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MrRissso [65]
3 years ago
11

An automobile can be considered to be mounted on four identical springs as far as vertical oscillations are concerned. The sprin

gs of a certain car are adjusted so that the oscillations have a frequency of 3.27 Hz. (a) What is the spring constant of each spring if the mass of the car is 1140 kg and the mass is evenly distributed over the springs? (b) What will be the oscillation frequency if five passengers, averaging 68.0 kg each, ride in the car with an even distribution of mass?
Physics
1 answer:
tresset_1 [31]3 years ago
7 0

Answer:

a) 120341 N/m

b) 1.432 Hz

Explanation:

Given that

Frequency of oscillations of spring, F = 3.27 Hz

Mass of the car, m = 1140 kg

Average mass of passengers, M = 68 kg

The mass on each spring,

m = 1/4 m,

m = 1140 / 4

m = 285 kg

The spring constant on each spring, k is

ω = √(k/m), also,

ω = 2πf, so that

2πf = √(k/m)

If we square both sides, then

4π²f² = k/m

k = 4π²f²m if we substitute, we have

k = 4 * 3.142² * 3.27² * 285

k = 120341 Nm

Since the mass of each person is 68 kg, then the total mass of all passengers would be

68 * 5 = 340 kg

Thus, the total mass of the oscillating system is,

mass of passengers + mass of car

340 kg + 1140 kg

M(total) = 1480 kg

To get the vibrating frequency, we substitute for values in the equation

2πf = √(k/m)

F = 1/2π * √(k/m)

F = 1/2π * √(120341 / 1480)

F = 1/2π * √81.31

F = 0.1591 * 9

F = 1.432 Hz

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Margarita [4]

The change in momentum is 5500 kg m/s

Explanation:

The change in momentum of an object is given by

\Delta p = m(v-u)

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In this problem, we have:

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v=28.0 m/s (final velocity)

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Therefore, the change in momentum is

\Delta p=(2.50\cdot 10^2)(28.0-6.0)=5500 kg m/s

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3 years ago
The resistance of resistor is greater for:
Monica [59]

Answer:

c: long and thin resistor.

Explanation:

The resistance of a resistor is given by:

R = ρ*L/A

where:

R = resistance

ρ = resistivity (depends on the material)

L =  length of the material

A = cross-sectional area of the material

We can see that the length is on the numerator, which means that if we increase the length, then the resistance is increased.

We also can see that the cross-sectional area is on the denominator, then if we increase the area (for example, with a ticker resistor) the resistance decreases.

Then if we want to maximize the resistance, we need to have a long and thin resistor, so the correct answer is c.

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(b) The particle displacement y of air molecules due to a sound wave is given by y 4m and w = = 0.008 cos wt sin kz. Where k - m
serious [3.7K]

The distance between two consecutive nodes and the amplitude after 0.56s are m/2 and 1.75×10^(-4) m respectively.

<h3>What's the distance between consecutive nodes of the displacement of air molecules?</h3>
  • Wavelength is the distance between two consecutive nodes or toughs or crests or anti-nodes.
  • So, distance between consecutive nodes = wavelength = 2π÷k

= 2π/(4π÷m)

= m/2

<h3>What's the amplitude after 0.56s of the displacement of air molecules?</h3>

Displacement after 0.56 s = 0.008×cos(50π×0.56s)

=1.75×10^(-4) m

Thus, we can conclude that the distance between consecutive nodes and displacement after 0.56 s are m/2 and 1.75×10^(-4) m respectively.

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Question: The particle displacement y of air molecules due to a sound wave is given by y=0.008coswtsinkz where k=4π÷m and w=50π rads/s.

Calculate:

I) the distance between 2 consecutive nodes

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8 0
1 year ago
Check all choices below that are correct. Increasing the frequency increases the current. Changing the frequency does not affect
Annette [7]

Answer:

the experyoeyv

Explanation:

3 0
3 years ago
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A counterflow double-pipe heat exchanger is used to heat water from 20°C to 80°C at a rate of 1.2 kg/s. The heating is to be com
bixtya [17]

Answer:L=109.16 m

Explanation:

Given

initial temperature =20^{\circ}C

Final Temperature =80^{\circ}C

mass flow rate of cold fluid \dot{m_c}=1.2 kg/s

Initial Geothermal water temperature T_h_i=160^{\circ}C

Let final Temperature be T

mass flow rate of geothermal water \dot{m_h}=2 kg/s

diameter of inner wall d_i=1.5 cm

U_{overall}=640 W/m^2K

specific heat of water c=4.18 kJ/kg-K

balancing energy

Heat lost by hot fluid=heat gained by cold Fluid

\dot{m_c}c(T_h_i-T_h_e)= \dot{m_h}c(80-20)

2\times (160-T)=1.2\times (80-20)

160-T=36

T=124^{\circ}C

As heat exchanger is counter flow therefore

\Delta T_1=160-80=80^{\circ}C

\Delta T_2=124-20=104^{\circ}C

LMTD=\frac{\Delta T_1-\Delta T_2}{\ln (\frac{\Delta T_1}{\Delta T_2})}

LMTD=\frac{80-104}{\ln \frac{80}{104}}

LMTD=91.49^{\circ}C

heat lost or gain by Fluid is equal to heat transfer in the heat exchanger

\dot{m_c}c(80-20)=U\cdot A\cdot (LMTD)

A=\frac{1.2\times 4.184\times 1000\times 60}{640\times 91.49}=5.144 m^2

A=\pi DL=5.144

L=\frac{5.144}{\pi \times 0.015}

L=109.16 m

6 0
3 years ago
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