1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
kodGreya [7K]
3 years ago
6

What possible scenarios may happen if you do the task without using PPE?

Engineering
1 answer:
ale4655 [162]3 years ago
8 0
Without PPE, employees are at risk of Cuts and punctures. Chemical burns. Electric shocks. Exposure to excessive noise or vibration.
You might be interested in
Need help with these 3 questions pls help
Burka [1]

Answer:

Explanation:

First one is passion i think

second one is true im not sure

Third one is Time managaemant

Hope this helps!

8 0
3 years ago
Which of these is the BEST description of
strojnjashka [21]

Answer:

i would say C but i may be wrong have a great day

Explanation:

3 0
3 years ago
Having an adequate ___________ provides a driver with the time and space necessary to react to potential and immediate hazards i
ra1l [238]

Answer:

Space Cushion

Explanation:

Having an adequate Space Cushion provides a driver with the time and space necessary to react to potential and immediate hazards in the driving scene.

8 0
2 years ago
Consider a 8-m-long, 8-m-wide, and 2-m-high aboveground swimming pool that is filled with water to the rim. (a) Determine the hy
Stolb23 [73]

Answer:

The hydrostatic force of 313920 N is acted on each wall of the swimming pool and this force is acted at 1 m from the ground. The hydrostatic force is quadruple if the height of the walls is doubled.

<u>Explanation:</u>

To calculate force on the walls of swimming pool whose dimensions are given as <em>8-m-long, 8-m-wide, and 2-m-high</em>. We know that formula for hydrostatic force is \text {hydrostatic force}=\text {pressure} \times \text {area,}=\rho g h \times(l \times h)  

\equiv \rho g h^{2} l, we know ρ=density of fluid=1000 g / c m^{3},

g=acceleration due to gravity=9.81 m / s^{2}, h=height of the pool=2 m and l=length of the pool=8 m.  

hydrostatic force on each wall=1000 \times 9.81 \times 2^{2} \times 8 = 313920 N.

<em>The distance at which hydrostatic force is acted is half of the height of the swimming pool. </em>

At 1 m from the ground this hydrostatic force is acted on each wall.  

The force is <em>quadruple if the height of the walls of the pool is doubled</em> this is because, the<em> height is doubled and taken as h=4 m</em> and substitute in the equation =\rho g h^{2} l = 1000 \times 9.81 \times 4^{2} \times 8 = 1255680 N. This is 4 times 313920 N.

5 0
3 years ago
Air at 1600 K, 30 bar enters a turbine operating at steady state and expands adiabatically to the exit, where the pressure is 2.
djyliett [7]

Solution :

The isentropic efficiency of the turbine is given as :

$\eta = \frac{\text{actual work done}}{\text{isentropic work done}}$

  $=\frac{m(h_1-h_2)}{m(h_1-h_{2s})}$

  $=\frac{h_1-h_2}{h_1-h_{2s}}$

The entropy relation for the isentropic process is given by :

$0=s^\circ_2-s^\circ_1-R \ln \left(\frac{P_2}{P_1}\right)$

$\ln \left(\frac{P_2}{P_1}\right)=\frac{s^\circ_2-s^\circ_1}{R}$

$ \frac{P_2}{P_1}=exp\left(\frac{s^\circ_2-s^\circ_1}{R}\right)$

$\left(\frac{P_2}{P_1}\right)_{s=constant}=\frac{P_{r2}}{P_{r1}}$

Now obtaining the properties from the ideal gas properties of air table :

At $T_1 = 1600 \  K,$

$P_{r1}=791.2$

$h_1=1757.57 \ kJ/kg$

Calculating the relative pressure at state 2s :

$\frac{P_{r2}}{P_{r1}}=\frac{P_2}{P_1}$

$\frac{P_{r2}}{791.2}=\frac{2.4}{30}$

$P_{r2}=63.296$

Obtaining the properties from Ideal gas properties of air table :

At $P_{r2}=63.296$,  $T_{2s}\approx 860 \ K$

Considering the isentropic relation to calculate the actual temperature at the turbine exit, we get:

  $\eta=\frac{h_1-h_2}{h_1-h_{2s}}$

$\eta=\frac{c_p(T_1-T_2)}{c_p(T_1-T_{2s})}$

$\eta=\frac{T_1-T_2}{T_1-T_{2s}}$

$0.9=\frac{1600-T_2}{1600-860}$

$T_2= 938 \ K$

So, at $T_2= 938 \ K$, $h_2=975.66 \ kJ/kg$

Now calculating the work developed per kg of air is :

$w=h_1-h_2$

  = 1757.57 - 975.66

  = 781 kJ/kg

Therefore, the temperature at the exit is 938 K and work developed is 781 kJ/kg.

4 0
3 years ago
Other questions:
  • a) A total charge Q = 23.6 μC is deposited uniformly on the surface of a hollow sphere with radius R = 26.1 cm. Use ε0 = 8.85419
    8·1 answer
  • According to the HERARCHY of hazard control what sequence is specified for protecting workers from chemical hazards assume that
    7·1 answer
  • The ___________ section of the bid package might include information concerning provisions for water and other utilities, sanita
    14·1 answer
  • Suppose there are 93 packets entering a queue at the same time. Each packet is of size 4 MiB. The link transmission rate is 1.4
    9·1 answer
  • import java.util.Scanner; public class FindSpecialValue { public static void main (String [] args) { Scanner scnr = new Scanner(
    11·1 answer
  • Water at 80°F and 30 psia is heated in a chamber by mixing it with saturated water vapor at 30 psia. If both streams enter the m
    13·1 answer
  • Whats a person who enters the electrical trade
    7·2 answers
  • Software piracy occurs when
    15·1 answer
  • (Laminar flow) A fluid flows through two horizontal pipes of equal length which are connected together to form a pipe of length
    6·1 answer
  • Which color is used to indicate that a wire will be energized when power is brought to the circuit
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!