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Anarel [89]
3 years ago
15

A typical cell phone consumes an average of about 1.00 W of electrical power and operates on 3.80 V. (a) What average current do

es the phone draw from its battery? (b) Calculate the energy stored in a fully charged battery if the phone requires charging after 5.00 hours of use.
Physics
1 answer:
dusya [7]3 years ago
4 0

Answer:

0.06925 A

18000 J

Explanation:

V = Voltage = 3.8 V

P = Power = 1 W

t = Time = 5 hours

Power is given by

P=V^2I\\\Rightarrow I=\dfrac{P}{V^2}\\\Rightarrow I=\dfrac{1}{3.8^2}\\\Rightarrow I=0.06925\ A

The average current the phone draws from its battery is 0.06925 A

Energy is given

E=Pt\\\Rightarrow E=1\times 5\times 3600\\\Rightarrow E=18000\ J

The energy stored in a fully charged battery is 18000 J

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A pipe is horizontal and carries oil that has a viscosity of 0.14 Pa s. The volume flow rate of the oil is 5.3 × 10-5 m3/s. The
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Answer:

Explanation:

Given

\mu =0.14 Pa.s

Volume Flow rate Q=5.3\times 10^{-5} m^3/s

Length L=37 m

radius r=0.58 cm

D=1.16 cm

According to Hagen-Poiseuille Equation

difference in Pressure is given

\Delta P=\frac{128\mu L\cdot Q}{\pi D^4}

P_{in}-P_{out}=\frac{128\mu L\cdot Q}{\pi D^4}

P_{in}=P_{out}+\frac{128\mu L\cdot Q}{\pi D^4}

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3 years ago
A proton is 0.9 meters away from a 1.4 C charge. What is the magnitude of the electric force between the proton and the charge
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Answer:

F = 2.49 x 10⁻⁹ N

Explanation:

The electrostatic force between two charged bodies is given by Colomb's Law:

F = \frac{kq_1q_2}{r^2}\\

where,

F = Electrostatic Force = ?

k = colomb's constant = 9 x 10⁹ N.m²/C²

q₁ = charge on proton = 1.6 x 10⁻¹⁹ C

q₂ = second charge = 1.4 C

r = distace between charges = 0.9 m

Therefore,

F = \frac{(9\ x\ 10^9\ N.m^2/C^2)(1.6\ x\ 10^{-19}\ C)(1.4\ C)}{(0.9\ m)^2}

<u>F = 2.49 x 10⁻⁹ N</u>

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Answer:

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