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3 years ago

HELP! Need the correct answer ASAP! Thanks

Engineering

2 answers:

Bess [88]3 years ago

7 0

Answer:

i think it might be answer b

Explanation:

den301095 [7]3 years ago

3 0

Answer:

I also think that it might be b

because these jobs are in there

aerospace engineering and operations technologist and teshnicians
agricultural technicians
air traffic controllers
air craft mechanics and service technicians

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Reverse engineering is not the process of analyzing an existing product to 5 points

Lunna [17]

true

Explanation:

I order for the engineering process to be reversed it had to developed an product and had to be proved what were used to form the product

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3 years ago

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Why is it important to have knowledge on abbreviations, sign and symbols in masonry servicing job in the construction business s

marta [7]

Maybe To know short cuts on the long run. And it is a way of coding

5 0

3 years ago

The velocity of the 7.7-kg cylinder is 0.49 m/s at a certain instant. What is its speed v after dropping an additional 1.28 m? T

alex41 [277]

Answer:

the velocity of the cylinder after dropping an additional 1.28 m is 1.15 m/s

Explanation:

Using the work energy system

$T_1 + U_{1-2} + T_2$

The initial kinetic energy $T_1$ is ;

$T_1 = \dfrac{1}{2}m_cv_1^2 + \dfrac{1}{2}I_o \omega^2$

$T_1 = \dfrac{1}{2}m_cv_1^2 + \dfrac{1}{2}(m_d \overline k^2)(\dfrac{v_1}{r_i})^2$

where;

$m_c$ = mass of the cylinder = 7.7 kg

$v_1 =$ initial velocity of the cylinder = 0.49 m/s

$I_o$ = moment of inertia of the drum about O

$m_d =$ mass of the drum = 10.5 kg

$r_i =$ radius of gyration = 0.3 m

$\omega$ = angular velocity of the drum

$T_1 = \dfrac{1}{2}(7.7)(0.49)^2 + \dfrac{1}{2}(10.5*(0.3)^2(\dfrac{0.49}{0.275})^2$

$T_1 = 2.426 \ J\\$

The final kinetic energy is also calculated as:

$T_2= \dfrac{1}{2}m_cv_2^2 + \dfrac{1}{2}(m_d \overline k^2)(\dfrac{v_2}{r_i})^2$

$T_2= \dfrac{1}{2}(7.7)(v_2^2)^2 + \dfrac{1}{2}(10.5*(0.3)^2(\dfrac{v_2}{0.275})^2$

$T_1 =10.10 v_2^2$

Similarly, The workdone by all the forces on the cylinder can be expressed as:

$U_{1-2} = m_cg(h) - (\dfrac{M}{r_i})h$

where;

g = acceleration due to gravity

h = drop in height of the cylinder

M = frictional moment at O

$U_{1-2} = 7.7*9.81*1.28 - 18.4(\dfrac{1.28}{0.275})$

$U_{1-2} =11.04 \ J$

Finally, using the work energy application;

$T_1 + U_{1-2} + T_2$

2.426 + 11.04 = 10.10 $v_2^2$

13.466 = 10.10 $v_2^2$

$v_2^2$ = $\dfrac{13.466}{10.10}$

$v_2^2$ = 1.333

$v_2 = \sqrt{1.333}$

$v_2 = 1.15 \ m/s$

Thus, the velocity of the cylinder after dropping an additional 1.28 m is 1.15 m/s

8 0

2 years ago

How many catalytic converters are in a ford ranger

fenix001 [56]

Answer:

up to 5

Explanation:

It depends on the year, you can also look on car forums for more detailed information.

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2 years ago

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Water enters a vertical jet with low velocity and a pressure of 350 kPa. What is the maximum height that the water can rise abov

Neporo4naja [7]

The maximum height that the water can rise above the jet is; 35 m

How to find the height pressure of water?

The formula for pressure here is;

P = P₀ + ρgh

where;

P₀ is the pressure at the fluid's surface.

ρ is the density of the fluid. (density of water = 1000 kg/m³)

g is the acceleration due to gravity = 10 m/s².

h is height of rise of fluid

Now, the pressure at water surface is zero because the pressure the atmosphere exerts on the water is equal to the pressure the water exerts on the atmosphere. Thus, P₀ = 0 Pa

We are given P = 350 kPa = 350000 Pa. Thus;

350000 = 0 + (1000 * 10 * h)

350000 = 10000h

h = 350000/10000

h = 35 meters

Read more about water pressure height at; brainly.com/question/14097489

#SPJ1

6 0

2 years ago