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Lyrx [107]
3 years ago
8

1. I drop a penny from the top of the tower at the front of Fort Collins High School and it takes 1.85 seconds to hit the ground

. Calculate the velocity in m/s after 1.10 seconds of freefall and calculate the velocity at impact in mi/hr.
Physics
1 answer:
ladessa [460]3 years ago
3 0

The acceleration of gravity on Earth is  9.8 m/s² .
The speed of a falling object keeps increasing smoothly,
in such a way that the speed is always 9.8 m/s faster than
it was one second earlier.

If you 'drop' the penny, then it starts out with zero speed. 
If you also start the clock at the same instant, then

         After  1.10 sec,  Speed = (1.10 x 9.8) = 10.78 meters/sec


         After  1.85 sec,  Speed = (1.85 x 9.8) = 18.13 meters/sec

But you want this second one given in a different unit of speed.
OK then:

     =  (18.13 meter/sec) x (3,600 sec/hr) x (1 mile/1609.344 meter)

     =    (18.13 x 3,600 / 1609.344)  (mile/hr)  =  40.56 mph  (rounded)

We did notice that in an apparent effort to make the question
sound more erudite and sophisticated, you decided to phrase
it in terms of 'velocity'.  We can answer it in those terms, if we
ASSUME that there is no wind, and the penny therefore doesn't
acquire any horizontal component of motion on its way down.

With that assumption in force, we are able to state unequivocally
and without fear of contradiction that each 'speed' described above ...
with the word 'downward' appended to it ... does become a 'velocity'.

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Give two mathematical examples of Newton's third law and how you get the solution​
bagirrra123 [75]

Answer:

1) Any particle moving in a horizontal plane slowed by friction, deceleration = 32 μ

2) The particle moving by acceleration = P/m - 32μ OR The external force = ma + 32μm

Explanation:

* Lets revise Newton’s Third Law:

- For every action there is a reaction, equal in magnitude and opposite

 in direction.

- Examples:

# 1) A particle moving freely against friction in a horizontal plane

- When no external forces acts on the particle, then its equation of

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∵ ∑ forces in direction of motion = mass × acceleration

∵ No external force

∵ The friction force (F) = μR, where μ is coefficient of the frictional force

   and R is the normal reaction of the weight of the particle on the

   surface

∵ The frictional force is in opposite direction of the motion

∴ ∑ forces in the direction of motion = 0 - F

∴ 0 - F = mass × acceleration

- Substitute F by μR

∴ - μR = mass × acceleration

∵ R = mg where m is the mass of the particle and g is the acceleration

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∴ - μ(mg) = ma ⇒ a is the acceleration of motion

- By divide both sides by m

∴ - μ(g) = a

∵ The acceleration of gravity ≅ 32 feet/sec²

∴ a = - 32 μ

* Any particle moving in a horizontal plane slowed by friction,

 deceleration = 32 μ

# 2) A particle moving under the action of an external force P in a

  horizontal plane.

- When an external force P acts on the particle, then its equation

 of motion is;

∵ ∑ forces in direction of motion = mass × acceleration

∵ The external force = P

∵ The friction force (F) = μR, where μ is coefficient of the frictional force

   and R is the normal reaction of the weight of the particle on the

   surface

∵ The frictional force is in opposite direction of the motion

∴ ∑ forces in the direction of motion = P - F

∴ P - F = mass × acceleration

- Substitute F by μR

∴ P - μR = mass × acceleration

∵ R = mg where m is the mass of the particle and g is the acceleration

  of gravity

∴ P - μ(mg) = ma ⇒ a is the acceleration of motion

∵ The acceleration of gravity ≅ 32 feet/sec²

∴ P - 32μm = ma ⇒ (1)

- divide both side by m

∴ a = (P - 32μm)/m ⇒ divide the 2 terms in the bracket by m

∴ a = P/m - 32μ

* The particle moving by acceleration = P/m - 32μ

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Answer:

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Explanation:

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As we know that weight of cube is given as

W = mg

W = \rho V g

here we know that

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Volume = L^3

Volume = 10^3 = 1000 cm^3

now the mass of the ice cube is given as

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now weight is given as

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Part b)

Weight of the liquid displaced must be equal to weight of the ice cube

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So here volume displaced is given as

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Part c)

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