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Nataliya [291]
3 years ago
9

The displacement of a certain object is described by y(t) = 23 sin 5t, where t is measured in seconds. Compute its period and it

s oscillation frequency in rad/sec and in hertz
Engineering
1 answer:
serious [3.7K]3 years ago
7 0

Answer:

5 rad/sec

0.796 Hz

1.256 seconds

Explanation:

Y(t) = 23sin5t ----1

Y(t) = aSin(w)t----2

w = 5 rad/sec

Then we get the Oscillation frequency from the formula below

w = 2πf

We make f to be the subject of this formula

f = w/2π

f = 5/2*3.24

f = 5/6.28

f = 0.796Hz

From the calculated frequency, we get the period = 1/f

= 1/0.796

= 1.256 seconds

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Answer:

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Explanation:

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4 years ago
An insulated piston-cylinder device contains 0.15 of saturated refrigerant-134a vapor at 0.8 MPa pressure. The refrigerant is no
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Answer:

Assumption:

1. The kinetic and potential energy changes are negligible

2. The cylinder is well insulated and thus heat transfer is negligible.

3. The thermal energy stored in the cylinder itself is negligible.

4. The process is stated to be reversible

Analysis:

a. This is reversible adiabatic(i.e isentropic) process and thus s_{1} =s_{2}

From the refrigerant table A11-A13

P_{1} =0.8MPa   \left \{ {{ {{v_{1}=v_{g}  @0.8MPa =0.025645 m^{3/}/kg } } \atop { {{u_{1}=u_{g}  @0.8MPa =246.82 kJ/kg } -   also  {{s_{1}=s_{g}  @0.8MPa =0.91853 kJ/kgK } } \right.

sat vapor

m=\frac{V}{v_{1} } =\frac{0.15}{0.025645} =5.8491 kg\\and \\\\P_{2} =0.2MPa  \left \{ {{x_{2} =\frac{s_{2} -s_{f} }{s_{fg }}=\frac{0.91853-0.15449}{0.78339}   = 0.9753 \atop {u_{2} =u_{f} +x_{2} }(u_{fg}) =  38.26+0.9753(186.25)= 38.26+181.65 =219.9kJ/kg \right. \\s_{1} = s_{2}

T_{2} =T_{sat @ 0.2MPa} = -10.09^{o}  C

b.) We take the content of the cylinder as the sysytem.

This is a closed system since no mass leaves or enters.

Hence, the energy balance for adiabatic closed system can be expressed as:

E_{in} - E_{out}  =ΔE

w_{b, out}  =ΔU

w_{b, out} =m([tex]u_{1} -u_{2)

w_{b, out}  = workdone during the isentropic process

=5.8491(246.82-219.9)

=5.8491(26.91)

=157.3993

=157.4kJ

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3 years ago
Draw a circuit diagram of one lamp controlled by one switch and show how insulation resistance test is carried out​
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3 years ago
After being purged with nitrogen, a low-pressure tank used to store flammable liquids is at a total pressure of 0.03 psig. (a) I
g100num [7]

Answer:

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Explanation:

8 0
3 years ago
A car hits a tree at an estimated speed of 10 mi/hr on a 2% downgrade. If skid marks of 100 ft. are observed on dry pavement (F=
Dafna11 [192]

Answer:

v_{o} = 22.703\,\frac{m}{s} \left(50.795\,\frac{m}{s}\right)

Explanation:

The deceleration of the car on the dry pavement is found by the Newton's Law:

\Sigma F = -\mu_{k,1}\cdot m\cdot g \cdot \cos \theta + m\cdot g \cdot \sin \theta = m\cdot a_{1}

Where:

a_{1} = (-\mu_{k,1}\cdot \cos \theta + \sin \theta)\cdot g

a_{1} = (-0.33\cdot \cos 1.146^{\textdegree}+\sin 1.146^{\textdegree})\cdot \left(9.807\,\frac{m}{s^{2}} \right)

a_{1} = -3.040\,\frac{m}{s^{2}}

Likewise, the deceleration of the car on the unpaved shoulder is:

a_{2} = (-\mu_{k,2}\cdot \cos \theta + \sin \theta)\cdot g

a_{2} = (-0.28\cdot \cos 1.146^{\textdegree}+\sin 1.146^{\textdegree})\cdot \left(9.807\,\frac{m}{s^{2}} \right)

a_{2} = -2.549\,\frac{m}{s^{2}}

The speed just before the car entered the unpaved shoulder is:

v_{o} = \sqrt{\left(4.469\,\frac{m}{s} \right)^{2}-2\cdot \left(-2.549\,\frac{m}{s^{2}} \right)\cdot (60.88\,m)}

v_{o} = 18.175\,\frac{m}{s}

And, the speed just before the pavement skid was begun is:

v_{o} = \sqrt{\left(18.175\,\frac{m}{s} \right)^{2}-2\cdot \left(-3.040\,\frac{m}{s^{2}} \right)\cdot (30.44\,m)}

v_{o} = 22.703\,\frac{m}{s} \left(50.795\,\frac{m}{s}\right)

8 0
3 years ago
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