Answer:
22.15 N/m
Explanation:
As we know potential energy = m*g*h
Potential energy of spring = (1/2)kx^2
m*g*h = (1/2)kx^2
Substituting the given values, we get -
(400)*(9.8)*(10) = (0.5)*(k)*(2.0^2)
k = 39200/2.645
k = 19600 N/m
For safety reasons, this spring constant is increased by 13 % So the new spring constant is
k = 19600 * 1.13 = 22148 N/m = 22.15 N/m
Answer:
__________________________________________________
No; the sample could not be aluminum;
since the density of aluminum, " 2.7 g/cm³ " , is NOT close enough to the density of the sample, " 3.04 g/cm³ " .
________________________________________________
Explanation:
________________________________________________
Density is expressed as "mass per unit volume" ;
in which:
"mass, "m", is expressed in units of "g" (grams); and:
"Volume, "V", is expressed in units of "cm³ " (such as in this problem); or in units of "mL" ;
__________________________________________________
{Note the exact conversion: " 1 cm³ = 1 mL " .}.
__________________________________________________
The formula for density: D = m/V ;
Given: The density of aluminum is: 2.7 g/cm³.
Given: A sample has a mass of 52.0 g ; and Volume of 17.1 cm³ ; could it be aluminum?
_________________________________________________________
Let us divide the mass of the sample by the volume of the sample;
by using the formula:
___________________________________________
D = m / V ;
and see if the value is at, or very close to "2.7 g/cm³ ".
If it is, then it could be aluminum.
____________________________________________________
The density for the sample:
D = (52.0 / 17.1) g/cm³ = 3.0409356725146199 g/cm³ ;
→round to "3 significant figures" ;
= 3.04 g/cm³ .
_______________________________________________
No; the sample could not be aluminum; since the density of aluminum,
"2.7 g/cm³ " is NOT close enough to the density of the sample,
"3.04 g/cm³ " .
____________________________________________________
The answer is A. meters (m)
