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kirill115 [55]
3 years ago
5

A tennis racket hits a tennis ball with a force of F=at−bt2, where a = 1200 N/ms , b = 370 N/ms2 , and t is the time (in millise

conds). The ball is in contact with the racket for 2.70 ms . If the tennis ball has a mass of 64.9 g , what is the resulting velocity of the ball, v, after the ball is hit by the racket? Express your answer numerically in meters per second to three significant fi
Physics
1 answer:
Veronika [31]3 years ago
4 0

Given Information:  

Mass of tennis ball = m = 64.9 g

Contact time = Δt = 2.70 ms

Force = F = at - bt²

a = 1200 N/ms

b = 370 N/ms²

Assuming that a and b is given as newton per milliseconds

Required Information:  

Velocity of the ball = Vf = ?  

Answer:  

Vf = 22.496 m/s

Explanation:  

The change in momentum of the ball is given by

Δp = -FΔt

Lets first find out the force acting on the tennis ball

F = aΔt - bΔt²

F = (1200)(2.70) - (370)(2.70)²

F = 542.7 N

Δp =  542.7*0.0027

Δp = 1.46 N.s

Δp = mVf - mVi

initial velocity is zero

Vf = Δp/m

Vf = 1.46/0.0649

Vf = 22.496 m/s

Therefore, the speed of the ball is 22.496 m/s

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A 1.50-m string of weight 0.0125 N is tied to the ceiling at its upper end, and the lower end supports a weight W. Ignore the ve
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Answer:

Explanation:

mass of string = .0125 / 9.8

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m = mass per unit length

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m = .85 x 10⁻³ kg/m

wave equation: y(x,t) = (8.50 mm)cos(172 rad/m x − 4830 rad/s t)

compare with equation of wave

y(x,t) = Acos(K x − ω t)

ω ( angular velocity ) = 4830 rad/s

k = 172 rad/m

Velocity = ω / k

= 4830/172 m /s

= 28.08 m /s

velocity of wave = \sqrt{\frac{W}{m } }

28.08 = \sqrt{\frac{W}{.85\times10^{-3} } }

788.48 =  W / .85 X 10⁻³

W = 670 x  10⁻³ N .

c ) wave length

wave length =2π  / k

= 2 x 3.14 / 172

= .0365 m

no of wave lengths over whole length of string

= 1.5 / .0365

= 41

d )

equation for waves traveling down the string

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PLEASE HELP
Sergeu [11.5K]

The vertical component of the initial velocity is v_0_y = \frac{y}{t} + \frac{1}{2} gt

The horizontal component of the initial velocity is v_0_x = \frac{x}{t}

The horizontal displacement when the object reaches maximum height is X = \frac{xy}{gt^2} + \frac{x}{2}

The given parameters;

the horizontal displacement of the object, = x

the vertical displacement of the object, = y

acceleration due to gravity, = g

time of motion, = t

The vertical component of the initial velocity is given as;

y = v_0_yt - \frac{1}{2} gt^2\\\\v_0_yt = y + \frac{1}{2} gt^2\\\\v_0_y = \frac{y}{t} + \frac{1}{2} gt

The horizontal component of the initial velocity is calculated as;

x = v_0_xt\\\\v_0_x = \frac{x}{t}

The time to reach to the maximum height is calculated as;

T = \frac{v_f_y -v_0_y}{-g} \\\\T = \frac{-v_0_y}{-g} \\\\T = \frac{v_0_y}{g} \\\\T =  \frac{1}{g}  (v_0_y)\\\\T = \frac{1}{g} (\frac{y}{t} + \frac{1}{2} gt)\\\\T = \frac{y}{gt} + \frac{1}{2} t

The horizontal displacement when the object reaches maximum height is calculated as;

X= v_0_x \times T\\\\X= \frac{x}{t} \times (\frac{y}{gt} + \frac{1}{2} t)\\\\X = \frac{xy}{gt^2} + \frac{x}{2}

Learn more here: brainly.com/question/20689870

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