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Elden [556K]
3 years ago
5

A complete series circuit consists of a 12.0 V battery, a 4.70 O resistor, and a switch. The internal resistance of the battery

is 0.30 O. The switch is open. What does an ideal voltmeter read when placed across the terminals of the switch?
12.0 V
9.40 V
2.40 V
zero
Physics
2 answers:
uysha [10]3 years ago
6 0

Answer:

12.0V is correct !!

Explanation:

RoseWind [281]3 years ago
4 0

With the switch open, there's no current in the circuit, and therefore
no voltage drop across any of the dissipative elements (the resistor
or the battery's internal impedance).  So the entire battery voltage
appears across the switch, and the voltmeter reads 12.0V .

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The SI unit of acceleration is the metre per second squared (m s−2); or "metre per second per second", as the velocity in metres per second changes by the acceleration value, every second.

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How many foot-pounds of work does it take to throw a baseball 90 mph? a baseball weighs 5 oz, or 0.3125 lb?
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Kinetic energy of the ball is (mv²) / 2, where m is the mass and v is the velocity

So plugging in the mass and the velocity into the kinetic energy expression, you get:

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A(n) _______ is a point level measuring system consisting of a circuit of two or more probes or electrodes, or an electrode and
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Answer:

Conductivity probe

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6 0
3 years ago
Suppose a shrimp has been put on the ground that has just been taken out of water.Now touch the shrimp from a distance by a stic
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4 0
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Read 2 more answers
Tennis balls traveling at greater than 100 mph routinely bounce off tennis rackets. At some sufficiently high speed, however, th
Kipish [7]

Answer:

Probability of tunneling is 10^{- 1.17\times 10^{32}}

Solution:

As per the question:

Velocity of the tennis ball, v = 120 mph = 54 m/s

Mass of the tennis ball, m = 100 g = 0.1 kg

Thickness of the tennis ball, t = 2.0 mm = 2.0\times 10^{- 3}\ m

Max velocity of the tennis ball, v_{m} = 200\ mph = 89 m/s

Now,

The maximum kinetic energy of the tennis ball is given by:

KE = \frac{1}{2}mv_{m}^{2} = \frac{1}{2}\times 0.1\times 89^{2} = 396.05\ J

Kinetic energy of the tennis ball, KE' = \frac{1}{2}mv^{2} = 0.5\times 0.1\times 54^{2} = 154.8\ m/s

Now, the distance the ball can penetrate to is given by:

\eta = \frac{\bar{h}}{\sqrt{2m(KE - KE')}}

\bar{h} = \frac{h}{2\pi} = \frac{6.626\times 10^{- 34}}{2\pi} = 1.0545\times 10^{- 34}\ Js

Thus

\eta = \frac{1.0545\times 10^{- 34}}{\sqrt{2\times 0.1(396.05 - 154.8)}}

\eta = \frac{1.0545\times 10^{- 34}}{\sqrt{2\times 0.1(396.05 - 154.8)}}

\eta = 1.52\times 10^{-35}\ m

Now,

We can calculate the tunneling probability as:

P(t) = e^{\frac{- 2t}{\eta}}

P(t) = e^{\frac{- 2\times 2.0\times 10^{- 3}}{1.52\times 10^{-35}}} = e^{-2.63\times 10^{32}}

P(t) = e^{-2.63\times 10^{32}}

Taking log on both the sides:

logP(t) = -2.63\times 10^{32} loge

P(t) = 10^{- 1.17\times 10^{32}}

6 0
3 years ago
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