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2 years ago

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A sample of gold has a volume of 2 cm3 and a mass of 38.6 grams. What would be the density, and three other properties of the sa

mple that would be considered physical properties.

1 answer:

lukranit [14]2 years ago

3 0

The density of the gold is calculated to be "19,300 kg/m³".

<u>Explanation:</u>

Given:

Volume = 2cm³

Mass = 38.6 grams.

To Find:

Density of the gold = ?

Solution:

Density is obtained by dividing mass of the sample by its volume and it is given in the units of kg/m³.

Mass in grams is converted into kg as,

1 g = 0.001 kg

38. 6 g = $\frac{38.6}{1000} = 0.0386 kg$

Now we have to convert cm³ to m³ as,

1 cm³ = 10⁻⁶ m³

2 cm³ = 2 × 10⁻⁶ m³

So Density = $\frac{0.0386 k g}{2 \times 10^{-6} m^{3}}=19,300 \mathrm{kg} / \mathrm{m}^{3}$

Physical properties of Gold:

Gold is a heavy metal.

It is Malleable and ductile.

It is a corrosion resistant element.

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(a) Greater

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So we have:

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So, the fundamental frequency of wire A is greater than the fundamental frequency of wire B.

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For standing waves on a string, the fundamental frequency is given by the formula:

$f_1 = \frac{v}{2L}$

where

v is the speed at which the waves travel back and forth on the wire

L is the length of the string

(c) Greater speed on wire A

We can solve the formula of the fundamental frequency for v, the speed of the wave:

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We know that the two wires have same length L. For wire A, $f_1 = 330 Hz$ , while for wave B, $f_B = 220 Hz$ , so we can write the ratio between the speeds of the waves in the two wires:

$\frac{v_A}{v_B}=\frac{2L(330 Hz)}{2L(220 Hz)}=\frac{3}{2}$

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1. A 9.4×1021 kg moon orbits a distant planet in a circular orbit of radius 1.5×108 m. It experiences a 1.1×1019 N gravitational

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Answer:

26 days

Explanation:

m = 9.4×1021 kg

r= 1.5×108 m

F = 1.1×10^ 19 N

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==> 1.1 × $10^{19}$ = (9.4 × $10^{21}$ × $v^{2}$ ) ÷ 1.5 × $10^{8}$

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Putting value of r and v from above in ;

T= 2πr ÷ v

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but

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1 year ago

Read 2 more answers

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t \approx 4.72 

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It will stay approx. 9.44 seconds in the air.

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1 year ago

Consider a spring that does not obey Hooke’s law very faithfully. One end of the spring is fixed. To keep the spring stretched o

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Answer:

a) $W=-0.0103125\ J$

b) $W=0.0059375\ J$

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Explanation:

Given:

Expression of force:

$F=kx-bx^2+cx^3$

where:

$k=100\ N.m^{-1}$

$b=700\ N.m^{-2}$

$c=12000\ N.m^{-3}$

$x$ when the spring is stretched

$x$ when the spring is compressed

hence,

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a)

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Now,

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$j=\frac{F}{x}$

$j=\frac{100\times (-0.05)-700\times (-0.05)^2+12000\times (-0.05)^3}{-0.05}$

$j=-8.25\ N.m^{-1}$

Now work done:

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$W=0.5\times -8.25\times (-0.05)^2$

$W=-0.0103125\ J$

b)

When compressing the spring by 0.05 m

we have, $x=0.05\ m$

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$j=\frac{F}{x}$

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c)

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1 year ago