That would depend on the man’s starting latitude. If 6.7 mi south of the north pole, for example, the resultant would be 6.7 mi due north of the starting point.
Her average speed would be 69.2 miles per hour
Explanation:
When you take the number of miles and divide it by the number of hours you get 69.2 (when rounded)
That is how fast she went per hour
Answer:
The charge is 
Explanation:
Given that,
Distance = 2.5 mm
Electric field = 800 NC
Length 
We need to calculate the linear charge density
Using formula of linear charge density


Put the value into the formula


We need to calculate the charge
Using formula of charge

Put the value into the formula


Hence, The charge is 
Answer:
Angular acceleration = 23.68 rad / s²
Explanation:
Given that,
acceleration = 9m/s²
Therefore acceleration of string is 9m/s²
since string is constant in length
cylinder of radius 38.0 cm = 0.38m
Angular acceleration = a / r
Angular acceleration = 9 / 0.38
= 23.68 rad / s²
Angular acceleration = 23.68 rad / s²
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