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Ivan
3 years ago
9

Which law of motion accounts for the following statement?

Physics
2 answers:
Mnenie [13.5K]3 years ago
6 0

The correct answer to the question is : The second law of Newton.

EXPLANATION:

As per the question, the ball is initially hit by a man.

The same ball is hit by a boy.

The ball moves faster for man than that of for the boy. It means the acceleration of ball in case of man is more as compared to the acceleration of the ball for the boy.

From Newton's second law, we know that force applied on a ball is equal to the product of mass with acceleration.

Mathematically force F = ma.

Here, m is the mass and a is the acceleration of the object.

Hence, more is the force, the more will be the acceleration.

As the ball is moving faster for man, he has applied more force on the ball.

Hence, the Newton's second law of motion accounts for the given statement.

julia-pushkina [17]3 years ago
5 0
That would be the second law
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At the same moment, one rock is dropped and one is theown downwand with an iniial velocily of 29 us frm op of a building that is
Helen [10]

Answer:

The thrown rock will strike the ground 2.42s earlier than the dropped rock.

Explanation:

<u>Known Data</u>

  • y_{i}=300m
  • y_{f}=0m
  • v_{iD}=0m/s
  • v_{iT}=-29m/s, it is negative as is directed downward

<u>Time of the dropped Rock</u>

We can use y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}, to find the total time of fall, so 0=300m-\frac{(9.8m/s^{2})t_{D}^{2}}{2}, then clearing for t_{D}.

t_{D}=\sqrt[2]{\frac{300m}{4.9m/s^{2}}} =\sqrt[2]{61.22s^{2}} =7.82s

<u>Time of the Thrown Rock</u>

We can use y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}, to find the total time of fall, so 0=300-29t_{T}-\frac{(9.8)t_{T}^{2}}{2}, then, 0=-4.9t_{T}^{2}-29t_{T}+300, as it is a second-grade polynomial, we find that its positive root is t_{T}=5.4s

Finally, we can find how much earlier does the thrown rock strike the ground, so t_{E}=t_{D}-t_{T}=7.82s-5.4s=2.42s

6 0
3 years ago
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}}

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                    = 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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