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enot [183]
2 years ago
11

What is the efficiency of an engine that puts out 250 J of work for every 500 J of heat put into it?

Physics
1 answer:
Oliga [24]2 years ago
5 0

If you put 300 J of heat into an engine with an efficiency of 0.35, how much work can be done? 3. How much energy must be put into an engine with an efficiency of 0.6 if 270 J of work are required? 4. An engine with an efficiency of 0.425 uses 1200 J of energy. Find the amount of energy wasted by the engine. 5. Calculate the efficiency of an engine operating between temperatures of 258 K and 600 K 6. An engine runs with its exhaust (cold) reservoir at a temperature of 200 K. To what temperature should the input (hot) temperature be set if an efficiency of 0.8 is desired? 7. Complete the following table of temperatures. Fahrenheit Celsius Kelvin 213 15 98.6 75 408

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Answer:

2Mg + O2 → 2MgO

Explanation:

this is what your looking for

8 0
3 years ago
If 300. mL of water are poured into the measuring cup, the volume reading is 10.1 oz . This indicates that 300. mL and 10.1 oz a
tigry1 [53]

Answer:

Milliliters to Ounces Conversions

some results rounded

mL    - fl oz

200.00 6.7628

200.01 6.7631

200.02 6.7635

200.03 6.7638

200.04 6.7642

200.05 6.7645

200.06 6.7648

200.07 6.7652

200.08 6.7655

200.09 6.7658

200.10 6.7662

200.11 6.7665

200.12 6.7669

200.13 6.7672

200.14 6.7675

200.15 6.7679

200.16 6.7682

200.17 6.7686

200.18 6.7689

200.19 6.7692

200.20 6.7696

200.21 6.7699

200.22 6.7702

200.23 6.7706

200.24 6.7709

mL fl oz

200.25 6.7713

200.26 6.7716

200.27 6.7719

200.28 6.7723

200.29 6.7726

200.30 6.7729

200.31 6.7733

200.32 6.7736

200.33 6.7740

200.34 6.7743

200.35 6.7746

200.36 6.7750

200.37 6.7753

200.38 6.7757

200.39 6.7760

200.40 6.7763

200.41 6.7767

200.42 6.7770

200.43 6.7773

200.44 6.7777

200.45 6.7780

200.46 6.7784

200.47 6.7787

200.48 6.7790

200.49 6.7794

mL fl oz

200.50 6.7797

200.51 6.7800

200.52 6.7804

200.53 6.7807

200.54 6.7811

200.55 6.7814

200.56 6.7817

200.57 6.7821

200.58 6.7824

200.59 6.7828

200.60 6.7831

200.61 6.7834

200.62 6.7838

200.63 6.7841

200.64 6.7844

200.65 6.7848

200.66 6.7851

200.67 6.7855

200.68 6.7858

200.69 6.7861

200.70 6.7865

200.71 6.7868

200.72 6.7872

200.73 6.7875

200.74 6.7878

mL fl oz

200.75 6.7882

200.76 6.7885

200.77 6.7888

200.78 6.7892

200.79 6.7895

200.80 6.7899

200.81 6.7902

200.82 6.7905

200.83 6.7909

200.84 6.7912

200.85 6.7915

200.86 6.7919

200.87 6.7922

200.88 6.7926

200.89 6.7929

200.90 6.7932

200.91 6.7936

200.92 6.7939

200.93 6.7943

200.94 6.7946

200.95 6.7949

200.96 6.7953

200.97 6.7956

200.98 6.7959

200.99 6.7963

Explanation:

5 0
3 years ago
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Step 2: Apply NEwton's second law Apply ∑Fy = may , what should ay be equal to, since the block doesn't move in the y direction
andrey2020 [161]

Answer:

∑Fy = 0, because there is no movement, N = m*g*cos (omega)

Explanation:

We can solve this problem with the help of a free body diagram where we show the respective forces in each one of the axes, y & x. The free-body diagram and the equations are in the image attached.

If the product of mass by acceleration is zero, we must clear the normal force of the equation obtained. The acceleration is equal to zero because there is no movement on the Y-axis.

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8 0
3 years ago
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