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AVprozaik [17]
2 years ago
9

An arrow is launched from a bow with an initial horizontal velocity of 40

Physics
1 answer:
Ivan2 years ago
4 0

Answer:

V = (Vx^2 + Vy^2)^1/2 = (40^2 + 62^2)^1/2

V = 73.8 m/s

tan theta = Vy / Vx = 62/40 = 1.55

theta = 57.2 deg

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Consider a long, cylindrical charge distribution of radius R with a uniform charge density rho. Find the electric field at dista
Flauer [41]

Answer:

E = p*r / 2*e_o

Explanation:

Given:

- Volume of cylinder V = pi*r^2*L

- Surface area A = 2*pi*r*L

- permittivity of space : e_o

Find:

Electric field E at distance r from the axis, where r < R.

Solution:

Step 1: Application of Gauss Law

- Form a Gaussian surface within the cylinder with r < R. Th cylinder has two surfaces i.e curved surfaces and end caps. Due to long charge distribution the flux through is zero, since the surface dA of end cap and E are at 90 degree angle to one another; hence, E . dA = E*dA*cos(90) = 0. For the curved surface we have:

              (surface integral) E.dA = Q_enclosed / e_o

Step 2: The charge enclosed (Q_enclosed) is function of r and proportional density:

              Q_enclosed = p*V

              Q_enclosed = p*pi*r^2*L

Step 3: The area of the curved surface:

              dA = 2*pi*r*L

Step 4: Compute E:

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              E = p*r / 2*e_o

5 0
3 years ago
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