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Natasha2012 [34]
2 years ago
14

A man leave his townA at 8 am for town B which is 384km away and he reach at4pm what is average speed in m/s

Physics
1 answer:
zlopas [31]2 years ago
5 0

Total distance covered = 384 Km

Total time taken to travel from A to B = 8 hours [from 8 am to 4 pm, there are 8 hours]

We know, Average speed = Total Distance Travelled/ Total Time Taken

Therefore, average speed = 384 Km/8 h = 384000m/8×60×60s =(384000/28800)m/s

= 13.3 m/s

Answer is 13.3 m/s

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faust18 [17]

Answer:

9)a

10) I think true

11)b

Explanation:

9)a. because it's told that the car is slowing down, the sum of the forces that are towards left, should be more than the ones that are towards right. if the car was gaining speed, "b" would have been correct. and if it was told that the car is moving without a change in the speed, "c" would have been correct.

10) if a moving object has a change of speed or direction, it would have an acceleration. now if a moving object experiences an unbalanced force, it'd either slow down, gain speed or change direction, and in all of the three possibilities it'd have an acceleration.

11) upward and downward forces are equal, and the sum of them would be 0N(because they have opposite directions). so they negate each other.

and the rightward force is 5N more than the leftward force. so the Net Force would be 5N.

-30+30-10+15=5N

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8 0
3 years ago
John and Linda are arguing about the definition of density. John says the density of an object is proportional to itsmass. Linda
NeX [460]

Answer:

They are both correct.

Explanation:

The density of an object is defined as the ratio of its mass to its volume. This implies that the density of the object is both proportional to the mass and also to the volume of the object. John only mentioned mass which is correct. Linda mentioned the second variable on which density depends which is the volume of the object.

Hence considering the both statements objectively, one can say that they are both correct.

8 0
3 years ago
Starting from your campsite you walk 3.0 km east, 6.0 km north, 1.0 km east, and then 4.0 km west. How far are you from your cam
Hatshy [7]
Think of it like a graph. You start at the origin which is (0,0).  go three to the east which now you are (3,0). Then, six to the north. Now, you are at (3,6).  1 to the east, ((4,6).  Then you go 4 to the west which is back tracking. So, you end at (0,6) which is saying you are now 6 km north from your campsite. 

Hope this helps!
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}}

                    = 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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Inessa05 [86]

Answer:

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Explanation:

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