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Elodia [21]
3 years ago
8

A certain part of an aircraft engine has a volume of 1.4 ft^3. (a) Find the weight of the piece when it is made of lead. (b) If

the same piece is made of aluminum, what is its weight? Determine how much weight is saved by using aluminum instead of lead.
Physics
1 answer:
nydimaria [60]3 years ago
6 0

Answer:

(a) 4400.83 N

(b) 1051.7 N

(c) 3349.13 N

Explanation:

V = 1.4 ft^3 = 0.0396 m^3

density of lead = 11.34 x 10^3 kg/m^3

density of aluminium = 2710 kg/m^3

(a) When lead is used

Weight = mass x gravity = volume x density x gravity

            = 0.0396 x 11.34 x 1000 x 9.8 = 4400.83 N

(b) when aluminium is used

Weight = mass x gravity = volume x density x gravity

           = 0.0396 x 2710 x 9.8 = 1051.7 N

(c) Difference in weight = 4400.83 - 1051.7 = 3349.13 N

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3 years ago
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What is the minimum value of the friction coefficient between the boxes that will keep them from slipping when the 100 N force i
CaHeK987 [17]

Answer:

The friction coefficient's minimum value will be "0.173".

Explanation:

The given query seems to be incomplete. Below is the attached file of the complete question.

According to the question,

(a)

The net friction force's magnitude will be:

⇒ F_{net}=ma

           =5\times 1.7

           =8.5 \ N

(b)

For m₃,

⇒ ma=\mu m_3 g

Or,

⇒    \mu=\frac{a}{g}

          =\frac{1.7}{9.8}

          =0.173

5 0
2 years ago
What is the slope if an object comes to a stop then remains at rest?
KIM [24]

It woul be 0 because it is not moving. It is staying at a constant rate.

3 0
3 years ago
How to I isolate t? (time value)
Vlad1618 [11]
Heya!!!

Answer to your question:

All you need to do is to compare both the equations...
On comparing, you'll get
d=6m
v=6.5m/s
a=-9.81m/s^2
we know,

6= 6.5t -0.5 (9.81)t^2
6=6.5t-4.905 t^2
4.905 t^2-6.5t +6=0
now you can use quadratic eq. to solve this.

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4 0
3 years ago
The density of molybdenum is 10.28 g/cm^3 and it crystallizes in the face centered cubic unit cell. Calculate the edge length of
Effectus [21]

The edge length of the unit cell at the given atomic mass and density of the molybdenum is 314.2 pm.

<h3>Volume of molybdenum</h3>

V = (zm/ρN)

where;

  • z is 2 for cubic unit cell
  • m is mass of the molybdenum
  • ρ is density of the molybdenum

V = (2 x 95.96) / (10.28 x 6.02 x 10²³)

V = 3.10 x 10⁻²³ cm³

<h3>Edge length of the unit cell</h3>

a³ = V

a = (V)^¹/₃

a = ( 3.10 x 10⁻²³)^¹/₃

a = 3.142 x 10⁻⁸ cm

a = 3.142 x 10⁻¹⁰ m

a = 314.2 x 10⁻¹² m

a = 314.2 pm

Thus, the edge length of the unit cell at the given atomic mass and density of the molybdenum is 314.2 pm.

Learn more about edge length here:

brainly.com/question/16673486

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4 0
1 year ago
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