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lora16 [44]
3 years ago
15

Give a real-life example of Newton's second law.

Physics
2 answers:
Juliette [100K]3 years ago
5 0

Answer:

Riding your bicycle is a  example of this law of motion at work. Your bicycle is the mass. Your leg muscles pushing pushing on the pedals of your bicycle is the force

Explanation:

Sladkaya [172]3 years ago
5 0
Examples of Newton's 2nd Law If you use the same force to push a truck and push a car, the car will have more acceleration than the truck, because the car has less mass. It is easier to push an empty shopping cart than a full one, because the full shopping cart has more mass than the empty one. Pushing a car is easier than pushing a truck with the same amount of force as the mass of the car is lesser than the mass of the truck. In golf game, acceleration of the golf ball is directly proportional to the force with which it is hit by the golf stick. Hope this helps! Mark brainly please
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(7) A 2500 lbm car moving at 25 mi/hr is accelerated at a constant rate of 15 ft/s2 up to a speed of 50 mi/hr. What is the force
balandron [24]

Answer:

The  force is  F =  1164.6\  lbf

The time is   \Delta t =  2.44 \  s

Explanation:

From the question we are told that

  The  mass of the car is  m  =  2500 \ lbm

   The  initial velocity of the car is u  =  25 \ mi/hr

   The final  velocity of the car is  v  =  50 \  mi/hr

  The acceleration is  a =  15 ft/s^2 =  \frac{15 *  3600^2}{ 5280} =  36818.2 \  mi/h^2

   

Generally the acceleration is mathematically represented as

      a =  \frac{v-u}{\Delta t}

=>   36818.2 =  \frac{50 - 25 }{ \Delta t}

=>   t = 0.000679 \  hr

converting to seconds

       \Delta t =  0.0000679 *  3600

=>     \Delta t =  2.44 \  s

Generally the force is mathematically represented as

        F  =  m * a

=>      F  =  2500 *  15

=>      F  =  37500 \ \frac{lbm *  ft}{s^2}

Now converting to foot-pound-second we have  

         F =  \frac{37500}{32.2}

=>        F =  1164.6\  lbf

7 0
3 years ago
Three importance of SI system​
diamong [38]

Answer:

Firstly they are, by design, easy to use in most scientific and engineering calculations; you only ever have to consider multiples of 10. If I’m given a measurement of 3.4 kilometres, I can instantly see that it’s 3′400 metres, or 0.0034 Megametres, or 3′400′000 millimetres. It’s not even necessary to use arithmetic, I just have to remember the definitions of the prefixes (“kilo” is a thousand, “megametre” is a million, “milli” is a thousandth) and shift the decimal point across to the left or the right. This is especially useful when we’re considering areas, speeds, energies, or other things that have multiple units; for instance,

1 metre^2 = (1000millimetre)^2 = 1000000 mm^2.

If we were to do an equivalent conversion in Imperial, we would have

1 mile^2 = (1760 yards)^2

and we immediately have to figure out what the square of 1760 is! However, the fact that SI is based on multiples of 10 has the downside that we can’t consider division by 3, 4, 8, or 12 very easily.

Secondly they are (mostly) defined in terms of things that are (or, that we believe to be) fundamental constants. The second is defined by a certain kind of radiation that comes from a caesium atom. The metre is defined in terms of the second and the speed of light. The kelvin is defined in terms of the triple point of water. The mole is the number of atoms in 12 grams of carbon-12. The candela is defined in terms of the light intensity you get from a very specific light source. The ampere is defined using the Lorentz force between two wires. The only exception is the kilogram, which is still defined by the mass of a very specific lump of metal in a vault in France (we’re still working on a good definition for that one).

Thirdly, most of the Imperial and US customary units are defined in terms of SI. Even if you’re not personally using SI, you are probably using equipment that was designed using SI.

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Answer:

Pascal

Explanation:

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