, but it cannot be determined whether it will reach the second plate. 

In fact, violet light has higher frequency than green light. This means that photons of violet light carry more energy than photons of green light (remember that the energy of a photon is proportional to its frequency: </span>E=hf)<span>, so when they hit the surface of the metal, more energy is transferred to the electrons. The electron was already emitted with green light, so it must be emitted also with violet light, given the more energy transferred. The electron will also have more kinetic energy when hit by violet light, however, we cannot determine if it will reach the second plate, since we don't know how much energy has been used to extract the electron from the metal (in fact, we don't know the work function of the metal, i.e. the energy needed to extract the electron)


3)  The correct answer is
</span><span>A. Violet light will cause electrons to be emitted at greater velocities than those removed by green light.

In fact, </span>violet light has higher frequency than green light. This means that photons of violet light carry more energy than photons of green light (remember that the energy of a photon is proportional to its frequency: E=hf), so when they hit the surface of the metal, more energy is transferred to the electrons. Therefore, the emitted electrons will have on average greater energy (and so, greater velocity) than those removed by green light.
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