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Angelina_Jolie [31]
3 years ago
5

Suppose a soup can is made from a sheet of steel19 which is .13 mm thick. If the can is 11 cm high and 6 cm in diameter, use dif

ferentials to estimate the mass of the can. The density of the steel being used is 8000 kg/m3
Physics
1 answer:
Flura [38]3 years ago
5 0

Answer:

The can mass is 0,00359 kg or 3,59 g

Explanation:

1. Relevant Data:

Steel thickness= 0.13 mm or 0.013 mm

h=11 cm

d=6 cm

ρ=800 kg/m^3

2. Calculate mass from densisty equation:

\rho=\frac{m}{v}, then m=\rho .v

We need to estimate the volume of the can to calculate the mass.

3. Estimate volume using differentials:

Cylinder volume equation is:

V=\frac{1}{4}\pi  d^{2}h

Considering that the can is an object with a hole inside, then we need to estimate the real volume of the sheet of steel.

Using differentials we have:

dV=\frac{1}{2}\pi  Dh (dD)

Then, we could say that dD=0.013 cm

Replacing the values of d, h and dD, we obtain:

dV=\frac{1}{6}\pi  (6 cm)(11 cm)(0,013 cm)

dV=0,4492 cm^3

4. Calculate the mass

Convert volume unit into m^3

0,4492 cm^3x\frac{1 m^3}{1x10^6 cm^3} =0,4492 x 10^-6 m^3

Calculate mass

m=\rho .v

m=8000 \frac{kg}{m^3}.0,4492 x10^-6 m^3

m=0,00359 kg =3,59 g

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2 years ago
A cannon fires a 0.2 kg shell with initial velocity vi = 9.2 m/s in the direction θ = 46 ◦ above the horizontal. The shell’s tra
Sedbober [7]

Answer:

∆h = 0.071 m

Explanation:

I rename angle (θ) = angle(α)

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Vy(t) and y(t)

We start by decomposing the speed in the direction ''y''

sin(\alpha) = \frac{Vyi}{Vi}

Vyi = Vi.sin(\alpha ) = 9.2 \frac{m}{s} .sin(46) = 6.62 \frac{m}{s}

Vy in this problem will follow this equation =

Vy(t) = Vyi -g.t

where g is the gravity acceleration

Vy(t) = Vyi - g.t= 6.62 \frac{m}{s} - (9.8\frac{m}{s^{2} }) .t

This is equation (1)

For Y(t) :

Y(t)=Yi+Vyi.t-\frac{g.t^{2} }{2}

We suppose yi = 0

Y(t) = Yi +Vyi.t-\frac{g.t^{2} }{2} = 6.62 \frac{m}{s} .t- 4.9\frac{m}{s^{2} } .t^{2}

This is equation (2)

We need the time in which Vy = 0 m/s so we use (1)

Vy (t) = 0\\0=6.62 \frac{m}{s} - 9.8 \frac{m}{s^{2} } .t\\t= 0.675 s

So in t = 0.675 s  → Vy = 0. Now we calculate the y in which this happen using (2)

Y(0.675s) = 6.62\frac{m}{s}.(0.675s)-4.9 \frac{m}{s^{2} }  .(0.675s)^{2} \\Y(0.675s) =2.236 m

2.236 m is the maximum height from the shell (in which Vy=0 m/s)

Let's calculate now the height for t = 0.555 s

Y(0.555s)= 6.62 \frac{m}{s} .(0.555s)-4.9\frac{m}{s^{2} } .(0.555s)^{2} \\Y(0.555s) = 2.165m

The height asked is

∆h = 2.236 m - 2.165 m = 0.071 m

6 0
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