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SCORPION-xisa [38]
3 years ago
11

For a body moving with simple harmonic motion state the equations to represent: i) Velocity ii) Acceleration iii) Periodic Time

iv) Frequency v) On a diagram show the positions of max and min values for Acceleration and Velocity and show using the equations why this is the case.

Engineering
1 answer:
max2010maxim [7]3 years ago
4 0

Answer with Explanation:

The general equation of simple harmonic motion is

x(t)=Asin(\omega t+\phi)

where,

A is the amplitude of motion

\omega is the angular frequency of the motion

\phi is known as initial phase

part 1)

Now by definition of velocity we have

v=\frac{dx}{dt}\\\\\therefore v(t)=\frac{d}{dt}(Asin(\omega t+\phi )\\\\v(t)=A\omega cos(\omega t+\phi )

part 2)

Now by definition of acceleration we have

a=\frac{dv}{dt}\\\\\therefore a(t)=\frac{d}{dt}(A\omega cos(\omega t+\phi )\\\\a(t)=-A\omega ^{2}sin(\omega t+\phi )

part 3)

The angular frequency is related to Time period 'T' asT =\frac{2\pi }{\omega }

where

\omega is the angular frequency of the motion of the particle.

Part 4) The acceleration and velocities are plotted below

since the maximum value that the sin(x) and cos(x) can achieve in their respective domains equals 1 thus the maximum value of acceleration and velocity is A\omega ^{2} and A\omega respectively.

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Neglecting the presence of friction, air drag, and other inefficiencies, how much gasoline is consumed when a 1300 kg automobile
koban [17]

Answer:

Explanation:

Given that, .

Mass of car is

M = 1300kg

Velocity of car

V = 80km/h = 80 × 1000/3600

V = 22.22m/s

Calculate the kinetic energy of the vehicle as follows:

K.E = ½ MV²

K.E = ½ × 1300 × 22.22²

K.E = 320,987.65 J

Given that,

Enthalpy is 45MJ / kg

h = 45MJ / kg

Then, enthalpy is given as.

Enthalpy = Energy / mass

h = E / m

45 × 10^6 = 320,987.65 / m

m = 320,987.65 / 45 × 10^6

m = 7.133 × 10^-3 kg

m = 7.133 mg

Also, given that, density is 680kg/m³

Density is given as

Density = mass / Volume

ρ = m / v

Then, v = m / ρ

v = 7.133 × 10^-3 / 680

v = 1.049 × 10^-5 m³

We know that

1mL = 10^-6 m³

Therefore,

v = 1.049 × 10^-5 m³ × 1mL / 10^-6m³

v = 10.49 mL

7 0
3 years ago
An alloy fin with a thermal conductivity of 200 W/ሺm ∙ Kሻ has a length of 2.5 cm and a thickness of 3.5 mm. The base of the fin
Svetllana [295]

Answer: heat flux into the fun is 21.714 mW/m^2

Explanation:

Heat flux Q = q/A

q = heat transfer rate W

A = area m^2

q = area * conductivity * temperature gradient

Temperature gradient = difference in temperature of the metal faces divided by the thickness.

Therefore Q = k * ( temp. gradient)

Q = 200 * ((400-20)/3.5*10^-2)

Q = 21714285.71 = 21.714 mW/m^2

Answer 2: convective heat transfer flux between fin and air

is 3800W/m^2

Explanation :

q = hA*(Ts-Ta)

h = convective heat transfer coefficient

Ts = temperature of fin

Ta = temperature of air

Q = q/A

Q = h(Ts-Ta)

Q = 10(400 - 20)

Q = 3800 W/m^2

5 0
4 years ago
A brass alloy is known to have a yield strength of 275 MPa (40,000 psi), a tensile strength of 380 MPa (55,000 psi), and an elas
Ugo [173]

Answer:

Computation of the load is not possible because E(test) >E(yield)

Explanation:

We are asked to ascertain whether or not it is possible to compute, for brass, the magnitude of the load necessary to produce an elongation of 7.0 mm (0.28 in.). It is first necessary/ important to compute the strain at yielding from the yield strength and the elastic modulus, and then the strain experienced by the test specimen. Then, if

E(test) is less than E(yield), deformation is elastic and the load may be computed. However is E(test) is greater than E(yield) computation/determination of the load is not possible even though defamation is plastic and we have neither a stress-strain plot or a mathematical relating plastic stress and strain. Therefore, we can compute these two values as:

Calculation of E(test is as follows)

E(test) = change in l/lo= Elongation produced/stressed tension= 7.0mm/267mm

=0.0262

Computation of E(yield) is given below:

E(yield) = σy/E=275Mpa/103 ×10^6Mpa= 0.0027

Therefore, we won't be able to compute the load because for computation to take place, E(test) <E(yield). In this case, E(test) is greater than E(yield).

7 0
3 years ago
Read 2 more answers
Ashrae standard 15 -2013 require that each machinery room activate an alarm and mechanical ventillation?
Kobotan [32]

ASHRAE Standard 15 - 1994 requires that each machinery room must activate an alarm and mechanical ventilation before refrigerant concentrations exceed the TLV-TWA (Threshold Limit Value - Time Weighted Average).

<h3>What is ASHRAE Standard 15 - 1994?</h3>
  • The key standards guiding refrigerant identification and usage developed by ASHRAE have been revised to comply with government regulations and achieve improved performance.
  • Standards 15 and 34 provide critical guidance to manufacturers, design engineers, and operators who must stay up to date on new air conditioning and refrigeration requirements.
  • Standard 34 describes a shorthand method of naming refrigerants and assigns safety classifications based on toxicity and flammability data, whereas Standard 15 establishes procedures for operating equipment and systems when those refrigerants are used.
  • Before refrigerant concentrations exceed the TLV-TWA, each machinery room must activate an alarm and mechanical ventilation, according to ASHRAE Standard 15 - 1994 (Threshold Limit Value - Time Weighted Average).

Therefore, ASHRAE Standard 15 - 1994 requires that each machinery room must activate an alarm and mechanical ventilation before refrigerant concentrations exceed the TLV-TWA (Threshold Limit Value - Time Weighted Average).

Know more about ASHRAE Standard here:

brainly.com/question/14483054

#SPJ4

The correct question is given below:
ASHRAE Standard 15 - 1994 requires that each machinery room must activate an alarm and mechanical ventilation before refrigerant concentrations exceed:

5 0
2 years ago
A coal-fired power plant equipped with a SO2 scrubber is required to achieve an overall SO2 removal efficiency of 85%. The exist
3241004551 [841]

Answer:

bypassed fraction B will be B= 0.105 (10.5%)

Explanation:

doing a mass balance of SO₂ at the exit

total mass outflow of SO₂ = remaining SO₂ from the scrubber outflow + bypass stream of SO₂

F*(1-er) =  Fs*(1-es) + Fb

where

er= required efficiency

es= scrubber efficiency

Fs and Fb = total mass inflow of  SO₂ to the scrubber and to the bypass respectively

F= total mass inflow of  SO₂

and from a mass balance at the inlet

F= Fs+ Fb

therefore the bypassed fraction B=Fb/F is

F*(1-er) =  Fs*(1-es) + Fb

1-er= (1-B)*(1-es) +B

1-er = 1-es - (1-es)*B + B

(es-er) = es*B

B= (es-er)/es = 1- er/es

replacing values

B= 1- er/es=1-0.85/0.95 = 2/19 = 0.105 (10.5%)

6 0
3 years ago
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