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kodGreya [7K]
3 years ago
10

4. Wheel grinders need be equipped with a(n):

Engineering
1 answer:
Bogdan [553]3 years ago
6 0
The answer is the B because that is the correct answer
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How many hours should it take an articulated wheel loader equipped with a 4-yd^3 bucket to load 3000 yd^3 of gravel (average mat
densk [106]

Answer:

17 hours 15 minutes

Explanation:

See attached picture.

4 0
3 years ago
Consider a W21x93. Determine the moment capacity of the beam. Assume the compression flange is not laterally braced and that the
OLga [1]

Answer:

The answer is "828.75"

Explanation:

Please find the correct question:

For W21x93 BEAM,

Z_x = 221.00 in^3 \\\\\to \frac{b_t}{2t_f} =4.53\\\\\to \frac{h}{t_w}=32.3

For A992 STREL,

F_y= 50\  ks

Check for complete section:

\to \frac{b_t}{2t_f} =4.53 < \frac{65}{\sqrt{f_y = 9.19}}\\\\\to \frac{h}{t_w} =32.3 < \frac{640}{\sqrt{f_y = 90.5}}

Design the strength of beam =\phi_b Z_x F_y\\\\

                                                =0.9 \times 221 \times 50\\\\=9945 \ in \ \ kips\\\\=\frac{9945}{12}\\\\= 828.75 \ft \ kips \\

8 0
3 years ago
Calculate the reluctance of a 4-meter long toroidal coil made of low-carbon steel with an inner radius of 1.75 cm and an outer r
My name is Ann [436]

Answer:

R = 31.9 x 10^(6) At/Wb

So option A is correct

Explanation:

Reluctance is obtained by dividing the length of the magnetic path L by the permeability times the cross-sectional area A

Thus; R = L/μA,

Now from the question,

L = 4m

r_1 = 1.75cm = 0.0175m

r_2 = 2.2cm = 0.022m

So Area will be A_2 - A_1

Thus = π(r_2)² - π(r_1)²

A = π(0.0225)² - π(0.0175)²

A = π[0.0002]

A = 6.28 x 10^(-4) m²

We are given that;

L = 4m

μ_steel = 2 x 10^(-4) Wb/At - m

Thus, reluctance is calculated as;

R = 4/(2 x 10^(-4) x 6.28x 10^(-4))

R = 0.319 x 10^(8) At/Wb

R = 31.9 x 10^(6) At/Wb

8 0
3 years ago
An engineer is testing the shear strength of spot welds used on a construction site. The engineer's null hypothesis at a 5% leve
lilavasa [31]

Answer:

b) The null hypothesis should be rejected.

Explanation:

The null hypothesis is  that the mean shear strength of spot welds is at least

3.1 MPa

H0: u ≥3.1 MPa  against the claim Ha: u< 3.1 MPa

The alternate hypothesis is  that the mean shear strength of spot welds is less than 3.1 MPa.

This is one tailed test

The critical region Z(0.05) < ± 1.645

The Sample mean= x`= 3.07

The number of welds= n= 15

Standard Deviation= s= 0.069

Applying z test

z= x`-u/s/√n

z= 3.07-3.1/0.069/√15

z= -0.03/0.0178

z= -1.68

As the calculated z= -1.68  falls in the critical region Z(0.05) < ± 1.645 the null hypothesis is rejected and the alternate hypothesis is accepted that the mean shear strength of spot welds is less than 3.1 MPa

8 0
2 years ago
Energy transfer in mechanical systems: During steady-state operation, a mechanical gearbox receives 70 KW of input power through
Degger [83]

Answer:

Heat transfer rate(Q)= 1.197kW

Power output(W)=68.803kW

3 0
3 years ago
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