Answer to Question #92136 in Physics for apon

Question #92136
Ques1: .Fielder ‘F’ threw a cricket ball at a constant speed of 15 m/s and hit the stumps from a distance 30 m. The batsman ‘B’ was 15 m from the stumps and had an initial speed of 6 m/s when ‘F’ threw the ball. If ‘B’ ran with an acceleration of 1.0 m/s2, determine if ‘B’ would be ‘run out’.?
Ques 2:. If a 25 kg load is applied on Spring A, it elongates 0.25 m, but it elongates 0.30 m after being attached to another Spring B in series. Determine the spring constant of both springs.
1
Expert's answer
2019-07-30T09:05:07-0400

1.


t=3015=2 st=\frac{30}{15}=2\ s

d=vt+0.5at2=(6)(2)+0.5(1)(2)2=14 m.d=vt+0.5at^2=(6)(2)+0.5(1)(2)^2=14\ m.

dd<D=15 m. Thus, ‘B’ would not be ‘run out’.

2.


kaxa=mgk_a x_a=mg

ka=mgxa=(25)(9.8)0.25=980Nmk_a=\frac{mg}{x_a}=\frac{(25)(9.8)}{0.25}=980\frac{N}{m}

K=mgX=(25)(9.8)0.3=817NmK=\frac{mg}{X}=\frac{(25)(9.8)}{0.3}=817\frac{N}{m}

So,


1K=1ka+1kb\frac{1}{K}=\frac{1}{k_a}+\frac{1}{k_b}

1817=1980+1kb\frac{1}{817}=\frac{1}{980}+\frac{1}{k_b}

kb=4900Nmk_b=4900\frac{N}{m}


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