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miss Akunina [59]
4 years ago
12

Supplies, clutter and __________ are two common tripping hazards in a shop environment.

Engineering
1 answer:
Andreyy894 years ago
6 0

Answer:

The correct option is;

Loose cords

Explanation:

Based on the 2011 Census of Occupational Injuries, which is published by the Bureau of Labor Statistics, one of the leading causes of injuries at work is slips, trips and falls which may lead long duration of time down and large huge amount of claims for compensation. Slips trips and falls are also comes fourth in the reasons of fatality at work

Trip hazards are hazards that causes trip and fall by stopping and locking the movement of the step of people walking along traffic lanes

Items that cause trips mainly include item used for work. The correct option is therefore loose cords, which should be kept under cable bridges for safety.

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A sports car has a drag coefficient of 0.29 and a frontal area of 20 ft2, and is travelling at a speed of 120 mi/hour. How much
Andrej [43]

Answer:

Power required to overcome aerodynamic drag is 50.971 KW

Explanation:

For explanation see the picture attached

4 0
3 years ago
A voltage regulator is to provide a constant DC voltage Vl=10V to a load Rl from a nominal Vcc=15V supply voltage. The load can
Luden [163]

Answer:

Beta values can be from the equation=change in Vcc/nominal Vcc

Beta=16-3/15=3/15=1/5=0.20

Maximum power=I^2*R=40 W

7 0
3 years ago
Coulomb's Law Two point charges experience an attractive force of 10.8 N when they are separated by 2.4 m. VWhat force in newton
SashulF [63]

Answer:

The force between the charges when the separation decreases to 0.7 meters equals 126.955 Newtons

Explanation:

We know that for two point charges of magnitude q_{1},q_{2} the magnitude of force between them is given by

F=\frac{k_{e}q_{1}\cdot q_{2}}{r^{2}}

where

k_{e} is constant

r is the separation between the charges

Initially when the charges are separated by 2.4 meters the force can be calculated as

F_{1}=\frac{k_{e}\cdot q_{1}q_{2}}{2.4^{2}}\\\\10.8=\frac{k_{e}\cdot q_{1}q_{2}}{2.4^{2}}\\\\\therefore k_{e}\cdot q_{1}q_{2}=10.8\times 2.4^{2}=62.208

Now when the separation is reduced to 0.7 meters the force is similarly calculated as

F_{2}=\frac{k_{e}\cdot q_{1}q_{2}}{0.7^{2}}

Applying value of the constant we get

F_{1}=\frac{62.208}{0.7^{2}}

Thus F_{2}=126.955Newtons

5 0
3 years ago
Our rule-of-thumb for presenting final results is to round to three significant digits or four if the first digit is a one. By t
olga_2 [115]

Answer:

To four significant digits = 2097 psi

Explanation:

<u>Applying the rule of thumb </u>

σ = Mc/I  ---- ( 1 )

M = 1835 Ibf in ,  I/c = 0.875 in^3

∴ c/l = 1 / 0.875 = 1.1429

back to equation 1

σ = 1835 * 1.1429 = 2097.2215 psi

To four significant digits = 2097 psi

4 0
3 years ago
Technician A says that rear-wheel drive vehicles usually get better traction than front-wheel drive vehicles. Technician B says
cluponka [151]
C is the correct answer
3 0
3 years ago
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