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4 years ago

9

A rod 16.0 cm long is uniformly charged and has a total charge of -25.0 µC. Determine the magnitude and direction of the electri

c field along the axis of the rod at a point 42.0 cm from its center.

1 answer:

Vlada [557]4 years ago

8 0

Answer:

-1.4x10^6N/C

Explanation:

Pls see attached file

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A boy on the beach holds a spherical balloon filled with air. at 10:00am, the temperature on the beach is 20°c and the balloon h

Anon25 [30]

For idea gases, volume is directly proportional to temperature. That is, an increase in temperature leads to increase in volume and vice versa.

Therefore,
V1/T1 = V2/T2 => T2 = (V2*T1)/V1

Assuming that the balloon is spherical in shape,

V= 4/3*pi*R^3.... In the formula for calculating T2, 4/3*pi cancels out.

R1 = 30/2 15 cm; R2 = 30.5/2 = 15.25 cm; T1 = 20+273.15 =293.15 K

Therefore,

T2 = (R2^3*T1)/R1^3 = (15.25^3*293.15)/15^3 = 308.05 K = 34.9 °C

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3 years ago

Water is discharged from a pipeline at a velocity v (in ft/sec) given by v = 1306p(1/2), where p is the pressure (in psi). If th

shepuryov [24]

Answer:

$a=38.5 ft/sec^{2}$

Explanation:

Note that acceleration is the rate change of velocity i.e

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Since the velocity is giving as a variable dependent on the pressure, we have to differentiate implicitly both side with respect to time,i.e

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if we substitute value for the pressure as giving in the question and also since the rate change of pressure is 0.354psi/sec, we have

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$a=\frac{dv}{dt}=653(36)^{-1/2}*0.354\\ a=38.5 ft/sec^{2}$

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3 years ago

Electricity from the local power company costs 7.22 cents/kWh. A homeowner wishes to pay a maximum of 25 cents per month for the

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You haven't said how much power the stereo uses. It matters !

Whatever that number is, the maximum hours per month is

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3 years ago

Read 2 more answers

The interior space of large box is kept at 30 C. The walls of the box are 3 m high and have a ‘sandwich’ construction consisting

White raven [17]

Answer:

$\frac{\dot Q}{A} =20.129\ W.m^{-2}$

$T_1=27.58\ ^{\circ}C$ & $T_2=2.41875\ ^{\circ}C$

Explanation:

Given:

interior temperature of box, $T_i=30^{\circ}C$

height of the walls of box, $h=3\ m$

thickness of each layer of bi-layered plywood, $x_p=1.25\ cm=0.0125\ m$

thermal conductivity of plywood, $k_p=0.104\ W.m^{-1}.K^{-1}$

thickness of sandwiched Styrofoam, $x_s=5\ cm=0.05\ m$

thermal conductivity of Styrofoam, $k_s=0.04\ W.m^{-1}.K^{-1}$

exterior temperature, $T_o=0^{\circ}C$

<u>From the Fourier's law of conduction:</u>

$\dot Q=\frac{dT}{(\frac{x}{kA}) }$

$\dot Q=\frac{dT}{R_{th} }$ ....................................(1)

<u>Now calculating the equivalent thermal resistance for conductivity using electrical analogy:</u>

$R_{th}=R_p+R_s+R_p$

$R_{th}=\frac{x_p}{k_p.A}+\frac{x_s}{k_s.A}+\frac{x_p}{k_p.A}$

$R_{th}=\frac{1}{A} (\frac{x_p}{k_p}+\frac{x_s}{k_s}+\frac{x_p}{k_p})$

$R_{th}=\frac{1}{A} (\frac{0.0125}{0.104}+\frac{0.05}{0.04}+\frac{0.0125}{0.104})$

$R_{th}=\frac{1.4904}{A}$ .....................(2)

Putting the value from (2) into (1):

$\dot Q=\frac{30-0}{\frac{1.4904}{A} }$

$\dot Q=\frac{30\ A}{1.4904}$

$\frac{\dot Q}{A} =20.129\ W.m^{-2}$ is the heat per unit area of the wall.

The heat flux remains constant because the area is constant.

<u>For plywood-Styrofoam interface from inside:</u>

$\frac{\dot Q}{A} =k_p.\frac{T_i-T_1}{x_p}$

$20.129=0.104\times \frac{30-T_1}{0.0125}$

$T_1=27.58\ ^{\circ}C$

&<u>For Styrofoam-plywood interface from inside:</u>

$\frac{\dot Q}{A} =k_s.\frac{T_1-T_2}{x_s}$

$20.129=0.04\times \frac{27.58-T_2}{0.05}$

$T_2=2.41875\ ^{\circ}C$

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3 years ago

In a tug of war, when one team is pulling with a force of 85 N and the other 40 N, what is the net

olya-2409 [2.1K]

85 N - 40 N = 45 N
And depending on direction the greater force is being pulled towards

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