Answer:
As point B is located inside the copper block so net electric field at point B is j.
Explanation:
Consider the figure attached below. The net electric field at location B,that is inside the copper block is zero because when a conductor is charged or placed in an electric field of external charges, net charge lies on the surface of conductor and there is no electric field inside the conductor. As point B is located inside the copper block so net electric field at point B is zero as well direction of net electric field at point B is zero.
Answer:
p waves travel faster than s waves and surface waves
Answer:
The magnetic field strength due to current flowing in the wire is9.322 x 10⁻⁶ T.
Explanation:
Given;
electric current, I = 21.3 A
distance of the magnetic field from the wire, R = 45.7 cm = 0.457 m
The strength of the resulting magnetic field at the given distance is calculated as;

Where;
μ₀ is permeability of free space = 4π x 10⁻⁷ T.m/A

Therefore, the magnetic field strength due to current flowing in the wire is 9.322 x 10⁻⁶ T.
Answer:
α= 15.4[rad/s^2], 500 [rev/min]
Explanation:
La aceleración angular constante esta definida por la siguiente ecuación:
![w=w_{0} + \alpha *t\\Donde:\\w=2000 [\frac{rev}{min}]*[\frac{1min}{60s} ] *[\frac{2\pi rad}{1 rev} ]=209.4[\frac{rad}{s} ]\\w_{0} =1500 [\frac{rev}{min}]*[\frac{1min}{60s} ] *[\frac{2\pi rad}{1 rev} ]=157.1[\frac{rad}{s} ]\\t=3.4[s]\\reemplazando:\\209.4-157.1=\alpha *(3.4)\\\\alpha =15.4[\frac{rad}{s^{2} } ]](https://tex.z-dn.net/?f=w%3Dw_%7B0%7D%20%2B%20%5Calpha%20%2At%5C%5CDonde%3A%5C%5Cw%3D2000%20%5B%5Cfrac%7Brev%7D%7Bmin%7D%5D%2A%5B%5Cfrac%7B1min%7D%7B60s%7D%20%5D%20%2A%5B%5Cfrac%7B2%5Cpi%20rad%7D%7B1%20rev%7D%20%5D%3D209.4%5B%5Cfrac%7Brad%7D%7Bs%7D%20%5D%5C%5Cw_%7B0%7D%20%3D1500%20%5B%5Cfrac%7Brev%7D%7Bmin%7D%5D%2A%5B%5Cfrac%7B1min%7D%7B60s%7D%20%5D%20%2A%5B%5Cfrac%7B2%5Cpi%20rad%7D%7B1%20rev%7D%20%5D%3D157.1%5B%5Cfrac%7Brad%7D%7Bs%7D%20%5D%5C%5Ct%3D3.4%5Bs%5D%5C%5Creemplazando%3A%5C%5C209.4-157.1%3D%5Calpha%20%2A%283.4%29%5C%5C%5C%5Calpha%20%3D15.4%5B%5Cfrac%7Brad%7D%7Bs%5E%7B2%7D%20%7D%20%5D)
Para poder llegar a las 2000 [revoluciones /min] estando en 1500 [rev/min], la cuchilla tuvo que incrementar su numero de revoluciones a 500 [rev/min]. Mientras aceleraba.