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3 years ago

When using fall arrest, free fall must be kept at or below how many feet

1 answer:

5 0

6 feet
Explanation: A few extra feet of free fall can significantly increase the arresting force on the employee, possibly to the point of causing injury. Because of this, the free fall distance should be kept at a minimum, and, as required by the standard, in no case greater than 6 feet (1.8 m).

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What is a microwave transmitter?a) A technology that uses active or passive tags in the form of chips or smart labels that can s

Blababa [14]

Answer:

b) Commonly used to transmit network signals over great distances.

Explanation:

The transmission of information or data by using microwave radio waves is known as microwave transmission. Microwave transmitter is commonly used to transmit network signals over great distances. It is an electronic device that transmits and receives radio frequency signals ranging from 1GHz to 100GHz.

The microwave transmitter has a wide range of applications and these includes, radio stations, television stations, mobile phones, radio astronomy, radar,

5 0

3 years ago

Read 2 more answers

Learning the key concepts of each approach is essential to successful management of a project. What type of unpredictability is

Levart [38]

Answer:

lemme write it down

Explanation:

hold down okay

3 0

3 years ago

Four race cars are traveling on a 2.5-mile tri-oval track. The four cars are traveling at constant speeds of 195 mi/h, 190 mi/h,

Snezhnost [94]

Answer:

Explanation:

1) The number of times, the car with the speed of 195 mph will cross the given point is equal to 30 minutes divided by the time taken by car to cross the 2.5 miles.

0 .5*195/2.5 = 39

Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times

and car with the speed of 180 mph crosses it 36 times

here, the time-mean speed, vt is given below,

vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

= 186.433 mph

and space mean speed is given by,

= (39+38+37+36)/(39/195+38/190+37/1850+36/180)

1) The number of times, the car with the speed of 195 mph will cross the given point is equal to 30 minutes divided by the time taken by car to cross the 2.5 miles.

0 .5*195/2.5 = 39

Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times

and car with the speed of 180 mph crosses it 36 times

here, the time-mean speed, vt is given below,

vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

= 186.433 mph

and space mean speed is given by,

= (39+38+37+36)/(39/195+38/190+37/1850+36/180)

=187.5 mph

2) There would be only four number of observations when the aerial photo is given, therefore time mean speed, vt in that condition will be calculated as

Vt = 195+190+185+180/4

= 187.5

Vs= 4/(1/195+1/190+1/185+1/180)

= 188.36 mph

2) There would be only four number of observations when the aerial photo is given, therefore time mean speed, vt, in that condition will be calculated as

Vt = 195+190+185+180/4

= 187.5

Vs= 4/(1/195+1/190+1/185+1/180)

= 188.36 mph

4 0

4 years ago

1. When the pneumatic device was demonstrated, explain something that was different than what you expected.

Vera_Pavlovna [14]

Answer:

1. Effect of air pressure

2. air- powered wheel chair

3. Pneumatic valves

Explanation:

1. In any pneumatic device, the mipact of air pressure to produce the moving effect on an heavy object is unexpected.

2. pneumatice demultiplexer when air in comprressed tank is allowed released to cause movement of the chair.

3. In industries, a pneumatic valve operates by force of air when actuated. A signal causes actuation of coil. When coil is energized, compressed high pressure air is allowe to enter in a small cylinder and cause operation of valve

5 0

3 years ago

Read 2 more answers

The time to half-maximum voltage is how long it takes the capacitor to charge halfway. Based on your experimental results, how l

satela [25.4K]

Answer:

Time taken for the capacitor to charge to 0.75 of its maximum capacity = 2 × (Time take for the capacitor to charge to half of its capacity)

Explanation:

The charging of a capacitor/the build up of its voltage follows an exponential progression and is given by

V(t) = V₀ [1 - e⁻ᵏᵗ]

where k = (1/time constant)

when V(t) = V₀/2

(1/2) = 1 - e⁻ᵏᵗ

e⁻ᵏᵗ = 0.5

In e⁻ᵏᵗ = In 0.5 = - 0.693

-kt = - 0.693

kt = 0.693

t = (0.693/k)

Recall that k = (1/time constant)

Time to charge to half of max voltage = T(1/2)

T(1/2) = 0.693 (Time constant)

when V(t) = 0.75

0.75 = 1 - e⁻ᵏᵗ

e⁻ᵏᵗ = 0.25

In e⁻ᵏᵗ = In 0.25 = -1.386

-kt = - 1.386

kt = 1.386

t = 1.386(time constant) = 2 × 0.693(time constant)

Recall, T(1/2) = 0.693 (Time constant)

t = 2 × T(1/2)

Hope this Helps!!!

3 0

3 years ago

Read 2 more answers