Answer:
The answer is D 100 newton
Explanation:
2.0m/s2 is d acceleration while the 50kg is the mass. Force = mass x acceleration. So f=50x2.so force is 100 newton
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)
Answer:
2.2 meters
Explanation:
Potential energy, PE created by a charge, q at a radius r from the charge source, Q, is expressed as:
![KE=\frac{kQq}{r}\ \ \ \ \ \ \ ...i](https://tex.z-dn.net/?f=KE%3D%5Cfrac%7BkQq%7D%7Br%7D%5C%20%20%20%20%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20...i)
is Coulomb's constant.
#The electric field,
at radius r is expressed as:
![E=\frac{kQ}{r^2}\ \ \ \ \ \ \ \ \ \ ...ii](https://tex.z-dn.net/?f=E%3D%5Cfrac%7BkQ%7D%7Br%5E2%7D%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20...ii)
From i and ii, we have:
![KE=Eqr](https://tex.z-dn.net/?f=KE%3DEqr)
![r=(KE)/Eq](https://tex.z-dn.net/?f=r%3D%28KE%29%2FEq)
#Substitute actual values in our equation:
![r=\frac{75J}{(7.2\times 10^{-5}C)(4.8\times 10^5 V/m)}\\\\=2.1701\approx2.2\ m](https://tex.z-dn.net/?f=r%3D%5Cfrac%7B75J%7D%7B%287.2%5Ctimes%2010%5E%7B-5%7DC%29%284.8%5Ctimes%2010%5E5%20V%2Fm%29%7D%5C%5C%5C%5C%3D2.1701%5Capprox2.2%5C%20m)
Hence, the distance between the charge and the source of the electric field is 2.2 meters
<span>two objects in contact with each other are the same temperature</span>
It is a chemical change. I always think of a chemical/physical change as if you could reverse it back as it started off, for example if you stepped on a can you can reverse the can back probably not exactly like it was before but you can still reverse it so this would be a physical change, and if you baked a pizza you could not reverse the dough and everything else back.