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tino4ka555 [31]
3 years ago
7

The tool that is used to measure clearances

Engineering
1 answer:
Alina [70]3 years ago
4 0

Answer:

Feeler gauges

Explanation:

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True or false? Don't break or crush mercury-containing lamps because mercury powder may be released.
mars1129 [50]

Answer:true

Explanation:

7 0
4 years ago
When is a handrail required for stairs?
Leviafan [203]

Answer:

after 8 stepshddnffuddbnggkbdbkloyr

5 0
3 years ago
A laboratory in the Y building keep a vacuum pressure of 0.1 kPa abs. What is the net force acting on the door considering the a
seropon [69]

Answer:

net force acting on the floor is 100 kN

Explanation:

Given data:

P_{vaccum} = 0.1 kPa

P_{atm} = 101.325 kPa

dimension of floor = 2 m \times 0.5 m

we know that

Net force can be calculated as follow

f_{net} = P_{vaccum} \times area

f_{net} = 0.1\times 10^3 \times 2\times 0.5

f_{net} = 0.1\times 10^3 \times 1

f_{net} = 100 kN

Therefore net force acting on the floor is 100 kN

7 0
4 years ago
The supplement file* that enclosed to this homework consists Time Versus Force data. The first column in the file stands for tim
AysviL [449]

Answer:

A.) 1mv = 2000N

B.) Impulse = 60Ns

C.) Acceleration = 66.67 m/s^2

Velocity = 4 m/s

Displacement = 0.075 metre

Absorbed energy = 60 J

Explanation:

A.) Using a mathematical linear equation,

Y = MX + C

Where M = (2000 - 0)/( 898 - 0 )

M = 2000/898

M = 2.23

Let Y = 2000 and X = 898

2000 = 2.23(898) + C

2000 = 2000 + C

C = 0

We can therefore conclude that

1 mV = 2000N

B.) Impulse is the product of force and time.

Also, impulse = momentum

Given that

Mass M = 30kg

Velocity V = 2 m/s

Impulse = M × V = momentum

Impulse = 30 × 2 = 60 Ns

C.) Force = mass × acceleration

F = ma

Substitute force and mass into the formula

2000 = 30a

Make a the subject of formula

a = 2000/30

acceleration a = 66.67 m/s^2

Since impulse = 60 Ns

From Newton 2nd law,

Force = rate of change in momentum

Where

change in momentum = -MV - (- MU)

Impulse = -MV + MU

Where U = initial velocity

60 = -60 + MU

30U = 120

U = 120/30

U = 4 m/s

Force = 2000N

Impulse = Ft

Substitute force and impulse to get time

60 = 2000t

t = 60/2000

t = 0.03 second

Using third equation of motion

V^2 = U^2 + 2as

Where S = displacement

4^2 = 2^2 + 2 × 66.67S

16 = 4 + 133.4S

133.4S = 10

S = 10/133.4

S = 0.075 metre

D.) Energy = 1/2 mV^2

Energy = 0.5 × 30 × 2^2

Energy = 15 × 4 = 60J

5 0
3 years ago
What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of 5 × 10-4
White raven [17]

Answer:

The question is a problem that requires the principles of fracture mechanics.

and we will need this equation below to get the Max. Stress that exist at the tip of an internal crack.

Explanation:

Max Stress, σ = 2σ₀√(α/ρ)

where,

σ₀  = Tensile stress = 190MPa = 1.9x10⁸Pa

α  = Length of the cracked surface = (4.5x10⁻²mm)/2 = 2.25x10⁻⁵m

ρ  = Radius of curvature of the cracked surface = 5x10⁻⁴mm = 5x10⁻⁷m

Max Stress, σ = 2 x  1.9x10⁸ x (2.25x10⁻⁵/5x10⁻⁷)⁰°⁵

Max Stress, σ = 2 x  1.9x10⁸ x 6.708 Pa

Max Stress, σ = 2549MPa

Hence, the magnitude of the maximum stress that exists at the tip of an internal crack = 2549MPa

7 0
4 years ago
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