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Eduardwww [97]
2 years ago
12

At one point in time, the price of gold was about $1400 per ounce, while that of silver was about $20 an ounce. The "ounce" in t

his case is the troy ounce, which is equal to 31.1035 g. (The more familiar avoirdupois ounce is equal to 28.35 g.) The density of gold is 19.3 g/cm3 and that of silver is 10.5 g/cm3.
A. If you found a spherical gold nugget worth $5 million, what would be its diameter?

ANSWER: ______________ cm

B. How much would a silver nugget of that size be worth?

ANSWER: $ ____________
Physics
1 answer:
madam [21]2 years ago
6 0

Answer:

Gold nugget diameter    22,24 cm

Silver nugget price   38860  $

Explanation:  

Gold price    1400  $ /ou            1  ounce  = 31.1035 grs

so  1400 /31.1035

Gold price is 45,01   $/grs.

If a nugget worth   5000000 $  then  5000000/ 45.01

a nugget mass  :  111086,43 grs

Now gold density  is  d = 19.3 grs/cm³

And she volume is  V = 4/*π*r³

d = m/V         V = m/d           V  =   111086.43/19.3    cm³

V = 5755,77 cm³

Now V of the sphere is  V =  5755,77 = 4/3*π*r³

r³  =3*5755,77 / 4π        r³  = 1374,79

r  =  11.12 cm      2r  =  22,24  =  Φ (sphere diameter)

B)  d (silver)  =  m/V       m = d*V

V = 5755,77 cm³     The same size the same volume

m =  10,5 * 5755,77 [grs/cm³ * cm³]       m  = 60435,59 grs

Silver nugget worth  :

20  $ /ou         20/31.1035  =   0.643  $ /grs

Price   0,643 *  60435,59  =  38860  $

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A 38 kg crate rests on a floor. A horizontal pulling force of 170 N is needed to start the crate
Mandarinka [93]

Answer:

0.456033049

Explanation:

F=\mu N where N=mg hence

F=\mu mg where m is mass of object, g is acceleration due to gravity whose value is taken as 9.81 m/s^{2}, \mu is the coefficient of static friction and F is the applied force.

Making \mu the subject we obtain

\mu=\frac {mg}{N} and substituting m for 38 Kg, g for 9.81 m/s^{2} and 170 N for  F we obtain

\mu=\frac{170} {38*9.81}=0.456033049

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2 years ago
An ice dancer with her arms stretched out starts into a spin with an angular velocity of 1.00 rad/s. Her moment of inertia with
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A) 0.957 J

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2 years ago
What volume in milliliters will 0.00922 g of h2 gas occupy at stp?
Nadusha1986 [10]
Assuming that this gas is in ideal state, we can use the relation that for every 1 mol of an ideal gas it would have a volume of 22.4 L. But before using this, relation we need to convert the number of grams of H2 into moles by using the molar mass of 2.02 g/mol.

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2 years ago
A 5.0-kg rock and a 3.0 × 10−4-kg pebble are held near the surface of the earth.(a)Determine the magnitude of the gravitational
a_sh-v [17]

Answer:

a). Determine the magnitude of the gravitational force exerted on each by the earth.

Rock: F = 49.06N

Pebble: F = 29.44N

(b)Calculate the magnitude of the acceleration of each object when released.

Rock: a =9.8m/s^{2}

Pebble:  a =9.8m/s^{2}

Explanation:

The universal law of gravitation is defined as:

F = G\frac{m1m2}{r^{2}}  (1)

Where G is the gravitational constant, m1 and m2 are the masses of the two objects and r is the distance between them.

<em>Case for the rock </em>m = 5.0 Kg<em>:</em>

m1 will be equal to the mass of the Earth m1 = 5.972×10^{24} Kg and since the rock and the pebble are held near the surface of the Earth, then, r will be equal to the radius of the Earth r = 6371000m.

F = (6.67x10^{-11}kg.m/s^{2}.m^{2}/kg^{2})\frac{(5.972x10^{24} Kg)(5.0 Kg)}{(6371000 m)^{2}}

F = 49.06N

Newton's second law can be used to know the acceleration.

F = ma

a =\frac{F}{m} (2)

a =\frac{(49.06 Kg.m/s^{2})}{(5.0 Kg)}

a =9.8m/s^{2}

<em>Case for the pebble </em>m = 3.0 Kg<em>:</em>

F = (6.67x10^{-11}kg.m/s^{2}.m^{2}/kg^{2})\frac{(5.972x10^{24} Kg)(3.0 Kg)}{(6371000 m)^{2}}

F = 29.44N

a =\frac{F}{m}

a =\frac{(29.44 Kg.m/s^{2})}{(3.0 Kg)}

a =9.8m/s^{2}

3 0
2 years ago
Read 2 more answers
The constant pressure molar heat capacity, C_{p,m}C p,m ​ , of nitrogen gas, N_2N 2 ​ , is 29.125\text{ J K}^{-1}\text{ mol}^{-1
antoniya [11.8K]

Answer:

Explanation:

Constant pressure molar heat capacity Cp = 29.125 J /K.mol

If Cv be constant volume molar heat capacity

Cp - Cv = R

Cv = Cp - R

= 29.125 - 8.314 J

= 20.811 J

change in internal energy = n x Cv x Δ T

n is number of moles , Cv is molar heat capacity at constant volume ,  Δ T is change in temperature

Putting the values

= 20 x 20.811 x 15

= 6243.3 J.

3 0
2 years ago
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