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Vika [28.1K]
3 years ago
6

A man is throwing a spear from the top of a cliff, 5 metres from ground, at a 20° angle towards a fish that is 35 metres away. a

t what Velocity must the person throw to hit the fish and how long will it take?​

Physics
1 answer:
svlad2 [7]3 years ago
6 0

Answer:

Horizontal component of the initial velocity=v ×cos20°

horizontal displacement= 35m

time taken. (t)=35÷(v×cos20°)

vertical component of the initial velocity= v×sin20°

vertical displacement= -5m(since it is opposite to the direction of the initial velocity. )

application of s=ut+1/2at×t. vertically,

-5=vsin20°×t -1/2 gt×t

-5= vsin20×35/vcos20-1/2×10×35×35/(vcos20×vcos20)

simplify further to obtain v and hence find the time taken

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An aluminum bar 600mm long, with diameter 40mm long has a hole drilled in the center of the bar.The hole is 30mm in diameter and
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Answer:

Total contraction on the Bar  = 1.22786 mm

Explanation:

Given that:

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Calculate the total contraction on the bar = ???

The relation used in  calculating the contraction on the bar is:

\delta L = \dfrac{P *L }{A*E}

The relation used in  calculating the total contraction on the bar can be expressed as :

Total contraction in the Bar = (contraction in part of bar without hole + contraction in part of bar with hole)

i.e

Total contraction on the Bar = \dfrac{P *L_1 }{A_1*E} +  \dfrac{P *L_2 }{A_2 *E}

Let's find the area of cross section without the hole and with the hole

Area of cross section without the hole is :

Using A = πd²/4

A = π (40)²/4

A = 1256.64 mm²

Area of cross section with the hole is :

A = π (40²-30²)/4

A = 549.78 mm²

Total contraction on the Bar = \dfrac{P *L_1 }{A_1*E} +  \dfrac{P *L_2 }{A_2 *E}

Total contraction on the Bar  = \dfrac{180 *10^3 \N  }{85*10^3 \ N/mm^2} [\dfrac{500}{1256.64}+ \dfrac{100}{549.78}]

Total contraction on the Bar  = 2.117( 0.398 + 0.182)

Total contraction on the Bar  = 2.117*(0.58)

Total contraction on the Bar  = 1.22786 mm

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