AYUSH NARAYAN S.

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Web Design

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Question:

Explain how can you align picture so that one may be higher or lower than the other?

AYUSH NARAYAN S.

Answer:

In order to align picture so that one may be higher or lower than the other, use the align statement in your IMG SRC tag. < IMG SRC = “http://www.xyz.com/chguy.gif” align=top > Also, you can use align=top or you can do align=middle/bottom

Chemistry

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Question:

What weight of AgCI will be precipitated when a solution containing 4.77 g of NaCI is added to a solution of 5.77 g of AgNO3?

AYUSH NARAYAN S.

Answer:

Write the balance chemical equation and use mole concept for limiting reagent. AgNO3 + NaCI՜ NaNO3 + AgCIO 170 g58.5 g 143.5 g From the given data, we find AgNO3 is limiting reagent as NaCI is in excess. since 170.0 g of AgNO3 precipitates AgCI = 143.5 g 5.77 g of AgNO3 precipitates AgCI = 143.5/170 * 5.77 = 4.87 g

Physics

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Question:

A 26-ton Navy jet requires an air speed of 280 ft/s for lift-off. Its own engine develops a thrust of 24,000 lb. The jet is to take off from an aircraft carrier with a 300-ft flight deck. What force must be exerted by the catapult of the carrier? Assume that the catapult and the jet’s engine each exert a constant force over the 300-ft takeoff distance.

AYUSH NARAYAN S.

Answer:

Concept:- Force acting on the body (F) is equal to the product of mas of the body (m) and acceleration of the body (a). So, F = ma Weight W of the object is equal to the mass m of the object times of the free fall acceleration g. W = mg So the mass m of the object would be, m = W/g To obtain force F in terms of weight W, substitute W/g for m in the equation F = ma, F = ma = (W/g) (a) Time t taken by a body to travel a distance x with average velocity vav will be, t = x/ vav The deceleration a is equal to the rate of change of velocity, a = Δv/Δt Solution:- To find the time t for the plane to travel 300 ft, substitute 300 ft for x and 140 ft /s for vav in the equation t = x/ vav, t = x/ vav = (300 ft)/(140 ft/s) = 2.14 s To obtain the acceleration a, substitute 280 ft/s for Δv and 2.14 s for Δt in the equation a = Δv/Δt, a = Δv/Δt = (280 ft/s)/(2.14 s) = 130 ft/s2 To find out the net force F on the plane, substitute 52,000 lb for W, 130 ft/s2 for a and 32 ft/s2 for free fall acceleration g in the equation F =(W/g) (a), F =(W/g) (a) = (52000 lb)( 130 ft/s2)/( 32 ft/s2) = 2.1×105 lb The force exerted by the catapult Fc is equal to the difference of the net force F on the plane and the thrust develop by the own engine T. So, Fc = F-T To obtain the force exerted by the catapult Fc, substitute 2.1×105 lb for F and 24,000 lb for T in the equation Fc = F-T, Fc = F-T = (2.1×105 lb) - (24,000 lb) = (2.1×105 lb) - (2.4×104 lb) =1.86×105 lb From the above observation we conclude that, the force exerted by the catapult Fc would be 1.86×105 lb.

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