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Otrada [13]
4 years ago
13

A block with mass ma = 14.0 kg on a smooth horizontal surface is connected by a thin cord that passes over a pulley to a second

block with mass mb = 6.0 kg which hangs vertically.determine the magnitude of the acceleration of the system.
Physics
1 answer:
Drupady [299]4 years ago
7 0
<span>We can assume that the horizontal surface has no friction and the pulley is massless. We can use Newton's second law to set up an equation. F = Ma F is the net force M is the total mass of the system a is the acceleration a = F / M a = (mb)(g) / (ma + mb) a = (6.0 kg)(9.80 m/s^2) / (6.0 kg + 14.0 kg) a = 58.8 N / 20 kg a = 2.94 m/s^2 The magnitude of the acceleration of the system is 2.94 m/s^2</span>
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The production of heat by metabolic processes takes place throughout the volume of an animal, but loss of heat takes place only
Sever21 [200]

To solve this problem we will apply the concepts related to the change in length in proportion to the area and volume. We will define the states of the lengths in their final and initial state and later with the given relationship, we will extrapolate these measures to the area and volume

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\text{Initial Length} = L

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The final measures

\text{Final Length} = L_f

\text{Final surface area} = 6L_f^2

\text{Final Volume} = L_f^3

Given,

\frac{(SA)_f}{(SA)_i} = 2

Now applying the same relation we have that

(\frac{L_f}{L_i})^2 = 2

\frac{L_f}{L_i} = \sqrt{2}

The relation with volume would be

\frac{(Volume)_f}{(Volume)_i} = (\frac{L_f}{L_i})^3

\frac{(Volume)_f}{(Volume)_i} = (\sqrt{2})^3

\frac{(Volume)_f}{(Volume)_i} = (2\sqrt{2})

\frac{(Volume)_f}{(Volume)_i} = 2.83

Volume of the cube change by a factor of 2.83

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4 years ago
Brad has three beakers of water. Each beaker contains 500 mL of water. The temperature of the water in the first beaker is 40°C.
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3 years ago
An automobile engine takes in 4000 j of heat and performs 1100 j of mechanical work in each cycle. (a) calculate the engine's ef
Semmy [17]
(a) The efficiency of an engine is defined as the ratio between the work done by the engine and the heat it takes in:
\eta= \frac{W}{Q_{in}}
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\eta= \frac{1100 J}{4000 J}=0.275 = 27.5 \%

(b) The heat taken by the machine is 4000 J; of this amount of heat, only 1100 J are converted into useful work. This means that the rest of the heat is wasted. Therefore, the wasted heat is the difference between the heat in input and the work done by the engine:
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3 years ago
Which of the following simple machines is designed primarily to redirect a force rather than reduce the effort required to move
lys-0071 [83]

Answer:

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Explanation:

7 0
3 years ago
A student pushed a 100 N bicycle over a distance of 15 m in 5 s. calculate the power generated.
Umnica [9.8K]

The catch in this one is:  We don't know how much <u>force</u> the student used to push the bike.  

It wasn't necessarily the 100N.  That's just the weight of the bike. But you know that you can push a car, a wagon, or a bicycle hard, you can push it not so hard, you can give it a little push, you can give it a big push, you can push it strong, you can push it weak, you can push it medium.  The harder you push, the more it'll accelerate, but it's completely up to you how hard you want to push.  That's what's so great about wheels !  That's why they were such a great invention ! This is where I made my biggest mistake. This guy came into my store one day and said he's got this great invention, it's definitely going to take off, it'll be a winner for sure, he called it a "wheel".  I looked at it, I turned it over and I looked on all sides. I thought it was too simple.  I didn't know then it was elegant. I threw him out.  I was so dumb.  I could have invested money in that guy, today I would have probably more than a hundred dollars.

Anyway, can we figure out how much force the student used to push with ?  Stay tuned:

-- The bike covered 15 meters in 5 seconds.  Its average speed during the whole push was (15m/5s) = 3 meters/sec.

-- If the bike started out with no speed, and its average speed was 3 m/s, then it must have been moving at 6 m/s at the end of the push.

-- If its speed increased from zero to 6 m/s in 5 seconds, then its acceleration was (6m/s / 5 sec) = 1.2 m/s²

-- The bike's weight is 100N.  

(mass) x (gravity) = 100N

Bikemass = (100N) / (9.8 m/s²)

Bikemass = 10.2 kilograms

-- F = m A

Force = (mass) x (acceleration)

Force = (10.2 kg) x (1.2 m/s²)

Force = 12.24 N

-- Work = (force) x (distance)

Work = (12.24 N) x (15 m)

Work = 183.67 Joules

-- Power = (work done) / (time to do the work)

Power = (183.67 joules) / (5 seconds)

<em>Power = 36.73 watts</em>

7 0
3 years ago
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