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3 years ago

HELP: I’ve been stuck on this problem for a while now.

Physics

1 answer:

Arlecino [84]3 years ago

3 0

Answer:

a.work done=force *displacement

=500N*46m

=23000 Joule

b.power=work done/time taken

=23000/25

=920 watt

c.GPE=m*g*h(m=mass,g=gravity due to acceleration,h=height)

=60kg*9.8m/s*14m

=8232 joule

Explanation:

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Speed of the block at the bottom of the incline: 5.42 m/s

The first part of the problem can be solved by using the law of conservation of energy. Since the ramp is frictionless, the initial gravitational potential energy of the block at the top of the ramp is converted into kinetic energy at the bottom:

$mgh = \frac{1}{2}mv^2$ (1)

where

m is the mass of the block

g = 9.8 m/s^2 is the acceleration of gravity

h is the initial height of the block

v is the speed of the block at the bottom

The initial height of the block is equal to the height of the ramp, so

$h=L sin \theta$ (2)

where

L = 3.00 m is the length of the ramp

$\theta=30^{\circ}$ is the angle of the ramp

Substituting (2) into (1) and re-arranging the equation, we find the speed

$2gL sin \theta = v^2$

$v=\sqrt{2gL sin \theta}=\sqrt{2(9.8)(3.00)sin 30^{\circ}}=5.42 m/s$

Coefficient of kinetic friction between the floor and the block: 0.3

In the second part of the motion, the block is slowed down by friction along the flat surface. According to the work-energy theorem, the work done by friction is equal to the change in kinetic energy of the block:

$W=\Delta K=K_f -K_i$

where

W is the work done by friction

Kf is the final kinetic energy of the block, which is zero since the block comes to rest

$K_i = \frac{1}{2}mv^2$ is the initial kinetic energy of the block, where

m = 10.0 kg is the mass of the block

v = 5.42 m/s is its initial speed

Substituting into the equation, we find

$W=-\frac{1}{2}mv^2=-\frac{1}{2}(10.0)(5.42)^2=-146.9 J$

and the work is negative, since the direction of the force of friction is opposite to the direction of motion of the block.

Now we can rewrite the work as the product between the force of friction and the displacement of the block:

$W=-F_f d = - \mu mg d$

where

$\mu$ is the coefficient of friction

d = 5.00 m is the displacement of the block

Solving for $\mu$ ,

$\mu = - \frac{W}{mgd}=-\frac{-146.9}{(10.0)(9.8)(5.00)}=0.3$

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Explanation:

The process feed is 1000 kg/h of syrup with 30% solids. 1/5 of this (200 kg/h) is diverted, and the remaining 800 kg/h is fed to the evaporator. The exit stream from the evaporator is mixed with the 200 kg/h and the resulting stream is 55% solids. This is fed to the dyer, and the final result is 5% solids.

a) The amount of solids going into the evaporator = the amount of solids coming out of the evaporator.

0.30 (800 kg/h) = xy

The solids going into the junction = the solids coming out of the junction.

0.30 (200 kg/h) + xy = 0.55 (200 kg/h + x)

Substitute.

0.30 (200 kg/h) + 0.30 (800 kg/h) = 0.55 (200 kg/h + x)

0.30 (1000 kg/h) = 0.55 (200 kg/h + x)

300 kg/h = 110 kg/h + 0.55x

x = 345.45 kg/h

The concentration is:

0.30 (800 kg/h) = y (345.45 kg/h)

y = 0.695

b) 200 + x = 545.45 kg/h

c) Solids going into the dryer = solids coming out of the dryer

0.55 (545.45 kg/h) = 0.95 z

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