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Neko [114]
3 years ago
5

Select the best answer to the questo

Engineering
2 answers:
Norma-Jean [14]3 years ago
7 0

Answer:

C

Explanation:

lana66690 [7]3 years ago
7 0
The correct one is “C”
Technician A
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The MOST common injury causing absence from work is
erastovalidia [21]

B

Now this is a guess sorry but i am positive on this guess

8 0
3 years ago
Read 2 more answers
WHAT IS THE EFFECT OF ICE ACCRETION ON THE LONGITUDINAL STABILITY OF AN AIRCRAFT?
soldier1979 [14.2K]

Answer:

The major effects of ice accretion on the aircraft is that it disturbs the flow of air and effects the aircraft's performance.

Explanation:

The ice accretion effects the longitudinal stability of an aircraft as:

1. The accumulation of ice on the tail of an aircraft results in the reduction the longitudinal stability and  the elevator's efficacy.

2. When the flap is deflected at 10^{\circ} with no power there is an increase in the longitudinal velocity.  

3. When the angle of attack is higher close to the stall where separation occurs in the early stages of flow, the effect of ice accretion are of importance.  

4. When the situation involves no flap  at reduced power setting results in the decrease in aircraft's longitudinal stability an increase in change in coefficient of pitching moment  with attack angle.

5 0
3 years ago
Torque is a twisting force. If the required torque applied on a 3 ft wrench is 45 ft·lb, what is the force that must be applied?
natka813 [3]

Answer:

15 lbs

Explanation:

assuming you push from the end of the wrench (3ft)

torque = force(distance)

force = torque/distance

(45 ft·lb)/(3 ft)= 15 lbs

8 0
4 years ago
5) A 80-kg man has a total foot imprint area of 480 cm2. Determine the pressure this man exerts on the ground if (a) he stands o
attashe74 [19]

Answer:

The pressure exerted by this man on ground

(a) if he stands on both feet is 8.17 KPa

(b) if he stands on one foot is 16.33 KPa

Explanation:

(a)

When the man stand on both feet, the weight of his body is uniformly distributed around the foot imprint of both feet. Thus, total area in this case will be:

Area = A = 2 x 480 cm²

A = 960 cm²

A = 0.096 m²

The force exerted by man on his area will be equal to his weight.

Force = F = Weight

F = mg

F = (80 kg)(9.8 m/s²)

F = 784 N

Now, the pressure exerted by man on ground will be:

Pressure = P = F/A

P = 784 N/0.096 m²

<u>P = 8166.67 Pa = 8.17 KPa</u>

(b)

When the man stand on one foot, the weight of his body is uniformly distributed around the foot imprint of that foot only. Thus, total area in this case will be:

Area = A = 480 cm²

A = 0.048 m²

The force exerted by man on his area will be equal to his weight, in this case, as well.

Force = F = Weight

F = mg

F = (80 kg)(9.8 m/s²)

F = 784 N

Now, the pressure exerted by man on ground will be:

Pressure = P = F/A

P = 784 N/0.048 m²

<u>P = 16333.33 Pa = 16.33 KPa</u>

4 0
3 years ago
Pipe (2) is supported by a pin at bracket C and by tie rod (1). The structure supports a load P at pin B. Tie rod (1) has a diam
Galina-37 [17]

Answer:

P_max = 25204 N

Explanation:

Given:

- Rod 1 : Diameter D = 12 mm , stress_1 = 110 MPa

- Rod 2: OD = 48 mm , thickness t = 5 mm , stress_2 = 65 MPa

- x_1 = 3.5 mm ; x_2 = 2.1 m ; y_1 = 3.7 m

Find:

- Maximum Force P_max that this structure can support.

Solution:

- We will investigate the maximum load that each Rod can bear by computing the normal stress due to applied force and the geometry of the structure.

- The two components of force P normal to rods are:

               Rod 1 : P*cos(Q)  

               Rod 2: - P*sin(Q)

where Q: angle subtended between x_1 and Rod 1 @ A. Hence,

               Q = arctan ( y_1 / x_1)

               Q = arctan (3.7 / 2.1 ) = 60.422 degrees.

- The normal stress in each Rod due to normal force P are:

               Rod 1 : stress_1 = P*cos(Q)  / A_1

               Rod 2: stress_2 = - P*sin(Q)  / A_2

- The cross sectional Area of both rods are A_1 and A_2:

               A_1 = pi*D^2 / 4

               A_2 = pi*(OD^2 - ID^2) / 4

- The maximum force for the given allowable stresses are:

               Rod 1: P_max =  stress_1 * A_1 / cos(Q)

                          P_max = (110*10^6)*pi*0.012^2 / 4*cos(60.422)

                          P_max = 25203.61848 N

               Rod 2: P_max =  stress_2 * A_2 / sin(Q)

                          P_max = (65*10^6)*pi*(0.048^2 - 0.038^2) / 4*sin(60.422)

                          P_max = 50483.4 N

- The maximum force that the structure can with-stand is governed by the member of the structure that fails first. In our case Rod 1 with P_max = 25204 N.

             

8 0
3 years ago
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