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3 years ago

What are the potential hazards relating to materials handling injuries?

Engineering

1 answer:

Ann [662]3 years ago

8 0

Answer:

it's says potential that something not moving.

incorrectly cutting ties or securing devices because

this is so dangerous thing I know because being electrocuted is here this for me.

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A heavy ball with a weight of 150 N is hung from the ceiling of a lecture hall on a 4.0-m-long rope. The ball is pulled to one s

shusha [124]

Answer:

The tension in the rope at the lowest point is 270 N

Explanation:

Given;

weight of the ball, W = 150 N

length of the rope, r = 4 m

velocity of the ball, v = 5.6 m/s

When the ball passes through the lowest point, the tension on the rope is the sum of weight of the ball and centripetal force.

T = W + F

Centripetal force, F = mv²/r

where;

m is the mass of the ball

m = W/g

m = 150 / 9.8 = 15.306 kg

Centripetal force, F = mv²/r

F = (15.306 x 5.6²)/4

F = 120 N

T = W + F

T = 150 + 120

T = 270 N

Therefore, the tension in the rope at the lowest point is 270 N

6 0

3 years ago

is released from under a vertical gate into a 2-mwide lined rectangular channel. The gate opening is 50 cm, and the flow rate in

SVEN [57.7K]

Answer:

hello your question is incomplete attached below is the complete question

answer: There is a hydraulic jump

Explanation:

First we have to calculate the depth of flow downstream of the gate

y1 = $C_{c} y_{g}$ ----------- ( 1 )

Cc ( concentration coefficient ) = 0.61 ( assumed )

Yg ( depth of gate opening ) = 0.5

hence equation 1 becomes

y1 = 0.61 * 0.5 = 0.305 m

calculate the flow per unit width q

q = Q / b ----------- ( 2 )

Q = 10 m^3 /s

b = 2 m

hence equation 2 becomes

q = 10 / 2 = 5 m^2/s

next calculate the depth before hydraulic jump y2 by using the hydraulic equation

answer : where y1 < y2 hence a hydraulic jump occurs in the lined channel

attached below is the remaining part of the solution

4 0

3 years ago

You are investigating surface hardening in iron using nitrogen gas. Two 5 mm thick slabs of iron are separately exposed to nitro

Luden [163]

Answer and Explanation:

The explanation is attached below

4 0

3 years ago

A 03-series cylindrical roller bearing with inner ring rotating is required for an application in which the life requirement is

-BARSIC- [3]

Answer:

$\mathbf{C_{10} = 137.611 \ kN}$

Explanation:

From the information given:

Life requirement = 40 kh = 40 $40 \times 10^{3} \ h$

Speed (N) = 520 rev/min

Reliability goal $(R_D)$ = 0.9

Radial load $(F_D)$ = 2600 lbf

To find C10 value by using the formula:

$C_{10}=F_D\times \pmatrix \dfrac{x_D}{x_o +(\theta-x_o) \bigg(In(\dfrac{1}{R_o}) \bigg)^{\dfrac{1}{b}}} \end {pmatrix} ^{^{^{\dfrac{1}{a}}$

where;

$x_D = \text{bearing life in million revolution} \\ \\ x_D = \dfrac{60 \times L_h \times N}{10^6} \\ \\ x_D = \dfrac{60 \times 40 \times 10^3 \times 520}{10^6}\\ \\ x_D = 1248 \text{ million revolutions}$

$\text{The cyclindrical roller bearing (a)}= \dfrac{10}{3}$

The Weibull parameters include:

$x_o = 0.02$

$(\theta - x_o) = 4.439$

$b= 1.483$

∴

Using the above formula:

$C_{10}=1.4\times 2600 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{1}{\dfrac{10}{3}}}$

$C_{10}=3640 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{3}{10}}$

$C_{10} = 3640 \times \bigg[\dfrac{1248}{0.9933481582}\bigg]^{\dfrac{3}{10}}$

$C_{10} = 30962.449 \ lbf$

Recall that:

1 kN = 225 lbf

∴

$C_{10} = \dfrac{30962.449}{225}$

$\mathbf{C_{10} = 137.611 \ kN}$

7 0

3 years ago

What 2 forces move the secondary piston ahead?

jekas [21]

Answer:

The primary piston activates one of the two subsystems. The hydraulic pressure created, and the force of the primary piston spring, moves the secondary piston forward.

5 0

3 years ago