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4 years ago

6

True or false a critique of hazwoper incidents that have occurred in the past year should not be included in hazwoper 8 hour ref

resher training

1 answer:

Jobisdone [24]4 years ago

7 0

Answer:

False

Explanation:

No matter if something happened in the past year or so, it still should be included for safety reasons so it wont happen again

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a solid circular shaft 150 mm diameter transmits a torque of 29820.586 N.m. find the maximum shear stress

bezimeni [28]

Answer:

45 Mpa

Explanation:

The maximum shear stress, $\tau_{max}$ is given by

$\tau_{max}=\frac {16T}{\pi d^{3}}$

Where T is the torque and d is the diameter.

Substituting 29820.586 N.m for T and 0.15m for d then we obtain

$\tau_{max}=\frac {16\times 29820.586}{\pi \times 0.15^{3}}=44999999.22\approx 45 Mpa$

4 0

4 years ago

Write a program that asks the user to input a vector of integers of arbitrary length. Then, using a for-end loop the program exa

ELEN [110]

Answer:

%Program prompts user to input vector

v = input('Enter the input vector: ');

%Program shows the value that user entered

fprintf('The input vector:\n ')

disp(v)

%Loop for checking all array elements

for i = 1 : length(v)

%check if the element is a positive number

if v(i) > 0

%double the element

v(i) = v(i) * 2;

%else the element is negative number.

else

%triple the element

v(i) = v(i) * 3;

end

end

%display the modified vector

fprintf('The modified vector:\n ')

disp(v)

4 0

4 years ago

Ket [755]

Answer:

mnfokfnfi3or

Explanation:

can you translate it into english.....

8 0

3 years ago

The expression with an AND operator ________ when the expression on its left is false and so it does not evaluate the expression

MatroZZZ [7]

It is the short circuit

3 0

3 years ago

Air flows at 45m/s through a right angle pipe bend with a constant diameter of 2cm. What is the overall force required to keep t

HACTEHA [7]

Answer:

b)1.08 N

Explanation:

Given that

velocity of air V= 45 m/s

Diameter of pipe = 2 cm

Force exerted by fluid F

$F=\rho AV^2$

So force exerted in x-direction

$F_x=\rho AV^2$

$F_x=1.2\times \dfrac{\pi}{4}\times 0.02^2\times 45^2$

F=0.763 N

So force exerted in y-direction

$F_y=\rho AV^2$

$F_y=1.2\times \dfrac{\pi}{4}\times 0.02^2\times 45^2$

F=0.763 N

So the resultant force R

$R=\sqrt{F_x^2+F_y^2}$

$R=\sqrt{0.763^2+0.763^2}$

R=1.079

So the force required to hold the pipe is 1.08 N.

3 0

3 years ago