(a) the weight of the fish is:
![W=mg=(3.5 kg)(9.81 m/s^2)=34.3 N](https://tex.z-dn.net/?f=W%3Dmg%3D%283.5%20kg%29%289.81%20m%2Fs%5E2%29%3D34.3%20N)
and this is the force that stretches the spring by
![x=3.7 cm=0.037 m](https://tex.z-dn.net/?f=x%3D3.7%20cm%3D0.037%20m)
. So, we can use Hook's law to find the constant of the spring:
![k= \frac{F}{x}= \frac{34.3 N}{0.037 m}=927.0 N/m](https://tex.z-dn.net/?f=k%3D%20%5Cfrac%7BF%7D%7Bx%7D%3D%20%5Cfrac%7B34.3%20N%7D%7B0.037%20m%7D%3D927.0%20N%2Fm%20%20)
(b) The fish is pulled down by 2.8 cm = 0.028 m more, so now the total stretch of the spring is
![x'=3.7 cm+2.8 cm=6.5 cm](https://tex.z-dn.net/?f=x%27%3D3.7%20cm%2B2.8%20cm%3D6.5%20cm)
But this is also the amplitude of the new oscillation, because this is the maximum extension the spring can get, so A=6.5 cm.
The angular frequency of oscillation is given by:
![\omega= \sqrt{ \frac{k}{m} }= \sqrt{ \frac{927.0 N/m}{3.5 kg} }=16.3 Hz](https://tex.z-dn.net/?f=%5Comega%3D%20%5Csqrt%7B%20%5Cfrac%7Bk%7D%7Bm%7D%20%7D%3D%20%5Csqrt%7B%20%5Cfrac%7B927.0%20N%2Fm%7D%7B3.5%20kg%7D%20%7D%3D16.3%20Hz%20%20)
and so the frequency is given by
Answer:
the sheets approach while the object is near
Explanation:
An electroscope is an apparatus that has two metal sheets attached, when these sheets have a charge and distribute evenly between them and the sheets repel.
When I approach an object charged with a counter (negative) charge, part of the charge of the electroscope moves near the charged external object, to neutralize the electric field, so as the charge on one of the plates decreases the electroscope has approached , as the objects are not touched the system remains in this configuration while the object is close. When the object is released, the electric field it creates disappears, so the positive charges repel inside the electroscope and the sheets repel to the initial position.
In short, the sheets approach while the object is near
1. Mary is 500 Meters away from her starting point because the hypotenuse represents the displacement of Mary from A to O.
2. Since Mary took the long way, she walked a total of 700 Meters. (Because she went from A to O to B, and the sum of 300 and 400 is 700).
3. Mary walked 140 Meters per minute, because S=D/T. so speed = 700 divided by 5, 140.
Answer:
590 ohms
Explanation:
Let the resistance of each resistor be 'R'.
Given:
Equivalent resistance of the three resistors in parallel ![(R_p)[/tex[ = 60 ohmWe know that, the equivalent resistance of three identical resistors in parallel is given as:[tex]R_p=\frac{R}{3}](https://tex.z-dn.net/?f=%28R_p%29%5B%2Ftex%5B%20%3D%2060%20ohm%3C%2Fp%3E%3Cp%3EWe%20know%20that%2C%20%3Cstrong%3Ethe%20equivalent%20resistance%20of%20three%20identical%20resistors%20in%20parallel%20is%20given%20as%3A%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3E%5Btex%5DR_p%3D%5Cfrac%7BR%7D%7B3%7D)
Plug in the given values and solve for 'R'. This gives,
![65=\frac{R}{3}\\\\R=65\times 3\\\\R=195\ ohm](https://tex.z-dn.net/?f=65%3D%5Cfrac%7BR%7D%7B3%7D%5C%5C%5C%5CR%3D65%5Ctimes%203%5C%5C%5C%5CR%3D195%5C%20ohm)
Now, the equivalent resistance of the three identical resistors in series is the sum of each of the resistors. Therefore,
![R_s=R+R+R\\\\R_s=3R\\\\R_s=3\times 195=585\ ohm](https://tex.z-dn.net/?f=R_s%3DR%2BR%2BR%5C%5C%5C%5CR_s%3D3R%5C%5C%5C%5CR_s%3D3%5Ctimes%20195%3D585%5C%20ohm)
As the question asks to round off to two significant figures, so we add 1 to the second significant figure because the number next to 8 is 5 which is greater than or equal to 5. So, the number becomes 590.
Hence, the equivalent resistance in series is 590 ohms.