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Liula [17]
4 years ago
13

You have a special light bulb with a very delicate wire filament. the wire will break if the current in it ever exceeds 1.50 a,

even for an instant. what is the largest root-mean-square current you can run through this bulb? 31.2 . a sinusoidal current
Physics
1 answer:
Montano1993 [528]4 years ago
8 0
The root mean square of the current is given by
I_{rms} =  \frac{I_0}{ \sqrt{2} }
where I_0 is the maximum value of the current.

In our problem, the maximum current allowed without breaking the filament is equal to I_0=1.50 A. Therefore, the largest root-mean-square current allowed without breaking the wire is
I_{rms}= \frac{1.50 A}{ \sqrt{2} }=1.06 A
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The earth rotates once per day about an axis passing through the north and south poles. True or False
alexandr402 [8]

Answer: False.

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While the revolution of the Earth around the gives rise to a year, as it takes 365 days for the Earth to go round the sun.

5 0
3 years ago
why do a circulatory system is important in meeting the needs of all cells throughout an animals body
valentinak56 [21]
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3 0
4 years ago
What type of animals does Dr. Grant study?
Rasek [7]
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Idk if this is related to what you ask but it might help.
7 0
3 years ago
Air flows through an adiabatic turbine that is in steady operation. The air enters at 150 psia, 900oF, and 350 ft/s and leaves a
Nonamiya [84]

Answer:

1486.5\frac{Btu}{s}

Explanation:

The inlet specific volume of air is given by:

v_1=\frac{RT_1}{P_1}\\\\v_1=\frac{(0.3704\frac{psia.ft^3}{lbm.R})(1360R)}{150psia}\\\\v_1=3.358\frac{ft^3}{lbm} \ \ \ \  \ \  \ \ \...i

The mass flow rates is expressed as:

\dot m=\frac{1}{v_1}A_1V_1\\\\\dot m=\frac{1}{3.358ft^3/psia}(0.1ft^2)(350ft/s)\\\\\dot m=10.42\frac{lbm}{s}

The energy balance for the system can the be expresses in the rate form as:

E_{in}-E_{out}=\bigtriangleup \dot E=0\\\\E_{in}=E_{out}\\\\\dot m(h_1+0.5V_1^2)=\dot W_{out}+\dot m(h_2+0.5V_2^2)+Q_{out}\\\\\dot W_{out}=\dot m(h_2-h_1+0.5(V_2^2-V_1^2))=-m({cp(T_2-t_1)+0.5(V_2^2-V_1^2)})\\\\\\\dot W_{out}=-(10.42lbm/s)[(0.25\frac{Btu}{lbm.\textdegree F})(300-900)\textdegree F+0.5((700ft/s)^2-(350ft/s)^2)(\frac{1\frac{Btu}{lbm}}{25037ft^2/s^2})]\\\\\\\\=1486.5\frac{Btu}{s}

Hence, the mass flow rate of the air is 1486.5Btu/s

5 0
3 years ago
if a 3-kg object has a momentum of 33 kg��m/s, what's its velocity? a. 99 m/s b. 36 m/s c. 11 m/s d. 9.1 m/s
horsena [70]
M=3kg
p=33kg.m/s
p=m*v
v=p/m
=33/3
=11m/s
thus option (c)
6 0
3 years ago
Read 2 more answers
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