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notsponge [240]
3 years ago
11

A merchant in Katmandu sells you a solid gold 6 kg statue for a very reasonable price. When you get home, you wonder whether or

not you got a bargain, so you lower the statue into a measuring cup and measure its volume. What volume will verify that it's pure gold?
Physics
1 answer:
Lilit [14]3 years ago
5 0

Answer:

0.000311 m³

Explanation:

The formula that connect volume, mass and density is given below

D = m/v ...................... Equation 1

Where D = Density of the gold, m = mass of the gold, v =  volume of the gold.

make v the subject of the equation

v = m/D .................. Equation 2

Given: m = 6 kg,

Note the density of pure gold = 19300 kg/m².

Substitute into equation 2

v = 6/19300

v = 0.000311 m³

Hence the volume that will verify that it's a pure gold = 0.000311 m³

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A horizontal spring with spring constant 130 N/m is compressed 17 cm and used to launch a 2.8 kg box across a frictionless, hori
olasank [31]

Explanation:

The given data is as follows.

        k = 130 N/m,       \Delta x = 17 cm = 0.17 m   (as 1 m = 100 cm)

     mass (m) = 2.8 kg

When the spring is compressed then energy stored in it is as follows.

             Energy = \frac{1}{2}kx^{2}

Now, spring energy gets converted into kinetic energy when the box is launched.

So,    \frac{1}{2}kx^{2} = \frac{1}{2}mv^{2}

   \frac{1}{2} \times 130 \times (0.17)^{2} = \frac{1}{2} \times 2.8 \times v^{2}

          v^{2} = \frac{3.757}{2.8}

                     = 1.34

                v = 1.15 m/sec

Now,

           Frictional force = \mu \times mg

                                    = 0.15 \times 2.8 \times 9.8

                                    = 4.116 N

Also,  Kinetic energy = work done by friction

           \frac{1}{2}mv^{2} = F_{f} \times d

           \frac{1}{2} \times 2.8 \times (1.15)^{2} = 4.116 \times d

             1.8515 = 4.116 \times d

                 d = 0.449 m

Thus, we can conclude that the box slides 0.449 m across the rough surface before stopping.

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Answer:

Explanation:

Given

Volume of bucket V=10\ gallon

Time taken to fill the bucket t=50\ s

so volume flow rate is \dot{V}=\frac{10}{50}=0.2\ gal/s

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v_{avg}=49.51\ ft/s

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