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omeli [17]
3 years ago
11

A spider begins to spin a web by first hanging from a ceiling by his fine, silk fiber. He has a mass of 0.025 kg and a charge of

3.4 µC. A second spider with a charge of 4.2 µC rests in her own web exactly 2.6 m vertically below the first spider.
(a) What is the magnitude of the electric field due to the charge on the second spider at the position of the first spider?
(b) What is the tension in the silk fiber above the first spider?
Physics
1 answer:
Rasek [7]3 years ago
8 0

Answer:

a) (5.59 × 10³) N/C

b) 0.226 N directed away from the spider.

Explanation:

a) Electric field, E, felt as a result of point charge, Q, at a distance, d away is given by

E = kQ/d²

So, magnitude of the electric field due to the charge on the second spider at the position of the first spider

Q = 4.2 µC = 4.2 × 10⁻⁶ C

k = Coulomb's constant = 8.99 × 10⁹ Nm²/C

d = 2.6 m

E = (8.99 × 10⁹ × 4.2 × 10⁻⁶)/2.6²

E = 5.59 × 10³ N/C

b) Tension in the silk fiber above the spider is the net force due to the weight of spider one and the force of repulsion due the two charges.

Force due to the two charges = Eq

where q now represents the charge of the first spider at the first point, feeling the electric field calculated in (a)

F = 5.59 × 10³ × 3.4 × 10⁻⁶ = 0.01901 N directed upwards. (That is, F = + 0.019 N)

Weight of the spider = mg = 0.025 × 9.8 = 0.245 N directed downwards. (That is, W = -0.245 N)

Net force, T = mg - F = 0.245 - 0.019 = 0.226 N (that is, 0.226 N, directed upwards, away from the spider).

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Speed of cart's might be less than the high speed after 5 seconds.

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Given that,

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The fan supplies a force to the cart. If a lower fan speed were used, less force would be applied. This would cause a slower change in the cart's speed. So, the cart would be rolling more slowly than x\ cm per second after 5 seconds. The speed of cart's might be less than x\ cm per second.

Force is needed

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C. for a motionless object to remain still

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Speed of cart's might be less than the high speed after 5 seconds.

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Which of the following is not a function of a simple machine?
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A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and t
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Answer:

The speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.

Explanation:

We can calculate the speed of the train using the Doppler equation:

f = f_{0}\frac{v + v_{o}}{v - v_{s}}        

Where:

f₀: is the emitted frequency

f: is the frequency heard by the observer  

v: is the speed of the sound = 343 m/s

v_{o}: is the speed of the observer = 0 (it is heard in the town)

v_{s}: is the speed of the source =?

The frequency of the train before slowing down is given by:

f_{b} = f_{0}\frac{v}{v - v_{s_{b}}}  (1)                  

Now, the frequency of the train after slowing down is:

f_{a} = f_{0}\frac{v}{v - v_{s_{a}}}   (2)  

Dividing equation (1) by (2) we have:

\frac{f_{b}}{f_{a}} = \frac{f_{0}\frac{v}{v - v_{s_{b}}}}{f_{0}\frac{v}{v - v_{s_{a}}}}

\frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - v_{s_{b}}}   (3)  

Also, we know that the speed of the train when it is slowing down is half the initial speed so:

v_{s_{b}} = 2v_{s_{a}}     (4)

Now, by entering equation (4) into (3) we have:

\frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - 2v_{s_{a}}}  

\frac{300 Hz}{290 Hz} = \frac{343 m/s - v_{s_{a}}}{343 m/s - 2v_{s_{a}}}

By solving the above equation for v_{s_{a}} we can find the speed of the train after slowing down:

v_{s_{a}} = 11.06 m/s

Finally, the speed of the train before slowing down is:

v_{s_{b}} = 11.06 m/s*2 = 22.12 m/s

Therefore, the speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.                        

I hope it helps you!                                                        

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a linear function has the same y-intercept as x + 4y equals 16 and it's graph contains the point (4,5). Find the slope of the li
navik [9.2K]

Answer:  \bold{\text{Slope (m)}=\dfrac{1}{4}}

<u>Explanation:</u>

A linear equation is of the form: y = mx + b   where

  • m is the slope
  • b is the y-intercept (where it crosses the y-axis)

x + 4y = 16

     4y = -x + 16

       y = -\dfrac{1}{4}x+\dfrac{16}{4}

       y=-\dfrac{1}{4}x+4

The y-intercept (b) = 4

Next, find the slope given point (4, 5) and b = 4

y=mx+b\\\\5=m(4)+4\\\\1=4m\\\\\dfrac{1}{4}=m\\\\\\\\\large\boxed{Slope (m)=\dfrac{1}{4}}

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