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Angelina_Jolie [31]
4 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]4 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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Explanation:

For projectile motion, use constant acceleration equation:

Δx = v₀ t + ½ at²

where Δx is the displacement,

v₀ is the initial velocity,

a is the acceleration,

and t is time.

Both objects are projected upward with velocity u.  The second object is thrown after a time t₀.

For the first object:

Δx = u t + ½ (-g) t²

Δx = u t − ½g t²

For the second object:

Δx = u (t−t₀) + ½ (-g) (t−t₀)²

Δx = u (t−t₀) − ½g (t−t₀)²

Assuming the objects meet, the displacements will be equal:

u t − ½g t² = u (t−t₀) − ½g (t−t₀)²

u t − ½g t² = u (t−t₀) − ½g (t² − 2tt₀ + t₀²)

u t − ½g t² = u t − u t₀ − ½g t² + g tt₀ − ½g t₀²

0 = -u t₀ + g tt₀ − ½g t₀²

0 = -u + g t − ½g t₀

g t = u + ½g t₀

t = u/g + t₀/2

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Answer:

I cannot do the laws of physics cuz I hate science but I'm just dance

Explanation:

<u>it's</u><u> it's true</u>

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3 years ago
If the time of the impact in a collision is extended by 4 times ,how much does the force of impact change
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You are working with a team that is designing a new roller coaster-type amusement park ride for a major theme park. You are pres
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Answer:

<em>The required constant friction force for the last 20 m is 6,862.8 N</em>

Explanation:

<u>Energy Conversion</u>

There are several ways the energy is manifested in our physical reality. Some examples are Kinetic, Elastic, Chemical, Electric, Potential, Thermal, Mechanical, just to mention some.

The energy can be converted from one form to another by changing the conditions the objects behave. The question at hand states some types of energy that properly managed, will make the situation keep under control.

Originally, the m=220 kg car is at (near) rest at the top of a h=101 m tall track. We can assume the only energy present at that moment is the potential gravitational energy:

E_1=mgh=220\cdot 9.8\cdot 101=217,756\ J

For the next x1=230 m, a constant friction force Fr1=350 N is applied until it reaches ground level. This means all the potential gravitational energy was converted to speed (kinetic energy K1) and friction (thermal energy W1). Thus

E_1=K_1+W_1

We can compute the thermal energy lost during this part of the motion by using the constant friction force and the distance traveled:

W_1=F_{r1}\cdot x_1=350\cdot 230=80,500\ J

This means that the kinetic energy that remains when the car reaches ground level is

K_1=E_1-W_1=217,756\ J-80,500\ J=137,256\ J

We could calculate the speed at that point but it's not required or necessary. That kinetic energy is what keeps the car moving to its last section of x2=20 m where a final friction force Fr2 will be applied to completely stop it. This means all the kinetic energy will be converted to thermal energy:

W_2=F_{r2}\cdot x_2=137,256

Solving for Fr2

\displaystyle F_{r2}=\frac{137,256}{20}=6,862.8\ N

The required constant friction force for the last 20 m is 6,862.8 N

3 0
3 years ago
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