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Rohit S.

An online and offline tutor having 4 years of tutoring experience.

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Mechanical Engineering

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Question:

Q2. The gas used in a gas engine trial was tested. The gas is supplied at 11 cm of water column.If the column on the barometer reads 760mm of Hg, find the absolute pressure of the gas?

Rohit S.

Answer:

So, we have been given that the barometric pressure or the atmospheric pressure is 760 mm Hg Now, So, h = 760 mm Hg, which means, $$p_{atm}$$ = ρgh = 13.6 X $$10^3$$ X 9.81 X 760/1000 = 101396.16 N/$$m^2$$ Now, the gas is supplied at 11 cm of water column so the gauge pressure can be found out from this as h = 11 cm So $$p_{gauge}$$ = ρgh = 1000*9.81*11/100 = 1079.1 N/$$m^2$$ Now the gauge always measures the pressure difference between the atmosphere and the absolute, so $$p_{gauge}$$ = $$p_{abs}$$ - $$p_{atm}$$ So, $$p_{abs}$$ = $$p_{gauge}$$ + $$p_{atm}$$ $$p_{abs}$$ = 1079.1 + 101396.16 = 102475.26 N/$$m^2$$ Sol.

Algebra

TutorMe

Question:

The length of a football field is 9/4 times its width. If the perimeter of the football field is 1040 feet, Find the length and width of the football field?

Rohit S.

Answer:

So here we have been given two facts: Given : 1) Length = 9/4 x Width 2) Perimeter = 1040 feet So, to solve this problem we will assume that let the length of the field be "l" feet and width of the field be "w"feet. This means that 1) l = 1040 x w , - let us name this equation as (i) 2) Perimeter = 1040 feet but we know perimeter = 2x (length + width) = 2 (l+w) 1040 = 2(l + w) , - let us name this equation as (ii) So, putting eqn. (i) in eqn (ii), we get 1040 = 2[( 9/4 x w) + w] 520 = 13/4 x w w = 520 x 4/13 w = 160 feet l = 9/4 w which means l = 9/4 x 160 l = 360 feet Solution : Length of football field = 360 feet Width of football field = 160 feet

Physics

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Question:

A ball attached to a weightless spring with a stiffness of 300 N/m is stretched to a length of 2 m from its mean position. The ball weighs 100 g and is released from this position. Find the velocity of the ball immediately after its release?

Rohit S.

Answer:

We can solve this problem by analysing the energies involved in the system. First you can see that when the ball is in the stretched position the ball only has potential energy. This Potential Energy can be found out by the equation PE = $$ 1/2 (kx^2)$$ where k= spring stiffness and x= extension of the spring from the mean position. Now, putting the values into the above formula we get: PE = 1/2 * 300* 2 = 300 J Now at the position of just the release, all the potential energy of the ball will get converted into Kinetic Energy, so at this point PE = KE KE = $$1/2 mv^2$$ So, if PE = KE it means, 300 = 1/2 * 0.1 kg * $$v^2$$ So, $$v^2$$ = 6000 v = 77.46 $$m^2/s$$

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