Water treatment plants have been in existence for almost 220 years
Answer:
a) 0.036 J b) 0.036J c) 0.036 d) 1.9m/s e) 0.18 m
Explanation:
Mass of the dart = 0.02kg, the spring was compressed to 6cm
Work needed to compress the spring = 1/2*k*x ^2 where k is the force constant of the spring in N/m, x is the distance it was compressed in m
Work needed to compress the spring = 0.5 * 20* 0.06^2 since 6cm = 6 / 100 = 0.06 m
Work needed to compress the spring = 0.036J
b) the total energy stored in the spring = the work done to compress the spring = 0.036J
c) kinetic energy of the dart as it leaves the the spring = elastic potential energy stored in the spring = the work done in compressing the = 0.036J using the law of conservation of energy; energy is neither created nor destroyed but transformed from one form to another.
d) 1/2mv^2 = 0.036
mv^2 = 0.036*2
v^2 = 0.036*2 / 0.02 = 3.6
v = √3.6 = 1.897 approx 1.9m/s
e) kinetic energy of the dart = work done against gravity to get the body to height h
Work done against gravity = potential energy conserved at height = -mgh g is negative because the motion is upward while gravity acts downward
0.036 = 0.02 * 9.81 * h
0.036 / ( 0.02*9.81) = h
h = 0.18 m
Answer:
(a) work required to lift the object is 1029 J
(b) the gravitational potential energy gained by this object is 1029 J
Explanation:
Given;
mass of the object, m = 35 kg
height through which the object was lifted, h = 3 m
(a) work required to lift the object
W = F x d
W = (mg) x h
W = 35 x 9.8 x 3
W = 1029 J
(b) the gravitational potential energy gained by this object is calculated as;
ΔP.E = Pf - Pi
where;
Pi is the initial gravitational potential energy, at initial height (hi = 0)
ΔP.E = (35 x 9.8 x 3) - (35 x 9.8 x 0)
ΔP.E = 1029 J
Density is calculated as mass per unit volume. In this case, since the material has a mass of 47 grams and we have the volume of 15 cm^3, we can simply divide the values:
Density = 47 grams / 15 cm^3 = 3.1 g/cm^3
Therefore, the material has a density of 3.1 g/cm^3
