Answer:
it can cause wear and tear and reduces efficiency as energy is lost.
Explanation:
Answer:
You know the saying "Opposites attract" well that is how you can remember that South and North Magnetic Poles connect.
Explanation:
Hope this helps ya
Answer:

Explanation:
As we know that there is no external torque on the system of two disc
then the angular momentum of the system will remains conserved
So we will have

now we have

also we have

now from above equation we have

now we have


Answer:
L = mp*v₀*(ms*D) / (ms + mp)
Explanation:
Given info
ms = mass of the hockey stick
uis = 0 (initial speed of the hockey stick before the collision)
xis = D (initial position of center of mass of the hockey stick before the collision)
mp = mass of the puck
uip = v₀ (initial speed of the puck before the collision)
xip = 0 (initial position of center of mass of the puck before the collision)
If we apply
Ycm = (ms*xis + mp*xip) / (ms + mp)
⇒ Ycm = (ms*D + mp*0) / (ms + mp)
⇒ Ycm = (ms*D) / (ms + mp)
Now, we can apply the equation
L = m*v*R
where m = mp
v = v₀
R = Ycm
then we have
L = mp*v₀*(ms*D) / (ms + mp)
Answer:
The correct answer is B
Explanation:
Let's calculate the electric field using Gauss's law, which states that the electric field flow is equal to the charge faced by the dielectric permittivity
Φ
= ∫ E. dA =
/ ε₀
For this case we create a Gaussian surface that is a sphere. We can see that the two of the sphere and the field lines from the spherical shell grant in the direction whereby the scalar product is reduced to the ordinary product
∫ E dA =
/ ε₀
The area of a sphere is
A = 4π r²
E 4π r² =
/ ε₀
E = (1 /4πε₀
) q / r²
Having the solution of the problem let's analyze the points:
A ) r = 3R / 4 = 0.75 R.
In this case there is no charge inside the Gaussian surface therefore the electric field is zero
E = 0
B) r = 5R / 4 = 1.25R
In this case the entire charge is inside the Gaussian surface, the field is
E = (1 /4πε₀
) Q / (1.25R)²
E = (1 /4πε₀
) Q / R2 1 / 1.56²
E₀ = (1 /4π ε₀
) Q / R²
= Eo /1.56
²
= 0.41 Eo
C) r = 2R
All charge inside is inside the Gaussian surface
=(1 /4π ε₀
) Q 1/(2R)²
= (1 /4π ε₀
) q/R² 1/4
= Eo 1/4
= 0.25 Eo
D) False the field changes with distance
The correct answer is B