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Brrunno [24]
3 years ago
15

Which health risk is acculipble to workers in the retail sector?

Physics
1 answer:
grigory [225]3 years ago
8 0

Answer:

D. injuries associated with lifting loads

Explanation:

The retail industry or store is the industry that involves the handling of their goods directly from the manufacturer and delivers goods to the consumer.

<em>The health risk is acculipble to workers in the retail sector is injuries associated with lifting loads because while manually handling heavy or awkward sized objects, which are common in the retail industry causes strains and sprains. </em>

So, the common activity in the retail store is lifting loads which leads to common injuries such as strains and sprains.

Hence, the correct answer is "D. injuries associated with lifting loads".

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MaRussiya [10]
In order to solve this problem, we must first find out the value of each line on the number line. However, we can make this problem more simple by ignoring every interval except for the ones between 0 and 6. There are three total intervals in between 0 and 6 (including 6 and excluding 0). Therefore, we can do 6/2, and get an interval value of 2. This means that each line adds a value of 2. Since the car is only one line past zero, we only have to add one value of 2. Since 0 + 2 = 2, our final answer is C. 2.

Hope this helps!
4 0
3 years ago
Read 2 more answers
The electric flux through a spherical surface is 1.4 ✕ 105 N · m2/C. What is the net charge (in C) enclosed by the surface?
Anit [1.1K]

Answer:

The  value is   Q_{net} =  1.239 *10^{-6} \  C

Explanation:

From the question we are told that

   The electric flux is \Phi =  1.4*10^{5} \  N\cdot m^2/C

     

Generally the net charge is mathematically represented as

    Q_{net} =  \Phi *  \epsilon_o

Here \epsilon_o is the permetivity of free space with value  

       \epsilon_o =  8.85*10^{-12}  \  \  m^{-3} \cdot kg^{-1}\cdot  s^4 \cdot A^2

So

   Q_{net} =  1.4*10^5 *  8.85*10^{-12}

=>   Q_{net} =  1.239 *10^{-6} \  C

8 0
3 years ago
A ball thrown horizontally from a point 30.0 m above the ground strikes the ground after travelling a horizontal distance of 18.
AURORKA [14]

The components of the ball's position r at time t are

\begin{cases}r_x(t)=v_{0x}t\\v_y(t)=30.0-4.9t^2\end{cases}

The ball stops 18.0 m from where it began, so that

\begin{cases}18.0=v_{0x}t\\0=30.0-4.9t^2\end{cases}

From the second equation, we can show that the ball travels for about t=2.47 seconds, which means it was initially thrown with a horizontal velocity of

18.0\,\mathrm m=v_{0x}(2.47\,\mathrm s)\implies v_{0x}=7.29\,\dfrac{\mathrm m}{\mathrm s}

7 0
4 years ago
If you push on the sides of a filled balloon, how does the gas pressure inside the balloon change?
stepan [7]
The pressure in the balloon increases.
4 0
3 years ago
An object moving in a straight line is decelerating. Which about its acceleration is necessarily true?
Varvara68 [4.7K]

Answer:

Its acceleration is certainly negative.

Explanation:

  • When an object moving in a straight line decelerates then its velocity certainly decreases.
  • Its acceleration is certainly negative.
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a=\frac{v_f-v_i}{t_f-t_i}

where:

v_f\ \&\ v_i are final and initial velocities of the object

t_f\ \&\ t_i are final and initial time of observation.

  • Its SI unit is m.s^{{-2}
3 0
3 years ago
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