In Rutherford's gold foil experiment, some of the positive particles would pass through the foil and some would bounce off. This led to a new theory that all of the positive subatomic particles were in the center of the atom instead of evenly spread throughout.
Answer:
The magnitude of the tension on the ends of the clothesline is 41.85 N.
Explanation:
Given that,
Poles = 2
Distance = 16 m
Mass = 3 kg
Sags distance = 3 m
We need to calculate the angle made with vertical by mass
Using formula of angle
![\tan\theta=\dfrac{8}{3}](https://tex.z-dn.net/?f=%5Ctan%5Ctheta%3D%5Cdfrac%7B8%7D%7B3%7D)
![\thta=\tan^{-1}\dfrac{8}{3}](https://tex.z-dn.net/?f=%5Cthta%3D%5Ctan%5E%7B-1%7D%5Cdfrac%7B8%7D%7B3%7D)
![\theta=69.44^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%3D69.44%5E%7B%5Ccirc%7D)
We need to calculate the magnitude of the tension on the ends of the clothesline
Using formula of tension
![mg=2T\cos\theta](https://tex.z-dn.net/?f=mg%3D2T%5Ccos%5Ctheta)
Put the value into the formula
![3\times9.8=2T\times\cos69.44](https://tex.z-dn.net/?f=3%5Ctimes9.8%3D2T%5Ctimes%5Ccos69.44)
![T=\dfrac{3\times9.8}{2\times\cos69.44}](https://tex.z-dn.net/?f=T%3D%5Cdfrac%7B3%5Ctimes9.8%7D%7B2%5Ctimes%5Ccos69.44%7D)
Hence, The magnitude of the tension on the ends of the clothesline is 41.85 N.
Answer:
jfjcgufnfhfufm TV fifnricnrhkddufnfif km fgkfkvntfmrugrhfifnh r
Answer:
Semantic memory is a category of long-term memory that involves the recollection of ideas, concepts and facts commonly regarded as general knowledge. Examples of semantic memory include factual information such as grammar and algebra.
Question:
A wire 2.80 m in length carries a current of 5.20 A in a region where a uniform magnetic field has a magnitude of 0.430 T. Calculate the magnitude of the magnetic force on the wire assuming the following angles between the magnetic field and the current.
(a)60 (b)90 (c)120
Answer:
(a)5.42 N (b)6.26 N (c)5.42 N
Explanation:
From the question
Length of wire (L) = 2.80 m
Current in wire (I) = 5.20 A
Magnetic field (B) = 0.430 T
Angle are different in each part.
The magnetic force is given by
![F=I \times B \times L \times sin(\theta)](https://tex.z-dn.net/?f=F%3DI%20%5Ctimes%20B%20%5Ctimes%20L%20%5Ctimes%20sin%28%5Ctheta%29)
So from data
![F = 5.20 A \times 0.430 T \times 2.80 sin(\theta)\\\\F=6.2608 sin(\theta) N](https://tex.z-dn.net/?f=F%20%3D%205.20%20A%20%5Ctimes%200.430%20T%20%5Ctimes%202.80%20sin%28%5Ctheta%29%5C%5C%5C%5CF%3D6.2608%20sin%28%5Ctheta%29%20N)
Now sub parts
(a)
![\theta=60^{o}\\\\Force = 6.2608 sin(60^{o}) N\\\\Force = 5.42 N](https://tex.z-dn.net/?f=%5Ctheta%3D60%5E%7Bo%7D%5C%5C%5C%5CForce%20%3D%206.2608%20sin%2860%5E%7Bo%7D%29%20N%5C%5C%5C%5CForce%20%3D%205.42%20N)
(b)
![\theta=90^{o}\\\\Force = 6.2608 sin(90^{o}) N\\\\Force = 6.26 N](https://tex.z-dn.net/?f=%5Ctheta%3D90%5E%7Bo%7D%5C%5C%5C%5CForce%20%3D%206.2608%20sin%2890%5E%7Bo%7D%29%20N%5C%5C%5C%5CForce%20%3D%206.26%20N)
(c)
![\theta=120^{o}\\\\Force = 6.2608 sin(120^{o}) N\\\\Force = 5.42 N](https://tex.z-dn.net/?f=%5Ctheta%3D120%5E%7Bo%7D%5C%5C%5C%5CForce%20%3D%206.2608%20sin%28120%5E%7Bo%7D%29%20N%5C%5C%5C%5CForce%20%3D%205.42%20N)