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4 years ago

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Multiple-Concept Example 8 presents an approach to problems of this kind. The hydraulic oil in a car lift has a density of 8.30

???? 10 2 kg/m3. The weight of the input piston is negligible. The radii of the input piston and output plunger are 7.70 ???? 10????3 m and 0.125 m, respectively. What input force F is needed to support the 24 500-N combined weight of a car and the output plunger, when (a) the bottom surfaces of the piston and plunger are at the same level, and (b) the bottom surface of the output plunger is 1.30 m above that of the input piston?

1 answer:

Lunna [17]4 years ago

4 0

Answer:

a. = 93 N

b. = 94.9 N

Explanation:

Answer:

Explanation:

from the question we were given the following:

density of oil (ρ) = 8.3 x $10^{2}$ $\frac{kg}{m^{3} }$

radius of the output plunger (R) = 0.125 m

radius of the input piston (r) = 7.70 x $10^{-3}$ m = 0.0077 m

output force (F2) = 24500 N

input force (F1) = ?

acceleration due to gravity (g) = 9.8 $\frac{m}{s^{2} }$

difference in height of the plunger and piston (h) = 1.3 m

we can find the required input force when the piston and the plunger are at the same height using the equation below

$\frac{F2}{A2}$ = $\frac{F1}{A1 }$

F1 = ( $\frac{F2}{A2 }$ ) x A1

where A1 is the area of the input piston and A2 is the area of the output plunger

Area = π x $radius^{2}$

A2 = π x $0.125^{2}$ = 0.049

A1 = π x $0.0077^{2}$ = 0.000186

recall that from above F1 = ( $\frac{F2}{A2 }$ ) x A1

F1 = ( $\frac{24500}{0.049 }$ ) x 0.000186

F1 = 93 N

we can find the required input force when the height of the piston and the plunger are 1.3 m apart using the equation below

P2 = P1 + ρgh

where P = pressure = $\frac{F}{ A }$

therefore the equation above now becomes

$\frac{F2}{ A2 }$ = $\frac{F1}{ A1 }$ + ρgh

F2 = ( $\frac{F1}{ A1 }$ + ρgh ) x A2

F2 = ( $\frac{24500}{ 0.000186 }$ + ( 8.3 x $10^{2}$ [tex] x 9.8 x 1.3 ) ) x 0.049

F2 = 94.9 N

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If your chunk of gold weighed 1 N in which case would you have the largest mass of gold?

kotykmax [81]

Ah ha ! Very interesting question.
Thought-provoking, even.

You have something that weighs 1 Newton, and you want to know
the situation in which the object would have the greatest mass.

Weight = (mass) x (local gravity)

Mass = (weight) / (local gravity)

Mass = (1 Newton) / (local gravity)

"Local gravity" is the denominator of the fraction, so the fraction
has its greatest value when 'local gravity' is smallest. This is the
clue that gives it away.

If somebody offers you 1 chunk of gold that weighs 1 Newton,
you say to him:

"Fine ! Great ! Golly gee, that's sure generous of you.
But before you start weighing the chunk to give me, I want you
to take your gold and your scale to Pluto, and weigh my chunk
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The local acceleration of gravity on Pluto is 0.62 m/s² ,
but on Earth, it's 9.81 m/s.

So if he weighs 1 Newton of gold for you on Pluto, its mass will be
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That's almost 3.6 pounds of gold, worth over $57,000 !

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4 years ago

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Strike441 [17]

Answer:

a.

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3 years ago

Read 2 more answers

Explain how polarization of a cell increases the cell's internal resistance.<br>(2<br>2.

Mandarinka [93]

Answer:

Explanation: The chemical action that occurs in the cell while the current is flowing causes hydrogen bubbles to form on the surface of the anode. This action is called POLARIZATION. Some hydrogen bubbles rise to the surface of the electrolyte and escape into the air, some remain on the surface of the anode. If enough bubbles remain around the anode, the bubbles form a barrier that increases internal resistance. When the internal resistance of the cell increases, the output current is decreased and the voltage of the cell also decreases.

A cell that is heavily polarized has no useful output. There are several methods to prevent polarization or to depolarize the cell.

One method uses a vent on the cell to permit the hydrogen to escape into the air. A disadvantage of this method is that hydrogen is not available to reform into the electrolyte during recharging. This problem is solved by adding water to the electrolyte, such as in an automobile battery. A second method is to use material that is rich in oxygen, such as manganese dioxide, which supplies free oxygen to combine with the hydrogen and form water.

A third method is to use a material that will absorb the hydrogen, such as calcium. The calcium releases hydrogen during the charging process. All three methods remove enough hydrogen so that the cell is practically free from polarization.

LOCAL ACTION

When the external circuit is removed, the current ceases to flow, and, theoretically, all chemical action within the cell stops. However, commercial zinc contains many impurities, such as iron, carbon, lead, and arsenic. These impurities form many small electrical cells within the zinc electrode in which current flows between the zinc and its impurities. Thus, the chemical action continues even though the cell itself is not connected to a load.

Local action may be prevented by using pure zinc (which is not practical), by coating the zinc with mercury, or by adding a small percentage of mercury to the zinc during the manufacturing process. The treatment of the zinc with mercury is called amalgamating (mixing) the zinc. Since mercury is many times heavier than an equal volume of water, small particles of impurities weighing less than mercury will float to the surface of the mercury. The removal of these impurities from the zinc prevents local action. The mercury is not readily acted upon by the acid. When the cell is delivering current to a load, the mercury continues to act on the impurities in the zinc. This causes the impurities to leave the surface of the zinc electrode and float to the surface of the mercury. This process greatly increases the storage life of the cell.

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3 years ago

Which of the following is not affect by muscles a blood temperature b digestion c hair growth d speech

Murrr4er [49]

Answer:

c

Explanation:

i would think c would be correct because

blood temperature IS affected by muscles

digestion IS affected by muscles

speech IS affected by muscles

so therefore <em>hair growth</em> IS NOT affected by muscles

<h2><u><em>Please mark Brainliest</em></u></h2>

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3 years ago

A 4-kg toy car with a speed of 5 m/s collides head-on with a stationary 1-kg car. After the collision, the cars are locked toget

mihalych1998 [28]

Kinetic energy lost in collision is 10 J.

<u>Explanation:</u>

Given,

Mass, $m_{1}$ = 4 kg

Speed, $v_{1}$ = 5 m/s

$m_{2}$ = 1 kg

$v_{2}$ = 0

Speed after collision = 4 m/s

Kinetic energy lost, K×E = ?

During collision, momentum is conserved.

Before collision, the kinetic energy is

$\frac{1}{2} m1 (v1)^2 + \frac{1}{2} m2(v2)^2$

By plugging in the values we get,

$KE = \frac{1}{2} * 4 * (5)^2 + \frac{1}{2} * 1 * (0)^2\\\\KE = \frac{1}{2} * 4 * 25 + 0\\\\$

K×E = 50 J

Therefore, kinetic energy before collision is 50 J

Kinetic energy after collision:

$KE = \frac{1}{2} (4 + 1) * (4)^2 + KE(lost)$

$KE = 40J + KE(lost)$

Since,

Initial Kinetic energy = Final kinetic energy

50 J = 40 J + K×E(lost)

K×E(lost) = 50 J - 40 J

K×E(lost) = 10 J

Therefore, kinetic energy lost in collision is 10 J.

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4 years ago