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