Answer:
The flux of the electric field through the surface is 3.24\times10^{3}\ Nm^/C[/tex].
Explanation:
Given that,
Area of cube = 48 cm²
Charge = 28.7 nC
We need to calculate the flux of the electric field through the surface
Using formula Gauss's law
The electric flux through any closed surface,
![\phi =\dfrac{q}{\epsilon_{0}}](https://tex.z-dn.net/?f=%5Cphi%20%3D%5Cdfrac%7Bq%7D%7B%5Cepsilon_%7B0%7D%7D)
Where, q = charge
Put the value into the formula
![\phi=\dfrac{28.7\times10^{-9}}{8.85\times10^{-12}}](https://tex.z-dn.net/?f=%5Cphi%3D%5Cdfrac%7B28.7%5Ctimes10%5E%7B-9%7D%7D%7B8.85%5Ctimes10%5E%7B-12%7D%7D)
![\phi =3.24\times10^{3}\ Nm^/C](https://tex.z-dn.net/?f=%5Cphi%20%3D3.24%5Ctimes10%5E%7B3%7D%5C%20Nm%5E%2FC)
Hence, The flux of the electric field through the surface is 3.24\times10^{3}\ Nm^/C[/tex].
Answer:
Explanation:
v = ω R
v is linear speed and ω is angular speed
ω = v / R
a ) Inner radius = 25 x 10⁻³ m
speed v = 1.25 m/s
ω = 1.25 / (25 x 10⁻³ )
= .05 x 10⁻³
= 5 x 10⁻⁵ rad / s
outer radius = 58 x 10⁻³ m
speed v = 1.25 m/s
ω = 1.25 / (58 x 10⁻³ )
= .0215 x 10⁻³
= 2.15 x 10⁻⁵ rad / s
b )
linear constant speed v = 1.25 m /s
time = 74 min = 74 x 60 s
distance tracked = speed x time
= 1.25 x 74 x 60
= 5550 m
c ) time given
= 74 min = 74 x 60 s
angular acceleration
= ( 2.15 - 5 ) x 10⁻⁵ / (74 x 60 )
= - 6.42 x 10⁻⁹ rad / s²
Answer:
Air flows from a high-pressure area to a low-pressure area.
Explanation:
Answer:
One arm row: Latissmus Dorsi (Lat)
Dumbell flys: Shoulders
Machine back press: pectorals
Push downs: Triceps
Explanation:
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My teacher said 36m when I asked her