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alexira [117]
3 years ago
13

Consider a spherical Gaussian surface and three charges: q1 = 1.60 μC , q2 = -2.61 μC , and q3 = 3.67 μC . Find the electric flu

x through the Gaussian surface if it completely encloses (a) only charges q1 and q2, (b) only charges q2 and q3, and (c) all three charges.

Physics
1 answer:
Olenka [21]3 years ago
4 0

Answer:

Explanation:

<h3>Guass Law: Also known as  "Gauss's flux theorem" is the total of the electric flux "φ" out of a closed surface is equal to the charge "Q" enclosed divided by the permittivity εο. Solution is attached.</h3>

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X rays are used in hospital to locate in the patients bones.If the x rays of wavelength of 2×10 to the power negative nine m tra
jasenka [17]

Answer:1.5×10 to the power of 17(unit-Hertz/H)

Explanation:V=F×Wavelength

F=V/Wavelength=3×10 to power/2×10 to power of -9=1.5×10 to power of 17

8 0
3 years ago
The inductor in a radio receiver carries a current of amplitude 200 mA when a voltage of amplitude 2.4 V is across it at a frequ
zzz [600]

Answer:

The value of the inductance is 1.364 mH.

Explanation:

Given;

amplitude current, I₀ = 200 mA = 0.2 A

amplitude voltage, V₀ = 2.4 V

frequency of the wave, f = 1400 Hz

The inductive reactance is calculated;

X_l = \frac{V_o}{I_o} \\\\X_l = \frac{2.4}{0.2} \\\\X_l =12 \ ohms

The inductive reactance is calculated as;

X_l = \omega L\\\\X_l = 2\pi fL\\\\L = \frac{X_l}{2 \pi f}

where;

L is the inductance

L = \frac{12}{2 \pi \times \ 1400} \\\\L = 1.364 \times \ 10^{-3} \ H\\\\L = 1.364 \ mH

Therefore, the value of the inductance is 1.364 mH.

7 0
3 years ago
A jetliner is traveling east from Salt Lake City to Washington DC, a distance of 1,850 miles. The jetliner travels at an average
maks197457 [2]
The jetliner is traveling against the wind. The net speed of the jetliner is
590 mph - 36 mph = 554 mph
The time it takes for the jetliner to arrive at the destination is
1850 miles / 554 mph = 3.34 hours
3 0
4 years ago
A ball is thrown downward at 12 m/s from a windowsill 35 m above the ground. At the same time, another ball is thrown upward at
wariber [46]

Answer:

The second ball lands 1.5 s after the first ball.

Explanation:

Given;

initial velocity of the ball, u = 12 m/s

height of fall, h = 35 m

initial velocity of the second, v = 12 m/s

Time taken for the first ball to land;

t = \sqrt{\frac{2h}{g} }\\\\t =\sqrt{ \frac{2*35}{9.8}}\\\\t = 2.67 \ s

determine the maximum height reached by the second ball;

v² = u² -2gh

at maximum height, the final velocity, v = 0

0 = 12² - (2 x 9.8)h

19.6h = 144

h = 144 / 19.6

h = 7.35 m

time to reach this height;

t_1 = \sqrt{\frac{2h}{g} }\\\\t_1 =  \sqrt{\frac{2*7.35}{9.8}}\\\\t_1 = 1.23 \ s

Total height above the ground to be traveled by the second ball is given as;

= 7.35 m + 35m

= 42.35 m

Time taken for the second ball to fall from this height;

t_2 = \sqrt{\frac{2h}{g} }\\\\t_2 = \sqrt{\frac{2*42.35}{9.8} }\\\\t_2 = 2.94 \ s

total time spent in air by the second ball;

T = t₁ + t₂

T = 1.23 s + 2.94 s

T = 4.17 s

Time taken for the second ball to land after the first ball is given by;

t = 4.17 s -  2.67 s

T = 1.5 s

Therefore, the second ball lands 1.5 s after the first ball.

4 0
3 years ago
A heat engine is operating on a Carnot cycle and has a thermal efficiency of 47 percent. The waste heat from this engine is reje
Fantom [35]

Answer:

Explanation:

Given that,

Efficiency of Carnot engine is 47%

η =47%=0.47

The wasted heat is at temp 60°F

TL=60°F

Rate of heat wasted is 800Btu/min

Therefore, rate of heat loss QL is

QL' = 800×60 =48000

The power output is determined from rate of heat obtained from the source and rate of wasted heat.

Therefore,

W' = QH' - QL'

Note QH' = QL' / (1-η)

W' = QL' / (1-η) - QL'

W'=QL' η / (1-η)

W'= 48000×0.47/(1-0.47)

W'=42566.0377 BTU

1 btu per hour (btu/h) = 0.00039 horsepower (hp)

Then, 42566.0377×0.00039

W'=16.6hp

Which is approximately 17hp

b. Temperature at source

Using ratio of wanted heat to temp

Then,

TH / TL = QH' / QL'

TH = TL ( QH' / QL')

Since, QH' = QL' / (1-η)

Then, TH= TL( QL' /QL' (1-η))

TH=TL/(1-η)

TL=60°F, let convert to rankine

°R=°F+459.67

TL=60+459.67

TL=519.67R

TH=519.67/(1-0.47)

TH=980.51R

Which is approximately 1000R

6 0
3 years ago
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