A beam has been fixed to the floor by the pin at B and the roller at A as shown in figure 1 below.

A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 5.0 mm (0.20 in

Sonja [21]

The load is 17156 N.

<u>Explanation:</u>

First compute the flexural strength from:

σ = FL / π $R^{3}$

= 3000 $\times$ (40 $\times$ 10^-3) / π (5 $\times$ 10^-3)^3

σ = 305 $\times$ 10^6 N / m^2.

We can now determine the load using:

F = 2σd^3 / 3L

= 2(305 $\times$ 10^6) (15 $\times$ 10^-3)^3 / 3(40 $\times$ 10^-3)

F = 17156 N.

7 0

3 years ago

The host at the end of the video claims that ___________ is crucial to his success as a driver. A. Reaction time B. A safe space

Snezhnost [94]

Answer:

answer is C. his seat belt

5 0

3 years ago

1. Design a circuit, utilizing set/reset coils where PB 1 starts Motor 1 and PB2 stops Motor 1. Pressing and releasing either pu

lianna [129]

Answer:

Circuit attached with explanation

Explanation:

Hi Dear,

A circuit is attached for your reference.

When you press "start" PB, the supply reaches the motor starter relay coil "M" that is also in parallel with the "start" PB which allows the motor to remain ON even when you release "start" PB as supply to relay coil is directly from supply "L" through "M".

To stop motor just press "stop" PB and the circuit breaks which de-energize the relay coil and the motor stops.

Hope this finds easy to you.

5 0

3 years ago

QUESTION 6

Aloiza [94]

It would be 2 Portfolio

3 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

4 years ago

A beam has been fixed to the floor by the pin at B and the roller at A as shown in figure 1 below.​

A beam has been fixed to the floor by the pin at B and the roller at A as shown in figure 1 below.