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Vlada [557]
3 years ago
13

The common electrical wall receptacle voltage in North America is often referred to as 120 volts AC. One hundred twenty volts is

: O the average voltage over many weeks of time. O the peak voltage from an AC wall receptacle. O the arithmetic mean of the voltage as it varies with time. O one-half the peak voltage. O the root mean square (rms) average of the voltage as it varies with time.
Physics
1 answer:
Vika [28.1K]3 years ago
5 0

Answer:

120 volts is the root mean square (rms) average of the voltage as it varies with time.

Explanation:

A. The average voltage over many weeks of time (false)

Reason: Average AC voltage over one cycle is cycle (from one peak to other) is zero and so over many weeks of time  it is zero.

B. The peak voltage from an AC wall receptacle (false)

Reason: The peak voltage of an AC source in North America is zero.

C. The arithmetic mean of the voltage as it varies with time (false)

Reason: Arithmetic mean AC voltage over one cycle is cycle (from one peak to other) is zero and so over many weeks of time  it is zero.

D.  One-half the peak voltage (false)

Peak voltage =170 Volts

One-half the peak voltage = 85 volts

E. The root mean square (rms) average of the voltage as it varies with time (True)

Reason:

The peak voltage and root mean square voltage are related by:

V_{rms}=\frac{V_{p}}{\sqrt{2} }\\\\V_{rms}=\frac{170}\sqrt{2}V_{rms}\\\\V_{rms}=120 Volts

Average value of voltage over one cycle is zero, so instead of calculating average voltage for AC peak voltage is first squared and the mean is calculated.

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A train is traveling away from you at 120 km/h. it blows its whistle, and you hear a tone of 0.400 kHz. Take the speed of sound
Vaselesa [24]

Answer:B)439.21 Hz

Explanation:

Given

velocity of train v=120 km/h\approx 120\times \frac{5}{18}=33.33 m/s

frequency of train f_a=0.400 kHz\approx 400 Hz

speed of sound c=340 m/s

frequency from Doppler effect can be calculated by

f_a=(\frac{c}{c+v})\cdot f

400=(\frac{340}{340+33.33})\cdot f

400=\frac{340}{373.33}\cdot f

f=400\times 1.098

f=439.21 Hz

5 0
3 years ago
Air is compressed from 14.7 psia and 60°F to a pressure of 150 psia while being cooled at a rate of 10 Btu/lbm by circulating wa
iren2701 [21]

Answer:

A. 6.36 lbm/s

b. T_2=341\textdegree F

Explanation:

a. Given the following information;

#Compressor inlet:

Air pressure,P_1=14.7psia,T_1=60\textdegree F

#Compressor outlet:

Air \ Pressure, P_2=150psia\\\\Air \ volume \ flow \ rate, \dot V_1=5000ft^3/min

#Cooling rate,

q_{out}=10Btu/lbm, \dot W=700hp

# From table A-1E

Gas constant of air R=0.3704\ psia.ft^3/lbm.R

Specific enthalpy at P_1=520R-h_1=124.27Btu/lbm

Using the mass balance:

\dot m_{in}-\dot m_{out}=\bigtriangleup \dot m_{sys}=0.0\\\\\dot m_{in}=\dot m_{out}\\\\v_1=\frac{RT_1}{P_1}\\\\v_1=\frac{0.3704\times520}{14.7}\\\\\\v_1=13.1026ft^3/lbm\\\\\dot m=\frac{5000/60}{13.1025}\\\\\dot m=6.36\ lbm/s

Hence, the mass flow rate of the air is 6.36lbm/s

b The specific enthalpy at the exit is defined as the energy balance on the system:

\dot m_{in}-\dot m_{out}=\bigtriangleup \dot m_{sys}=0.0\\\\\dot m_{in}=\dot m_{out}\\\\\dot W_{in}+ \dot mh_1=\dot Q_{out}+\dot m_2\\\\h_2=\frac{\dot W_{in}-\dot Q_{out}}{\dot m}+h_1\\\\h_2=\frac{700\times 0.7068-10\times 6.36}{6.36}+124.27\\\\h_2=192.06\ Btu/lbm\\\\#at \ h_2=182.06Btu/lbm, T_2=801R=341\textdegree F

Hence, the temperature at the compressor exit T_2=341\textdegree F

5 0
4 years ago
A 89 kg bobsled is pushed along a horizontal surface by two athletes. After the bobsled is pushed a distance of 3.8 m starting f
shutvik [7]

Answer:

Net force, F = 205.59 N

Explanation:

Given that,

Mass of a bobsled, m = 89 kg

Initial speed of the bobsled, u = 0

Distance travelled, d = 33.8 m

Final speed of the bobsled, v = 12.5 m/s

To find,

The magnitude of the net force

Solution,

Let a is the acceleration of the bobsled. Using the third equation of kinematics as :

v^2-u^2=2ad

(12.5)^2=2a\times 33.8

a=2.31\ m/s^2

Let F is the magnitude of net force acting on the bobsled. It can be calculated as :

F=ma

F=89\ kg\times 2.31\ m/s^2

F = 205.59 N

Therefore, the magnitude of the net force acting on the bobsled is 205.59 N.

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