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garri49 [273]
3 years ago
5

A ball is thrown at a 60.0° angle above the horizontal across level ground. It is thrown from a height of 2.00 m above the groun

d with a speed of 20.0 m/s and experiences no appreciable air resistance. The time the ball remains in the air before striking the ground is closest to_____________.a. 16.2 s.b. 3.07 s.c. 3.32 s.d. 3.53 s.e. 3.64 s.

Physics
1 answer:
klio [65]3 years ago
7 0

Answer:

Option E is correct.

Time the ball remains in the air before striking the ground is closest to 3.64 s

Explanation:

yբ = yᵢ + uᵧt + gt²/2

yբ = 0

yᵢ = 2 m

uᵧ = u sinθ = 20 sin 60 = 17.32 m/s

g = -9.8 m/s², t = ?

0 = 2 + 17.32t - 4.9t²

4.9t² - 17.32t - 2 = 0

Solving the quadratic equation,

t = 3.647 s or t = -0.1112 s

time is a positive variable, hence, t = 3.647 s. Option E.

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3 years ago
A current I flows down a wire of radius a.
Helga [31]

Answer:

(a) K = \frac{I}{2\pi a}

(b) J = \frac{I}{2\pi as}

Explanation:

(a) The surface current density of a conductor is the current flowing per unit length of the conductor.

                                   K = \frac{dI}{dL}

Considering a wire, the current is uniformly distributed over the circumferenece of the wire.

                                   dL = 2\pi r

The radius of the wire = a

                                    dL = 2\pi a

The surface current density K = \frac{I}{2\pi a}

(b) The current density is inversely proportional

                                     J \alpha  s^{-1}    

                                     J = \frac{k}{s}           ......(1)

k is the constant of proportionality

                                     I = \int\limits {J} \, dS

                                     I = J \int\limits \, dS     ........(2)

substituting (1) into (2)

                                     I = \frac{k}{s} \int\limits\, dS

                                     I = k \int\limits^a_0 \frac{1}{s}  {s} \, dS

                                     I = 2\pi k\int\limits\, dS

                                     I = 2\pi ka

                                     k = \frac{I}{2\pi a}

substitute J = \frac{k}{s}

                                     J = \frac{I}{2\pi as}

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3 years ago
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The answer would be C. (:
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3 years ago
Read 2 more answers
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