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Zinaida [17]
3 years ago
13

Gold has a density of 19.32 g/cm3. what is the volume of a sample of gold with a mass of 27.63 grams?

Physics
1 answer:
schepotkina [342]3 years ago
4 0
Answer:  The volume is:  " 1.430 cm³ " .
_____________________________
Explanation:  
________________________
Volume, " V  = ? ; (unknown, we need to solve for this).
Density,  "D  = 19.32 g / cm³ ; 
mass , "m" = 27.63 g ; 
_____________________________________________
The formula for density is:  
____________________________________________
  D = m / V ;   Divide each side of the equation by:  "(1/m)" ;
                    to isolate " V" on one side of the equation; and to solve for "V" ; 
____________________________________________
1/m)*D = (1/m) * (m / V) ;
_______________________________
    to get:   
_______________________________
       D/m = ; ↔  1 / V ;  Take the reciprocal of EACH SIDE; to isolate "V" on each side of the equation:    
_____________________________
      m / D = V/1 ↔ V = m / D ;
_____________________________
    V = m / D ; 
_______________________________
         Now, plug in our given values for mass, "m" ; and Density, D";              to solve for "Volume, V " ;
_____________________________________
    V  =  m / D  =  (27.63 g)  ÷  (<span>19.32 g / cm</span>³)  ;                
 
                       =  (27.63 g) * (1 cm³ / 19.32 g) ;
 
                       =   (27.63 ÷ 19.32)  cm³  ;
  
                       =   1.4301242236024845 cm³ ;

                                              → Round to "4 significant figures" ; 
                       =   1.430 cm³ .
____________________________________________
   The volume is:  " 1.430 cm³ " .
______________________________________________
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E) is described by all of these

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B = magnitude of field vector

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The correct option is E.

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3 years ago
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Amiraneli [1.4K]

After the collision the magnitude of the momentum of the system is Mv

Given:

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Answer:

A) mass = 3121.58 kg

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Explanation:

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Since it started from rest, initial speed is zero.

Using Newton's equation of motion we have,

S = ut + 0.5at^2

S = distance covered = 6.75 m

t = time = 3 s

a = acceleration upwards

u = initial velocity = 0

Substituting values, we have,

6.75 = 0(3) + (0.5 x a x 3^2)

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From the image below we solve from

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T = 3440.37 + 22500 = 25940.37 N (this is the tension on the rope)

On the other side,

mg - T = ma

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(9.81 - 1.5)m = 25940.37

8.31m = 25940.37

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See image below

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