All categories

3 years ago

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?

1 answer:

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) ÷ (19.32 g / cm³) ;

= (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 " .
______________________________________________

You might be interested in

Suppose a rocket ship accelerates upwards with acceleration equal in magnitude to twice the magnitude of g (we say that the rock

pashok25 [27]

Answer:

a) $s_a=98100\ m$ is the height where the rocket stops accelerating and its fuel is finished and starts decelerating while it still continues to move in the upward direction.

b) $v_a=1962\ m.s^{-1}$ is speed of the rocket going when it stops accelerating.

c) $H=294300\ m$

d) $t_T=544.95\ s$

e) Zero, since the average velocity is the net displacement per unit time and when the rocket strikes back the earth surface the net displacement is zero.

Explanation:

Given:

acceleration of rocket, $a=2g=2\times 9.81=19.62\ m.s^{-2}$

time for which the rocket accelerates, $t_a=100\ s$

For the course of upward acceleration:

using eq. of motion,

$s_a=ut+\frac{1}{2}at_a^2$

where:

$u=$ initial velocity of the rocket at the launch $=0$

$s_a=$ height the rocket travels just before its fuel finishes off

so,

$s_a=0+\frac{1}{2}\times 19.62\times 100^2$

a) $s_a=98100\ m$ is the height where the rocket stops accelerating and its fuel is finished and starts decelerating while it still continues to move in the upward direction.

Now the velocity of the rocket just after the fuel is finished:

$v_a=u+at_a$

$v_a=0+19.62\times 100$

b) $v_a=1962\ m.s^{-1}$ is speed of the rocket going when it stops accelerating.

After the fuel is finished the rocket starts to decelerates. So, we find the height of the rocket before it begins to fall back towards the earth.

Now the additional height the rocket ascends before it begins to fall back on the earth after the fuel is consumed completely, at this point its instantaneous velocity is zero:

using equation of motion,

$v^2=v_a^2-2gh$

where:

$g=$ acceleration due to gravity

$v=$ final velocity of the rocket at the top height

$0^2=1962^2-2\times 9.81\times h$

$h=196200\ m$

c) So the total height at which the rocket gets:

$H=h+s$

$H=196200+98100$

$H=294300\ m$

Time taken by the rocket to reach the top height after the fuel is over:

$v=v_a+g.t$

$0=1962-9.81t$

$t=200\ s$

Now the time taken to fall from the total height:

$H=v.t'+\frac{1}{2}\times gt'^2$

$294300=0+0.5\times 9.81\times t'^2$

$t'=244.95\ s$

Hence the total time taken by the rocket to strike back on the earth:

$t_T=t_a+t+t'$

$t_T=100+200+244.95$

$t_T=544.95\ s$

Zero, since the average velocity is the net displacement per unit time and when the rocket strikes back the earth surface the net displacement is zero.

8 0

2 years ago

The earth's magnetic field deflects the flow of current from?

alexgriva [62]

I don't really know the answer but maybe north pole and south pole?

7 0

3 years ago

An object is five focal lengths from a concave mirror.how do the object and image heights compare?

enot [183]

An object distance is presented as s = 5f and we know that the mirror equation relates the image distance to the object distance and the focal length.

The mirror equation is 1/f = 1/s + 1/s’ where the variable f stands for the focal length of the mirror. Variable (s) represents the distance between the mirror surface and the object and the variable (s’) represents the distance between the mirror surface and the image.

In addition, a concave mirror will have a positive focal length (f) and a convex mirror will have a negative focal length (f).

Now, we then have 1/f = 1/5f + 1/s’ which is s’ = 5f/4

Then we get the magnification ratio that expresses the size or amount of magnification or reduction of the object or image and to get the magnification, we use this equation: M= s’/s

M= 5f/4x5f

s’ = 1/4s

Therefore, the image height is one fourth of the object height

7 0

3 years ago

On vacation at the beach, you bring a radio to the beach. The radio can pick up several different radio stations.Which is the be

Harrizon [31]

D) waves are used to transmit the rail signal though the air. these waves are encoded at different frequencies for different stations

7 0

3 years ago

Read 2 more answers

Bob has been Springfield's top meteorologist for ten years. He received data from satellite imaging and computer models that a h

djyliett [7]

The correct letter of the answer would be letter b, Bob is using his personal knowledge of the local area to make his forecast. It does not look like he's ignoring the computer for he's been the top meteorologist for ten years, meaning to say, he would use his knowledge about the local area and tell them his knowledge about the weather.

7 0

3 years ago

Read 2 more answers