What is the least count of screw gauge?<br> (a) 0.01 cm<br> (b) 0.001 cm<br> (c) 0.1 cm<br> (d) 1 mm

The ________ is the part of the drill press that holds and rotates the cutting tool.

lana66690 [7]

Answer:

Spindle

Explanation:

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8 0

3 years ago

Read 2 more answers

Pls help and solve this problem. I don't have an idea to solve this.

lions [1.4K]

Answer:

MIS HIEVOSTES bbbbbbbbbbbb MIS HUEVOTES

7 0

2 years ago

Velocity components in an incompressible flow are: v = 3xy + x^2 y: w = 0. Determine the velocity component in the x-direction.

cupoosta [38]

Answer:

Velocity component in x-direction $u=-\frac{3}{2}x^2-\frac{1}{3}x^3$ .

Explanation:

v=3xy+ $x^{2}$ y

We know that for incompressible flow

$\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0$

$\frac{\partial v}{\partial y}=3x+x^{2}$

So $\frac{\partial u}{\partial x}+3x+x^{2}=0$

$\frac{\partial u}{\partial x}= -3x-x^{2}$

By integrate with respect to x,we will find

$u=-\frac{3}{2}x^2-\frac{1}{3}x^3$ +C

So the velocity component in x-direction $u=-\frac{3}{2}x^2-\frac{1}{3}x^3$ .

3 0

3 years ago

I wish to have a computer whose machine-level instructions are all 32 bits each. If I want to have all instructions of the form

melisa1 [442]

Answer:

Maximum number that can be represented by 13 bits is 8192 Instructions

Explanation:

number of instructions = 1000

number of bits = log(1000) x number of register

= 6 bits

Since the complete instruction must have 32 bits, then

remaining number of bits = 32 - 6 = 236

number of registers in instruction = 2

number of bits per register = 26/2 = 13

Maximum number that can be represented by 13 bits = $2^{n}$

= 2¹³ = 8192

4 0

3 years ago

An interior beam supports the floor of a classroom in a school building. The beam spans 26 ft. and the tributary width is 16 ft.

saul85 [17]

Answer:

a. $L_o$ = 40 psf

b. L ≈ 30.80 psf

c. The uniformly distributed total load for the beam = 812.8 ft./lb

d. The alternate concentrated load is more critical to bending , shear and deflection

Explanation:

The given parameters of the beam the beam are;

The span of the beam = 26 ft.

The width of the tributary, b = 16 ft.

The dead load, D = 20 psf.

a. The basic floor live load is given as follows;

The uniform floor live load, = 40 psf

The floor area, A = The span × The width = 26 ft. × 16 ft. = 416 ft.²

Therefore, the uniform live load, $L_o$ = 40 psf

b. The reduced floor live load, L in psf. is given as follows;

$L = L_o \times \left ( 0.25 + \dfrac{15}{\sqrt{k_{LL} \cdot A_T} } \right)$

For the school, $K_{LL}$ = 2

Therefore, we have;

$L = 40 \times \left ( 0.25 + \dfrac{15}{\sqrt{2 \times 416} } \right) = 30.80126 \ psf$

The reduced floor live load, L ≈ 30.80 psf

c. The uniformly distributed total load for the beam, $W_d$ = b × $W_{D + L}$ =

∴ $W_d$ = = 16 × (20 + 30.80) ≈ 812.8 ft./lb

The uniformly distributed total load for the beam, $W_d$ = 812.8 ft./lb

d. For the uniformly distributed load, we have;

$V_{max}$ = 812.8 × 26/2 = 10566.4 lbs

$M_{max}$ = 812.8 × 26²/8 = 68,681.6 ft-lbs

$v_{max}$ = 5×812.8×26⁴/348/EI = 4,836,329.333/EI

For the alternate concentrated load, we have;

$P_L$ = 1000 lb

$W_{D}$ = 20 × 16 = 320 lb/ft.

$V_{max}$ = 1,000 + 320 × 26/2 = 5,160 lbs

$M_{max}$ = 1,000 × 26/4 + 320 × 26²/8 = 33,540 ft-lbs

$v_{max}$ = 1,000 × 26³/(48·EI) + 5×320×26⁴/348/EI = 2,467,205.74713/EI

Therefore, the loading more critical to bending , shear and deflection, is the alternate concentrated load

7 0

2 years ago

What is the least count of screw gauge? (a) 0.01 cm (b) 0.001 cm (c) 0.1 cm (d) 1 mm