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tamaranim1 [39]
3 years ago
6

What are the 5 basic types of propulsion systems​

Engineering
1 answer:
Basile [38]3 years ago
5 0
Propeller , turbine engine
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You can help build a safe work environment by using your knowledge of violence prevention strategies to spot what?
Ostrovityanka [42]

Answer:

warning signs

Explanation:

give directions on your surroundings

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3 years ago
Air at a temperature of 333 °C flows with a velocity of 10 m/s over a flat plate 0.5m long. Estimate the cooling rate per unit w
Rama09 [41]

The answer & explanation for this question is given in the attachment below.

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4 years ago
What Are 2 Properties electromagnets have that permanent magnets do not?
NikAS [45]

Answer:The distinction between a permanent magnet and an electromagnet is essentially one in how the field is created, not the properties of the field afterwards. So electromagnets still have two poles, still attract ferromagnetic materials, and still have poles that repel other like poles and attract unlike poles.

Explanation:

4 0
3 years ago
Read 2 more answers
Based on experimental observations, the acceleration of a particle is defined by the relation a = –(0.1 + sin x/b), where a and
Allisa [31]

Answer:

a) v = +/- 0.515 m/s

b) x = -0.098164 m

c)  v = +/- 1.005 m/s

Explanation:

Given:

The relationship for the acceleration is given as follows:

                            a = - (0.1 + sin(x/b))

Where,  b = 0.98

- IVP is v = 1 m/s @ x = 0

Find:

(a) the velocity of the particle when x = -1 m

(b) the position where the velocity is maximum

(c) the maximum velocity.

Solution:

- We will compute the velocity by integrating a by dt.

                          a = v*dv / dx =  - (0.1 + sin(x/0.98))

- Separate variables:

                          v*dv = - (0.1 + sin(x/0.98)) . dx

-Integrate from v = 1 m/s to v and @ x = 0 to x:

                         0.5*(v^2) = - (0.1*x - 0.98*cos(x/0.8)) - 0.98 + 0.5

                         0.5*v^2 =  0.98*cos(x/0.98) - 0.1*x - 0.48

- Evaluate at, x = -1

                         0.5*v^2 = 0.98 cos(-1/0.98) + 0.1 -0.48

                         v = sqrt (0.2651155)

                         v = +/- 0.515 m/s

- v = v_max when a = 0. Set the given expression to zero and solve for x:

                          -0.1 = sin(x/0.98)

                           x = -0.98*0.1002

                           x = -0.098164 m

- Hence now evaluate velocity through the derived expression:

                           v^2 = 1.96 cos(-0.098164/0.98) -0.96 -0.2*-0.098164

                           v = sqrt (1.0098)

                           v = +/- 1.005 m/s

4 0
3 years ago
An inventor claims to have developed a power cycle operating between hot and cold reservoirs at 1350 K and 295 K, respectively,
Art [367]

Answer:

efficiency = 0.71999  <    0.78148  

efficiency = 0.7999   ≈     0.78148  

Explanation:

given data

Hot reservoir Temp Th = 1350 K

Cold reservoir Temp Tc = 295 K

heat transfer rate  = 150,000 kJ/h

solution

first we get here maximum possible efficiency that is express as

maximum possible efficiency = 1 -  \frac{Tc}{Th}   ..........1

put here value  and we get

maximum possible efficiency = 1 - \frac{295}{1350}

maximum possible efficiency = 0.78148  

and

efficiency = \frac{Wout}{Qin}   .................2

here Qin = \frac{150000}{3600}  = 41.667

so put here value in equation 2 we get

efficiency = \frac{30}{41.667}  

efficiency = 0.71999  <    0.78148  

and

efficiency = \frac{33.33}{41.667}  

efficiency = 0.7999   ≈     0.78148  

claim might be true or not because there always be looser

5 0
4 years ago
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