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Kaylis [27]
3 years ago
5

Which example best demonstrates how unbalanced forces change the speed of an object's motion?

Physics
1 answer:
exis [7]3 years ago
4 0

I don’t see a picture but unbalanced forces could be two boys pushing with a combined force of 400 Newton’s but the surface of what the box is laying on being 600 meaning since the ground is creating a higher force in the form of friction it will slow the boys down. When forces are unbalanced it means that the object can not be still or moving at a constant speed  when one force is greater by a significant amount the object either slows quickly or accelerates fast depending on which factor is greater.

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the height of seven falls in colorado is 5/2 the height of twin falls in idahi. the sum of the two heights is 420 ft. find the h
Masja [62]
Let height of twin falls = x
height of seven falls = 2.5x

x + 2.5x = 420
3.5x = 420
x = 420/3.5 = 120

so twin falls = x = 120 ft
seven falls = 2.5x = 300 ft
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Most radiation exposure comes from
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How does this diagram demonstrate the law of superposition?
Mandarinka [93]

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Explanation:

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2 years ago
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How much work is required to compress 5.05 mol of air at 19.5°C and 1.00 atm to one-eleventh of the original volume by an isothe
Rus_ich [418]

Explanation:

(a)  For an isothermal process, work done is represented as follows.

             W = -nRT ln(\frac{V_{2}}{V_{1}})

Putting the given values into the above formula as follows.

        W = -nRT ln(\frac{V_{2}}{V_{1}})

             = - 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})

             = -12280.82 \times ln (0.09)

             = -12280.82 \times -2.41

             = 29596.78 J

or,         = 29.596 kJ       (as 1 kJ = 1000 J)

Therefore, the required work is 29.596 kJ.

(b) For an adiabatic process, work done is as follows.

         W = \frac{P_{1}V^{\gamma}_{1}(V^{1-\gamma}_{2} - V(1-\gamma)_{1})}{(1 - \gamma)}

              = \frac{-nRT_{1}(11^{\gamma - 1} - 1)}{1 - \gamma}

              = \frac{-5.05 \times 8.314 J/mol K \times 292.5 (11^{1.4 - 1} - 1)}{1 - 1.4}

              = 49.41 kJ

Therefore, work required to produce the same compression in an adiabatic process is 49.41 kJ.

(c)   We know that for an isothermal process,

               P_{1}V_{1} = P_{2}V_{2}

or,       P_{2} = \frac{P_{1}V_{1}}{V_{2}}

                    = 1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})

                    = 11 atm

Hence, the required pressure is 11 atm.

(d)   For adiabatic process,  

          P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}

or,       P_{2} = P_{1} (\frac{V_{1}}{V_{2}})^{1.4}

                    = 1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})^{1.4}

                    = 28.7 atm

Therefore, required pressure is 28.7 atm.

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Write a list of at least five scientific methods.
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1- You should always have a question for your experiment.
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3- Propose a hypothesis (educated guess on what you believe the outcome of the experiment will be)
4- Design and perform an experiment to test your hypothesis (include independent and dependent variable)
5- Record observations and analyze what the data means.
6- Conclude whether you need to accept or reject your hypothesis, which accepting means your hypothesis was right and rejected is if it was wrong. 
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