Answer:
1. Density = 1200[kg/m^3]; 2. Volume= 0.005775[m^3], mass= 15.59[kg]
Explanation:
1. We know that the density is defined by the following expression.
![Density = \frac{mass}{volume} \\where:\\mass=90[kg]\\volume=0.075[m^{3} ]\\density=\frac{90}{0.075} \\density=1200[\frac{kg}{m^{3} }]](https://tex.z-dn.net/?f=Density%20%3D%20%5Cfrac%7Bmass%7D%7Bvolume%7D%20%5C%5Cwhere%3A%5C%5Cmass%3D90%5Bkg%5D%5C%5Cvolume%3D0.075%5Bm%5E%7B3%7D%20%5D%5C%5Cdensity%3D%5Cfrac%7B90%7D%7B0.075%7D%20%5C%5Cdensity%3D1200%5B%5Cfrac%7Bkg%7D%7Bm%5E%7B3%7D%20%7D%5D)
2. First we need to convert the units to meters.
wide = 35[cm] = 35/100 = 0.35[m]
long = 11 [dm] = 11 decimeters = 11/10 = 1.1[m]
Thick = 15[mm] = 15/1000 = 0.015[m]
Now we can find the density using the expression for the density.
![density= \frac{mass}{volume} \\where:\\volume = wide*long*thick\\volume=0.35*1.1*0.015 = 0.005775[m^3]\\\\mass= density*volume = 2700*0.005775 = 15.59[kg]](https://tex.z-dn.net/?f=density%3D%20%5Cfrac%7Bmass%7D%7Bvolume%7D%20%5C%5Cwhere%3A%5C%5Cvolume%20%3D%20wide%2Along%2Athick%5C%5Cvolume%3D0.35%2A1.1%2A0.015%20%3D%200.005775%5Bm%5E3%5D%5C%5C%5C%5Cmass%3D%20density%2Avolume%20%3D%202700%2A0.005775%20%3D%2015.59%5Bkg%5D)
Answer: 3.92 N.
Explanation:
Your box weighs 400g, or 0.4kg. In order to lift it, you need to overcome the force of gravity. F = ma, and acceleration due to gravity is -9.8 m/s^2. So gravity acts on the box with a force of 0.4 kg * -9.8 m/s^2 = -3.92 N. A force of +3.92 N is required to overcome this.
Answer:
1) Determine the domain of the following functions: d ... 3) If g(x) = x + 3 and f(x)= x² – 2x, find the value of f(g(a)). ... 6) Given the graph of f(x) to the right, determine: ... 8) Given f(x)= x? and g(x)= 2* The inverse of g is a function, but the inverse off is ... -3(x-1)= -5 4 (-3). -3x+ 3 = y. 10) The graph of a function f (x) is given at the.
Explanation:
The amount of work done in emptying the tank by pumping the water over the top edge is 163.01* 10³ ft-lbs.
Given that, the tank is 8 feet across the top and 6 feet high
By the property of similar triangles, 4/6 = r/y
6r = 4y
r = 4/6*y = 2/3*y
Each disc is a circle with area, A = π(2/3*y)² = 4π/9*y²
The weight of each disc is m = ρw* A
m = 62.4* 4π/9*y² = 87.08*y²
The distance pumped is 6-y.
The work done in pumping the tank by pumping the water over the top edge is
W = 87.08 ∫(6-y)y² dy
W = 87.08 ∫(6y³ - y²) dy
W = 87.08 [6y⁴/4 - y³/3]
W = 87.08 [3y⁴/2- y³/3]
The limits are from 0 to 6.
W = 87.08 [3*6⁴/2 - 6³/3] = 87.08* [9*6³ - 2*36] = 87.08(1872) = 163013.76 ft-lbs
The amount of work done in emptying the tank by pumping the water over the top edge is 163013.76 ft-lbs.
To know more about work done:
brainly.com/question/16650139
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To solve the problem it is necessary to apply energy conservation.
By definition we know that kinetic energy is equal to potential energy, therefore
PE = KE
Where,
m = mass
g = gravitaty constat
v = velocity
h = height
Re-arrange to find h,
Replacing with our values
Therefore the correct answer is C.
