41 to 44 A cork floats in water in a bucket resting on the floor of an elevator.

Question

41 to 44

A cork floats in water in a bucket resting on the floor of an elevator. The elevator then accelerates upward. During the acceleration: A) the cork is immersed more B) the cork is immersed less C) the cork is immersed the same amount D) at first the cork is immersed less but as the elevator speeds up it is immersed more E) at first the cork is immersed more but as the elevator speeds up it is immersed less A Boston Red Sox baseball player catches a ball of mass m that is moving toward him with speed v. While bringing the ball to rest, his hand moves back a distance d. Assuming constant deceleration, the horizontal force exerted on the ball by his hand is: A) mv/d B) mvd C)mv^2/d D) 2mv/d E) mv^2/(2d) A 5.0-kg cart is moving horizontally at 6.0 m/s. In order to change its speed to 10.0 m/s, the net work done on the cart must be: A) 40 J B) 90 J C) 160 J D) 400 J E) 550 J A non-viscous incompressible liquid is flowing through a horizontal pipe of constant cross-section. A) is zero B) depends on the fluid velocity D) depends on the cross-sectional area of the pipe E) depends on the heights of the pipe A phonograph record is dropped onto a freely spinning turntable. Then:

Explanation / Answer

41.) The fraction of the cork immersed is dependent on its density and relative density to be more specific. It has not got any dependency on the acceleration due to gravity. When the elevator accelerates in the upward direction, the acceleration sure increases and hence the cork's weight. But we also need to note that the weight of the water (and hence the water displaced by the immersion of the cork) also increases accordingly and hence the fraction of the cork submerged will not change. So, effectively the cork is immersed te same amount.

42.) acceleration of the ball while coming to rest is a = (02 - v2 ) /2d

a = - v2/2d (negative sign indicates that it is retardation)

Now the force applied on the ball = mass x acceleration = m x ( - v2/2d ) = - mv2/2d

-ve sign just indicaties that the force is acting in the direction opposite to the direction of the motion of the ball.

So, the horizontal force exerted by the hand is mv2/2d

43.) The net work done in this case must be equal to the change in the Kinetic energy of the cart.

Work done = KEfinal - KEinitial  = 0.5 m vf2  -  0.5 m vi2     

where vf , the final velocity is 10 m/s and vi , the initial velocity is 6 m/s

So, Work done W = 0.5 m ( vf2 - vi2 ) = 0.5 x 5 ( 102 - 62 ) = 0.5 x 5 x 36 = 90 Joules

44.) According to the Bernoulli equation :

P + 0.5v2 + gh = constant (across a streamline)

where g is the acceleration due to gravity which is constant

and , the density is also constant (incompressible liquid)

According to the equation of continuity, since the area of cross-section doesn't change, the velocity also will not change and remain constant.

and h, the height is also constant since it is a HORIZONTAL pipe.

So, essentially if we write Bernoulli equation as

P1 + 0.5v12 + gh1 = P2 + 0.5v22 + gh2

then as per our arguments, v1 = v2   , h1 = h2  and g and are anyways constants.

So, P1 must be equal to P2

Hence the pressure drop along the pipeline P1 - P2 = 0

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