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gulaghasi [49]
3 years ago
14

Describe 4 common workplace practices that are risk factors for injury

Physics
1 answer:
Aleonysh [2.5K]3 years ago
7 0
High task repitition, forceful exertions, repetitive or sustained awkward posture
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If the observed test value of a hypothesis test is outside of the established critical value(s), a researcher would __________.
sashaice [31]
I just had this question, the awnser is A.

6 0
3 years ago
Read 2 more answers
A thin rod of length 0.75 m and mass 0.42 kg is suspended
MrRissso [65]

Answer:

a)  K = 0.63 J, b)  h = 0.153 m

Explanation:

a) In this exercise we have a physical pendulum since the rod is a material object, the angular velocity is

         w² = \frac{m g d}{I}

where d is the distance from the pivot point to the center of mass and I is the moment of inertia.

The rod is a homogeneous body so its center of mass is at the geometric center of the rod.

              d = L / 2

the moment of inertia of the rod is the moment of a rod supported at one end

              I = ⅓ m L²

we substitute

            w = \sqrt{\frac{mgL}{2}  \ \frac{1}{\frac{1}{3} mL^2} }

            w = \sqrt{\frac{3}{2}  \ \frac{g}{L} }

            w = \sqrt{ \frac{3}{2} \ \frac{9.8}{0.75}  }

            w = 4.427 rad / s

an oscillatory system is described by the expression

              θ = θ₀ cos (wt + Φ)

the angular velocity is

             w = dθ /dt

             w = - θ₀ w sin (wt + Ф)

In this exercise, the kinetic energy is requested in the lowest position, in this position the energy is maximum. For this expression to be maximum, the sine function must be equal to ±1

In the exercise it is indicated that at the lowest point the angular velocity is

           w = 4.0 rad / s

the kinetic energy is

           K = ½ I w²

           K = ½ (⅓ m L²) w²

           K = 1/6 m L² w²

           K = 1/6 0.42 0.75² 4.0²

           K = 0.63 J

b) for this part let's use conservation of energy

starting point. Lowest point

             Em₀ = K = ½ I w²

final point. Highest point

             Em_f = U = m g h

energy is conserved

             Em₀ = Em_f

             ½ I w² = m g h

             ½ (⅓ m L²) w² = m g h

             h = 1/6 L² w² / g

             h = 1/6 0.75² 4.0² / 9.8

             h = 0.153 m

5 0
2 years ago
John recently suffered a blow to his head. Since then, he finds it difficult to comprehend what others say to him. He also finds
katovenus [111]

Answer: ​Wernicke's aphasia

Explanation:

John recently suffered a blow to his head. Since then, he finds it difficult to comprehend what others say to him. He also finds it difficult to express his thoughts and cannot seem to find the right words to say while speaking. However, he can speak freely with proper syntax. In this scenario, John is most likely suffering from Wernicke's aphasia.

Wernicke's aphasia occurs when the leftward side of the middle of the brain is damaged or has been altered. An individual who suffers from Wernicke's aphasia will have difficulty in speaking in meaningful and coherent sentences or may have difficulty in understanding the speech of others.

6 0
3 years ago
Suppose that you connect the terminals of two batteries of different emfs positive to positive and negative to negative (opposin
gulaghasi [49]

Answer:

Answer is explained in the explanation section below.

Explanation:

This question is very basic and easy. The answer to this question is:

Answer: If both batteries are connected we would get less amount of charge as compared to connected a single battery.

Reasoning:

If both batteries are connected in a manner of positive terminal to positive terminal and negative terminal to negative terminal then a capacitor is added to charge it from the batteries then, total electromotive force (emf) would decrease.

As a result, the capacitor added would get less amount of charge stored. But capacitor added will get more amount of charge stored when a single battery is connected.

7 0
2 years ago
This illustration shows two opposing forces pulling on a wagon. Which description best describes how the wagon will move?
Talja [164]

Answer:

The wagon will move to the right.

Explanation:

From the question given above, the following data were obtained:

Force applied to the left (Fₗ) = 10 N

Force applied to the right (Fᵣ) = 30 N

Direction of the wagon =.?

To determine the direction in which the wagon will move, we shall determine the net force acting on the wagon. This can be obtained as follow:

Force applied to the left (Fₗ) = 10 N

Force applied to the right (Fᵣ) = 30 N

Net force (Fₙ) =?

Fₙ = Fᵣ – Fₗ

Fₙ = 30 – 10

Fₙ = 20 N to the right

From the calculations made above, the net force acting on the wagon is 20 N to the right. Hence the wagon will move to the right.

8 0
2 years ago
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