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Cloud [144]
4 years ago
9

A bicycle tire is spinning counterclockwise at 2.70 rad/s. During a time period At 1.90 s, the tire is stopped and spun in the o

pposite (clockwise) direction also at 2.70 rad/s. Calculate the change in the tire's angular velocity Aw and the tire's average angular acceleration aav. (Indicate the direction with the signs of your answers.)
(a) the change in the tire's angular velocity Aco (in rad/s) rad/s
(b) the tire's average angular acceleration aav (in rad/s2) rad/s
Physics
1 answer:
Naddik [55]4 years ago
7 0

Answer:

a) -5.40 rad/s

b) -2.842 rad/s²

Explanation:

The direction is important in dealing with such questions. Clockwise is considered negative and counterclockwise is considered positive

a) Δω = final angular velocity - initial angular velocity

          = -2.70 rad/s - 2.70 rad/s

          = -5.40 rad/s

b) ∝ = Δω/Δt = (-5.40 rad/s)/1.90s = -2.842 rad/s²

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Answer:

Check attachment for complete question

Question

Find a unit vector in the direction in which

f increases most rapidly at P and give the rate of change of f

in that direction; Find a unit vector in the direction in which f

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that direction.

f (x, y, z) = x²z e^y + xz²; P(1, ln 2, 2).

Explanation:

The function, z = f(x, y,z), increases most rapidly at (a, b,c) in the

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Given that

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At the point P(1, In2, 2)

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The rate of change of f at point P is.

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2. F increases most rapidly in the positive direction of -∇f

∇f=- (df/dx i + df/dy j +df/dz k)

∇f=-(2xze^y+z²)i - (x²ze^y) j - (x²e^y + 2xz)k

At the point P(1, In2, 2)

Then,

∇f= -(2×1×2×e^In2+2²)i -(1²×2×e^In2)j -(1²e^In2+2×1×2)

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Then, unit vector

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Then, |∇f|= √ 12²+4²+6²

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Then,

Unit vector

V=-(12i+4j+6k)/14

V= - 6/7 i - 2/7 j - 3/7 k

This is the increasing unit vector

The rate of change of f at point P is.

|∇f|= √ 12²+4²+6²

|∇f|= 14

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