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3 years ago

QUESTION 1 If you do 1,000 J of work on a system and remove 500 J of heat from it, what is the change in its internal energy?

2 answers:

7 0

1. U = Q + W
U = -500 + 1000
U = 500 J

2. The first law of thermodynamic is about the law of conservation of energy where energy in should be equal to energy out.

3. It is the windmill that does not transform energy from heat to mechanical instead it is the transforms the opposite.

4. In a heat engine, work is used to transfer thermal energy from a hot reservoir to a cold one.

5. 5.00 × 10^4 J - 2.00 × 10^4 J = 3.00 × 10^4 J

6. To increase the work done, we raise the temperature of the cold reservoir.

denpristay [2]3 years ago

4 0

Sallyablow has them all correct except 4 and 6.

4. D
6. C

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Because the snow suddenly gets too slushy, you decide to carry your 100-n sled the rest of the way home. How much work do you do

soldier1979 [14.2K]

Work = (weight) x (distance lifted)

= (100 N) x (0.5 m)

= 50 joules.

If you carry the sled level all the way to your house, you're not moving it
against gravity.

Although you're huffing and puffing and sweating by the time you reach
your house, still, according to the Physics definition of work, you have
done NO work on the sled all the way home.

5 0

3 years ago

If a 140 lb. climber saved her potential energy as she descended from Mt. Everest (Elev. 29,029 ft) to Kathmandu (Elev. 4,600 ft

mamaluj [8]

Answer: 3217.79 hours.

Explanation:

Given, A 140 lb. climber saved her potential energy as she descended from Mt. Everest (Elev. 29,029 ft) to Kathmandu (Elev. 4,600 ft).

Power = 0.4 watt

Mass of climber = 140 lb

= 140 x 0.4535 kg [∵ 1 lb= 0.4535 kg]

⇒ Mass of climber (m) = 63.50 kg

Let $h_1=29,029\ ft= 8848.04\ m\ \ \ \ [ 1 ft=0.3048\ m ]$ and $h_2= 4,600 ft = 1402.08\ m$

Now, Energy saved = $mg(h_1-h_2)=(63.50)(9.8)(8848.04-1402.08)=4633620.91\ J$

$\text{Power}=\dfrac{\text{energy}}{\text{time}}\\\\\Rightarrow 0.4=\dfrac{4633620.91}{\text{time}}\\\\\Rightarrow\ \text{time}=\dfrac{4633620.91}{0.4}\approx11584052.28\text{ seconds}\\\\=\dfrac{11584052.28}{3600}\text{ hours}\ \ \ [\text{1 hour = 3600 seconds}]\\\\=3217.79\text{ hours}$

Hence, she can power her 0.4 watt flashlight for 3217.79 hours.

5 0

3 years ago

Vector C has a magnitude of 24.6 m and points in the − y ‑ direction. Vectors A and B both have positive y ‑ components, and mak

frez [133]

Answer:

A= 61.35

B= -44.40

Explanation:

1. Using the components method we have:

$A_{x}= A cos \alpha\\B_{x}= B cos \beta\\C_{x}= 0\\\\A_{y}= A sin \alpha\\B_{y}= B sin \beta\\C_{y}= 24.6\\$

Considering that the vector sum $A+B+C=0$ , then:

$|V|=\sqrt{V_{x}^{2} +V_{y}^{2} }=0$

Then:

$V_{x} ^{2} =0; V_{x} =0\\V_{y} ^{2} =0; V_{y} =0$

It means the value of x and y component is 0.

2. Determinate the equations that describe each component:

$V_{x}= A cos \alpha -B cos \beta=0 (1)\\V_{y}= A sin \alpha +B sin \beta - C=0 (2)$

Form Eq. (1):

$A=B \frac{cos \beta}{cos \alpha} (3)$

Replacing A in Eq. (2):

$(B\frac{cos \beta}{cos \alpha})(sin \alpha)+ B sin\beta-C=0\\(B\frac{cos \beta}{cos \alpha})(sin \alpha)+ B sin\beta-=C\\\\B(\frac{cos \beta. sin \alpha}{cos \alpha}+ sin\beta)=C\\B=C(\frac{cos \beta. sin \alpha}{cos \alpha}+ sin\beta)^{-1} (4)$

Replacing values of C, α and β in (4):

$B= 24.6 (\frac{(cos 27.7)(sin 44.9)}{cos 44.9}+sin 27.7)^{-1} \\B= -44.4$

Replacing value of B in (3)

$A=-44.40\frac{cos 27.7}{sin 49.9} \\A= 61.35$

5 0

3 years ago

An object, with mass 32 kg and speed 26 m/s relative to an observer, explodes into two pieces, one 5 times as massive as the oth

Brut [27]

Answer:

$\Delta K = 2164.053\,J$

Explanation:

Let consider the observer as an inertial reference frame. The object is modelled after the Principle of Momentum Conservation:

$(32\,kg)\cdot (26\,\frac{m}{s} ) = (5.333\,kg)\cdot (0\,\frac{m}{s} )+(26.665\,kg )\cdot v$

The speed of the more massive piece is:

$v = 31.202\,\frac{m}{s}$

The kinetic energy added to the system is:

$\Delta K = \frac{1}{2}\cdot [(5.333\,kg)\cdot (0\,\frac{m}{s} )^{2}+(26.665\,kg )\cdot (31.202\,\frac{m}{s} )^{2}]-\frac{1}{2}\cdot (32\,kg)\cdot (26\,\frac{m}{s} )^{2}$