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QveST [7]
3 years ago
14

If the compound swivel base is set on 60 degrees at the lathe centerline index, how many degrees will the reading be at the cros

s slide index? A. 45 B. 150 C. 30 D. 90
Engineering
1 answer:
Effectus [21]3 years ago
6 0

Answer:

C.30°

Explanation:

Given that compound swivel base is set on 60° at the lathe center line index.

We need to find reading on cross slide index

We know that relationship between center line index and cross slide index in angle 2∝=β

Where ∝ Angle of swivel and β is the reading on cross line index.

So by using above  relationship between center line index and cross slide index  

2∝=60°⇒∝=30°

So our option is C.

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What is photosynthesis​
sashaice [31]

\sf{\underline{\underline{~~~~~~~~~~~~~~photosynthesis~~~~~~}}}

Photosynthesis is a chemical reaction that takes place inside a plant, producing food for the plant to survive.Carbon dioxide, water and light are all needed for photosynthesis to take place.Photosynthesis happens in the leaves of a plant.

8 0
3 years ago
Read 2 more answers
The rate of energy transfer by work is called power. a)-True b)-False
Pie

Answer:

Yes the statement is true.

Explanation:

Power is defined as the rate at which energy is transferred by an object on account of work done.

Mathematically

Power=\frac{dE}{dt}

An object that does work loses it's energy while an object on which work is done gains energy.

Power is often dependent on the type of energy transfer thus we have Electrical Power, Mechanical Power depending on the type of energy involved in the system.

Concept of power is important since it gives us a measure of how fast energy can be derived to given to a system.

5 0
3 years ago
Could I please get help with this​
alex41 [277]

Answer:

1.I_{xc} = 7.161458\overline 3 in.⁴

I_{yc} = 36.661458\overline 3 in.⁴

Iₓ = 28.6458\overline 3 in.⁴

I_y = 138.6548\overline 3 in.⁴

2. I_{xc} = 114.\overline 3 in.⁴

I_{yc} = 37.\overline 3 in.⁴

Iₓ = 457.\overline 3 in.⁴

I_y = 149.\overline 3 in.⁴

3. The maximum deflection of the beam is 2.55552 inches

Explanation:

1. The height of the beam having a rectangular cross section is h = 2.5 in.

The breadth of the beam, is = 5.5 in.

The moment of inertia of a rectangular beam through its centroid is given as follows;

I_{xc} = b·h³/12 = 5.5 × 2.5³/12 = 1375/192 = 7.161458\overline 3

I_{xc} = 7.161458\overline 3 in.⁴

I_{yc} = h·b³/12 = 2.5 × 5.5³/12 = 6655/192 = 36.661458\overline 3

I_{yc} = 36.661458\overline 3 in.⁴

The moment of inertia about the base is given as follows;

Iₓ = b·h³/3 = 5.5 × 2.5³/3 = 625/24 = 28.6458\overline 3

Iₓ = 28.6458\overline 3 in.⁴

I_y = h·b³/3 = 2.5 × 5.5³/3 = 6655/48= 138.6548\overline 3

I_y = 138.6548\overline 3 in.⁴

2. The height of the beam having a rectangular cross section is h = 7 in.

The breadth of the beam, b = 4 in.

The moment of inertia of a rectangular beam through its centroid is given as follows;

I_{xc} = b·h³/12 = 4 × 7³/12 = 114.\overline 3

I_{xc} = 114.\overline 3 in.⁴

I_{yc} = h·b³/12 = 7 × 4³/12 = 37.\overline 3

I_{yc} = 37.\overline 3 in.⁴

The moment of inertia about the base is given as follows;

Iₓ = b·h³/3 = 4 × 7³/3 = 457.\overline 3

Iₓ = 457.\overline 3 in.⁴

I_y = h·b³/3 = 2.5 × 5.5³/3 = 149.\overline 3

I_y = 149.\overline 3 in.⁴

3. The deflection, \delta _{max}, of a simply supported beam having a point load at the center is given as follows;

\delta_{max} = \dfrac{W \times L^3}{48 \times E \times I}

The given parameters of the beam are;

The length of the beam, L = 22 ft. = 264 in.

The applied load at the center, W = 750 lbs

The modulus of elasticity for Cedar = 10,000,000 psi

The height of the wood, h = 3 in.

The breadth of the wood, b = 5 in.

The moment of inertia of the wood, I_{xc} = b·h³/12 = 5 × 3³/12 = 11.25 in.⁴

By plugging in the given values, we have;

\delta_{max} = \dfrac{750 \times 264^3}{48 \times 10,000,000 \times 11.25} = 2.55552

The maximum deflection of the beam, \delta _{max} = 2.55552 inches

5 0
3 years ago
In a tensile test on a steel specimen, true strain is 0.171 at a stress of 263.8 MPa. When true stress is 346.2 MPa, true strain
swat32

Answer:n=0.973

Explanation:

Given

When True strain\left ( \epsilon _T_1\right )=0.171

at \sigma _1=263.8 MPa

When True stress\left ( \sigma _2\right )=346.2 MPa

true strain \left ( \epsilon _T_2\right )=0.226

We know

\sigma =k\epsilon ^n

where \sigma=True stress

\epsilon=true strain

n=strain hardening exponent

k=constant

Substituting value

263.8=k\left ( 0.171\right )^n------1

346.2=k\left ( 0.226\right )^n-----2

Divide 1 & 2 to get

\frac{346.2}{263.8}=\left ( \frac{0.226}{0.171}\right )^n

1.312=\left ( 1.3216\right )^n

Taking Log both side

ln\left ( 1.312\right )=nln\left ( 1.3216\right )

n=0.973

6 0
3 years ago
To ensure that a vehicle crash is inelastic, vehicle safety designers add crumple zones to vehicles. A crumple zone is a part of
spin [16.1K]

Answer:

Explanation:

Answer: With crumple zones at the front and back of most cars, they absorb much of the energy (and force) in a crash by folding in on itself much like an accordion. ... As Newton's second law explains force = Mass x Acceleration this delay reduces the force that drivers and passengers feel in a crash.Sep 30, 2020

5 0
3 years ago
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