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Aleonysh [2.5K]
3 years ago
13

Guiana dolphins are one of the few mammals able to detect electric fields. In a test of sensitivity, a dolphin was exposed to th

e variable electric field from a pair of charged electrodes. The magnitude of the electric field near the sensory organs was measured by detecting the potential difference between two measurement electrodes located 1.0 cm apart along the field lines. The dolphin could reliably detect a field that produced a potential difference of 0.50 mV between these two electrodes.
1. What is the corresponding electric field strength?
Physics
1 answer:
iren [92.7K]3 years ago
8 0

Answer:

0.05 V/m

Explanation:

V = Potential difference that is possible for the dolphin to detect = 0.5 mV

d = Distance between electrodes = 1 cm

Electric field strength is given by

E=\dfrac{V}{d}

\Rightarrow E=\dfrac{0.5\times 10^{-3}}{1\times 10^{-2}}

\Rightarrow E=0.05\ V/m

The corresponding electric field strength is 0.05 V/m

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Say I have a series circuit with 20v and four 65 ohm resistors, what is the current in each resistor?
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Data:

E = 20 V
R_{1} = 65\Omega
R_{2} = 65\Omega
R_{3} = 65\Omega
R_{4} = 65\Omega

<span>Now that we have all the values ​​we need properly identified, simply calculate the equivalent total resistance of the circuit and the intensity of the total electric current using the Ohm's Law:

</span>R_{T} =  R_{1} + R_{2} + R_{3} + R_{4}
R_{T} = 65 + 65 + 65+ 65
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<span>Like this:
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I_{T} =  \frac{E}{ R_{T} }

I_{T} = \frac{20}{ 260 }
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\boxed{\boxed{I_{T} \approx 0,07A}} 
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<span>The intensity of the total electric current 
</span>\boxed{\boxed{I_{T} \approx 0,07A}} 

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3 years ago
At the intersection of Texas Avenue and University Drive,
Zielflug [23.3K]

Answer:

  • The initial speed of the truck is 21.93 m/s, and the initial speed of the car is 19.524 m/s  

Explanation:

We can use conservation of momentum to find the initial velocities.

Taking the unit vector \hat{i} pointing north and \hat{j} pointing east, the final velocity will be

\vec{V}_f = 16.0 \frac{m}{s} \ ( \ cos(24.0 \°) \ , \ sin (24.0 \°) \ )

\vec{V}_f = ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )

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\vec{P}_f = (m_{car}+ m_{truck}) * V_f

\vec{P}_f = (950 \ kg \ + 1900 \ kg \ ) *  ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )

\vec{P}_f = (2.850 \ kg \ ) *  ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )

\vec{P}_f = ( \ 41,658.45 \frac{ kg \ m}{s} \ , \ 18,547.8 \frac{kg \ m}{s} \ )

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\vec{P}_0= \vec{P}_f

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 \vec{P}_0= \vec{P}_f 

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\left \{ {{m_{truck} \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}  } \atop {m_{car} \ v_{car}=18,547.8 \frac{kg \ m}{s} }} \right.

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m_{truck} \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}

1900 \ kg \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}

v_{truck} = \frac{41,658.45 \frac{ kg \ m}{s}}{1900 \ kg}

v_{truck} = \frac{41,658.45 \frac{ kg \ m}{s}}{1900 \ kg}

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950 \ kg \ v_{car}=18,547.8 \frac{kg \ m}{s}

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Re-arranging the formula, we find:
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