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fiasKO [112]
3 years ago
15

Review Multiple-Concept Example 7 in this chapter as an aid in solving this problem. In a fast-pitch softball game the pitcher i

s impressive to watch, as she delivers a pitch by rapidly whirling her arm around so that the ball in her hand moves in a circle. In one instance, the radius of the circle is 0.602 m. At one point on this circle, the ball has an angular acceleration of 64.1 rad/s2 and an angular speed of 13.0 rad/s. (a) Find the magnitude of the total acceleration (centripetal plus tangential) of the ball. (b) Determine the angle of the total acceleration relative to the radial direction.
Physics
1 answer:
soldi70 [24.7K]3 years ago
6 0

Answer:

108.8 m /s

θ = 21°

Explanation:

Centripetal acceleration = ω² R

=( 13 )² x .602

= 101.74 m/s²

tangential acceleration a_t = angular acceleration x R

= 64.1 x .602

= 38.58 m /s²

Total acceleration R

R² = ( 101.74 )² + ( 38.58 )²

R = 108.8 m /s

angle required be θ

Tanθ = tangential accn  / radial accn

= 38.58  / 101.74

θ = 21°

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Answer:

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3 years ago
A railroad car moves under a grain elevator at a constant speed of 3.20 m/s. Grain drops into the car at the rate of 240 kg/min.
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Answer:

F = 768 N                  

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We need to find the magnitude of force needed to keep the car moving constant speed. The relation between the momentum and the force is given by :

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3 years ago
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(b)  if the force on it is doubled and its mass is halved, the new acceleration will be four times as great.

The force on an object is determined by applying Newton's second law of motion;

F = ma

\frac{F_1}{m_1a_1} = \frac{F_2}{m_2a_2}

(a)

when the force is halved, F₂ = 0.5F₁,

mass is doubled, m₂ = 2m₁

\frac{F_1}{m_1a_1} = \frac{0.5F_1}{2m_1a_2} \\\\2m_1a_2F_1 = 0.5F_1 m_1a_1\\\\2a_2 = 0.5a_1\\\\a_2 = \frac{0.5a_1}{2} = \frac{a_1}{2 \times 2} = \frac{a_1}{4} \\\\a_2 = \frac{1}{4} (a_1)

Thus, If the force is halved and the mass of the object is doubled, the acceleration will be one-fourth as great.

(b)

when the force is doubled, F₂ = 2F₁,

mass is halved, m₂ = 0.5m₁

\frac{F_1}{m_1 a_1} = \frac{2F_1}{0.5m_1 a_2} \\\\0.5m_1a_2 F_1 = 2F_1m_1a_1\\\\0.5a_2 = 2a_1\\\\a_2 = \frac{2a_1}{0.5} \\\\a_2 = 4(a_1)

Thus, if the force on it is doubled and its mass is halved, the new acceleration will be four times as great.

Learn more here:brainly.com/question/19887955

3 0
2 years ago
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