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3 years ago

15

The specific heat of a liquid Y is 4.2 cal/g degrees Celsius. A sample grams of this liquid at 137 K is heated to 265K. The liqu

id absorbs. 7.82 kcal. What is the sample of liquid in grams?

1 answer:

sergiy2304 [10]3 years ago

8 0

Answer:

54

Explanation:

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When did the ball fly farther—when it hit the moving bat or the bat that did not move?

8_murik_8 [283]

Answer:

when it hit the moving bat

Explanation:

force equals mass times acceleration which means the moving bat will add more force to the ball.

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3 years ago

A vertical spring gun is used to launch balls into the air. If the spring is compressed by 4.9 cm, the ball of mass 5.5 g is lau

AleksandrR [38]

We know, by conservation of energy :

$\dfrac{kx^2}{2}=mgh$

Therefore,

$\dfrac{x_1^2}{x_2^2}=\dfrac{h_1}{h_2}$

Putting given values, we get :

$\dfrac{x_1^2}{x_2^2}=\dfrac{h_1}{h_2}\\\\\dfrac{4.9^2}{x_2^2}=\dfrac{50.2}{2\times 50.2}\\\\x_2^2=2\times 4.9^2\\\\x_2 = 4.9\times \sqrt{2}\\\\x_2=6.93\ cm$

Therefore, the spring be compressed to 6.93 cm to send the ball twice as high.

Hence, this is the required solution.

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2 years ago

Sliva [168]

The stopwatch will be the most useful in determining the kinetic energy of a 50 g battery- powered car traveling a distance of 10 m.

<h3>What is kinetic energy?</h3>

Kinetic energy is the energy of a body possessed due to motion.

This means that for an object to possess kinetic energy, it must be in motion.

The kinetic energy is measured in Joules, which is a product of the mass of the substance and the time taken to travel a distance.

A stopwatch is an instrument used to measure time as one of the components of kinetic energy.

Therefore, the stopwatch will be the most useful in determining the kinetic energy of a 50 g battery- powered car traveling a distance of 10 m.

Learn more about kinetic energy at: brainly.com/question/12669551

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2 years ago

A catamaran with a mass of 5.44×10^3 kg is moving at 12 knots. How much work is required to increase the speed to 16 knots? (One

Andre45 [30]

The work that is required to increase the speed to 16 knots is 14,176.47 Joules

If a catamaran with a mass of 5.44×10^3 kg is moving at 12 knots, hence;

5.44×10^3 kg = 12 knots

For an increased speed to 16knots, we will have:

x = 16knots

Divide both expressions

$\frac{5.44 \times 10^3}{x} = \frac{12}{16}\\12x = 16 \times 5.44 \times 10^3\\x = 7.23\times 10^3kg\\$

To get the required work done, we will divide the mass by the speed of one knot to have:

$w=\frac{7230}{0.51}\\w= 14,176.47Joules$

Hence the work that is required to increase the speed to 16 knots is 14,176.47 Joules

Learn more here: brainly.com/question/25573786

8 0

2 years ago

A heat engine with 0.300 mol of a monatomic ideal gas initially fills a 1000 cm3 cylinder at 500 K . The gas goes through the fo

LuckyWell [14K]

Complete Question:

A heat engine with 0.300 mol of a monatomic ideal gas initially fills a 1000 cm3 cylinder at 500 K . The gas goes through the following closed cycle: - Isothermal expansion to 5000 cm3. - Isochoric cooling to 400 K . - Isothermal compression to 1000 cm3. - Isochoric heating to 500 K .

a) what is the work for one cycle

b) what is the thermal efficiency

Answer:

a) Work done for 1 cycle = 402.13

b) Thermal efficiency = 20%

Explanation:

Number of moles, n = 0.300 mol

Initial Volume, V₁ = 1000 cm³

Temperature, T = 500 K

Isothermal expansion to 5000 cm³

Final volume, V₂ = 5000 cm³

R = 8.314 J/ mol.K

Work done, W = nRT ln(V₂/V₁)

W = (0.3 * 8.314 * 500) * ln(5000/1000)

W = 1247.1 * ln5

W₁ = 2007.13 J

Isochoric cooling

In an Isochoric process, volume is constant i.e. V₂ = V₁ = V

W = nRT ln(V/V)

But ln(V/V) = ln 1 = 0

Work done, W₂ = 0 Joules

Isothermal Compression to 1000 cm³

V₂ = 1000 cm³

V₁ = 5000 cm³

W = nRT ln(V₂/V₁)

W = 0.3 * 8.314 * 400 ln(1000/5000)

W₃ = -1605 J

Isochoric heating to 500 K

Since there is no change in volume, no work is done

W₄ = 0 J

a) Work done for 1 cycle

W = W₁ + W₂ + W₃ + W₄

W = 2007.13 + 0 + 0 -1605+0

W = 402.13 Joules

b) Thermal efficiency

Thermal efficiency = (Net workdone for 1 cycle)/(Heat absorbed)

Heat absorbed = Work done due to thermal expansion = 2007.13 J

Thermal efficiency = 402.13/2007.13

Thermal efficiency = 0.2

Thermal efficiency = 0.2 * 100% = 20 %

3 0

3 years ago