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It's very wonderful to help you today!
Answer:
The resistance of the wire after being heated is approximately 1.15 times the resistance before being heated.
Explanation:
The resistance of the wire goes up as the temperature goes up.
By Ohm's law, we have the relationship between the resistance of a conductor R, the current flowing through it, I, and the voltage across it, V,
![V=R~x~I](https://tex.z-dn.net/?f=V%3DR~x~I)
As described, the current through the wire decreases so, to maintain the same voltage, the resistance must increase when the wire is heated. Let's say R1 is the resistance before heating the wire and R2 is the resistance after,
![V=R_{1} x~I_{1} =R_{2}~} x~I_{2}](https://tex.z-dn.net/?f=V%3DR_%7B1%7D%20x~I_%7B1%7D%20%3DR_%7B2%7D~%7D%20%20x~I_%7B2%7D)
Let's solve for R2,
![R_{2} =\frac{I_{1} }{I_{2} }~x~R_{1}](https://tex.z-dn.net/?f=R_%7B2%7D%20%3D%5Cfrac%7BI_%7B1%7D%20%7D%7BI_%7B2%7D%20%7D~x~R_%7B1%7D)
In this case, I1 = 0.3A and I2 = 0.26A,
![R_{2} =\frac{0.3A}{0.26A} ~x~R_{1} =1.15R_{1}](https://tex.z-dn.net/?f=R_%7B2%7D%20%3D%5Cfrac%7B0.3A%7D%7B0.26A%7D%20~x~R_%7B1%7D%20%3D1.15R_%7B1%7D)
Hence, the <u><em>resistance of the wire after being heated is approximately 1.15 times the resistance before being heated.</em></u>
Reference point is the answere
Answer:
Explanation:Required formula is
W=F *S
W=work =?
F=force =5N
S=displacement =0.5m
please feel free to ask if you have any questions
Answer:
I think is 3 I am not sure
Answer:
The new speed must be ![\frac{V_0}{\sqrt{2}}](https://tex.z-dn.net/?f=%5Cfrac%7BV_0%7D%7B%5Csqrt%7B2%7D%7D)
Explanation:
In order for the satellite to be on a stable orbit around the planet, the gravitational attraction must be equal to the centripetal force that keeps the satellite in circular motion:
![G \frac{Mm}{R^2}= m\frac{V_0^2}{R}](https://tex.z-dn.net/?f=G%20%5Cfrac%7BMm%7D%7BR%5E2%7D%3D%20m%5Cfrac%7BV_0%5E2%7D%7BR%7D)
where G is the gravitational constant, M is the mass of the planet, m is the mass of the satellite, V0 the speed of the satellite at distance R from the center of the planet.
We can re-write V0, the initial satellite speed, by re-arranging the equation:
![V_0 = \sqrt{\frac{GM}{R}}](https://tex.z-dn.net/?f=V_0%20%3D%20%5Csqrt%7B%5Cfrac%7BGM%7D%7BR%7D%7D)
Now, if we want the satellite to orbit at a distance of 2R, the new tangential speed must be:
![V' = \sqrt{\frac{GM}{2R}}=\frac{1}{\sqrt{2}} \sqrt{\frac{GM}{R}}= \frac{V_0}{\sqrt{2}}](https://tex.z-dn.net/?f=V%27%20%3D%20%5Csqrt%7B%5Cfrac%7BGM%7D%7B2R%7D%7D%3D%5Cfrac%7B1%7D%7B%5Csqrt%7B2%7D%7D%20%5Csqrt%7B%5Cfrac%7BGM%7D%7BR%7D%7D%3D%20%5Cfrac%7BV_0%7D%7B%5Csqrt%7B2%7D%7D)