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k0ka [10]
3 years ago
11

A standard 1 kilogram weight is a cylinder 54.0 mm in height and 55.0 mm in diameter. what is the density of the material

Physics
1 answer:
denis-greek [22]3 years ago
5 0

The radius of the cylinder is equal to half the diameter:

r=\frac{d}{2}=\frac{55.0 mm}{2}=27.5 mm

The volume of the cylinder is given by:

V=\pi r^2 h=\pi (27.5 mm)^2 (54.0 mm)=1.28 \cdot 10^5 mm^3

where h is the heigth of the cylinder. Converting into meters,

V=1.28 \cdot 10^{-4} m^3

And the density of the material will be given by the ratio between the mass and the volume:

d=\frac{m}{V}=\frac{1 kg}{1.28 \cdot 10^{-4} m^3}=7812.5 kg/m^3

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Two charged objects have a repulsive force of 0.040 N. If the distance separating the objects is doubled, then what is the new f
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Answer:

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Explanation:

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On the Apollo 14 mission to the moon, astronaut Alan Shepard hit a golf ball with a golf club improvised from a tool. The free-f
aliya0001 [1]

Answer:

15.3 s and 332 m

Explanation:

With the launch of projectiles expressions we can solve this problem, with the acceleration of the moon

    gm = 1/6 ge

    gm = 1/6  9.8 m/s² = 1.63 m/s²

We calculate the range

    R = Vo² sin 2θ  / g

    R = 25² sin (2 30) / 1.63

    R= 332 m

We will calculate the time of flight,

   Y = Voy t – ½ g t2  

   Voy = Vo sin θ

When the ball reaches the end point has the same initial  height Y=0

0 = Vo sin  t – ½  g t2

0 = 25 sin (30)  t – ½ 1.63 t2

0= 12.5 t –  0.815 t2

We solve the equation

0= t ( 12.5 -0.815 t)

 t=0 s

t= 15.3 s

The value of zero corresponds to the departure point and the flight time is 15.3 s

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R2 = 25² sin (2 30) / 9.8

R2 = 55.2 m

R/R2 = 332/55.2

R/R2 = 6

Therefore the ball travels a distance six times greater on the moon than on Earth

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