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boyakko [2]
2 years ago
13

Platinum has a density of 21 g/cm3. a platinum ring is placed in a graduated cylinder that contains water. the water level rises

from 5.0 ml to 5.3 ml when the ring is added. what is the mass of the ring?
Physics
2 answers:
sergey [27]2 years ago
8 0

Answer:

The mass of the ring is 6.3 kg.

Explanation:

It is given that,

Density of platinum, d=21\ g/cm^3

A platinum ring is placed in a graduated cylinder that contains water. the water level rises from 5.0 ml to 5.3 ml when the ring is added. the volume of the ring is, (5.3 - 5) mL = 0.3 mL

Volume of the ring, V=0.3\ cm^3

We need to find the mass of the ring. The density of any substance is given by :

d=\dfrac{m}{V}

m=d\times V

m=21\ g/cm^3\times 0.3\ cm^3

m = 6.3 kg

So, the mass of the ring is 6.3 kg. Hence, this is the required solution.

Oliga [24]2 years ago
6 0
The little lump of platinum that you dropped into the graduated cylinder has

         (5.3 - 5.0) = 0.3 mL  =  0.3 cm³ of volume.

That's nothing but a metal shaving, a little sliver, maybe like
a 1/4-inch piece cut out of a metal paper-clip.  I'm surprised
that you can read the graduated cylinder that precisely.

Its mass is

           (0.3 cm³) x (21 gm/cm³)  =  6.3 grams  =  0.0063 kg .
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7 0
3 years ago
A 45 N girl sits on a bench 0.6 meters off the ground. How much work is done on the bench?
ycow [4]

Answer: 27 joules

Explanation:

Work is done when force is applied on the bench over a distance. it is measured in joules.

Workdone = force x distance

= 45 N x 0.6 metres

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6 0
3 years ago
A surgical microscope weighing 200 lb is hung from a ceiling by four springs with stiffness 25 lb/ft. The ceiling has a vibratio
Nikitich [7]

Answer:

If there is no damping, the amount of transmitted vibration that the microscope experienced is   = 5.676*10^{-3} \ mm

Explanation:

The motion of the ceiling is y = Y sinωt

y = 0.05 sin (2 π × 2) t

y = 0.05 sin 4 π t

K = 25 lb/ft  × 4  sorings

K = 100 lb/ft

Amplitude of the microscope  \frac{X}{Y}= [\frac{1+2 \epsilon (\omega/ W_n)^2}{(1-(\frac{\omega}{W_n})^2)^2+(2 \epsilon  \frac{\omega}{W_n})^2}]

where;

\epsilon = 0

W_n = \sqrt { \frac{k}{m}}

= \sqrt { \frac{100*32.2}{200}}

= 4.0124

replacing them into the above equation and making X the subject of the formula:

X = Y * \frac{1}{\sqrt{(1-(\frac{\omega}{W_n})^2)^2})}}

X = 0.05 * \frac{1}{\sqrt{(1-(\frac{4 \pi}{4.0124})^2)^2})}}

X = 5.676*10^{-3} \ mm

Therefore; If there is no damping, the amount of transmitted vibration that the microscope experienced is   = 5.676*10^{-3} \ mm

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3 years ago
A crate filled with delicious junk food rests on a horizontal floor. Only gravity and the support force of the floor act on it,
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The words "... as shown ..." tell us that there's a picture that goes along
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2 years ago
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