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3 years ago

What is the voltage measured with a voltmeter across a wire in a circuit?

Physics

2 answers:

Law Incorporation [45]3 years ago

6 0

<h3><u>Answer;</u></h3>

<em>The voltage is zero</em>

<h3><u>Explanation;</u></h3>

<em><u>Voltage is the potential difference between two points in a circuit or the electromotive force in a given circuit and is expressed in volts (V). </u></em>

Voltage is measured by a device called voltmeter that is always connected in parallel with the current source.

<em><u>When voltage is measured between two points on a wire in a circuit with zero or no resistance in between the two points then the voltage will be zero</u></em><em>. </em>From the ohm's Law, voltage is given the product of resistance and current and thus if the resistance is zero then the voltage will be zero.

Volgvan3 years ago

6 0

C) The voltage is zero.

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A proton orbits a long charged wire, making 1.80 ×106 revolutions per second. The radius of the orbit is 1.20 cm What is the wir

Fantom [35]

Answer:

linear charge density = -9.495 × $10^{-34}$ C/m

Explanation:

given data

revolutions per second = 1.80 × $10^{6}$

radius = 1.20 cm

solution

we know that when proton to revolve around charge wire then centripetal force is require to be in orbit of radius around provide by electric force

- q × E = m × w² × r ..................1

- 9 × $10^{9}$ × $\frac{2*linear\ charge\ density}r}$ q = m × w² × r ............2

and w = $\frac{2*\pi}{T}$

w = $\frac{d\theta }{dt}$

w = 1.80 × $10^{6}$ × $\frac{2*\pi}{1}$

w = 11304000 rad/s

so here from equation 2

- 9 × $10^{9}$ × $\frac{2*linear\ charge\ density}{0.012}$ 1.80 × $10^{6}$ = 1.672 × $10^{-27}$ × 11304000² × 0.0120

linear charge density = -9.495 × $10^{-34}$ C/m

8 0

3 years ago

James and John dive from an overhang into the lake below. James simply drops straight down from the edge. John takes a running s

liraira [26]

Answer:

Both of them reach the lake at the same time.

Explanation:

We have equation of motion s = ut + 0.5at²

Vertical motion of James : -

Initial velocity, u = 0 m/s

Acceleration, a = g

Displacement, s = h

Substituting,

s = ut + 0.5 at²

h = 0 x t + 0.5 x g x t²

$t_{James}=\sqrt{\frac{2h}{g}}$

Vertical motion of John : -

Initial velocity, u = 0 m/s

Acceleration, a = g

Displacement, s = h

Substituting,

s = ut + 0.5 at²

h = 0 x t + 0.5 x g x t²

$t_{John}=\sqrt{\frac{2h}{g}}$

So both times are same.

Both of them reach the lake at the same time.

3 0

3 years ago

Temperature and pressure of a region upstream of a shockwave are 295 K and 1.01* 109 N/m². Just downstream the shockwave, the te

seraphim [82]

Answer:

change in internal energy 3.62*10^5 J kg^{-1}

change in enthalapy 5.07*10^5 J kg^{-1}

change in entropy 382.79 J kg^{-1} K^{-1}

Explanation:

adiabatic constant $\gamma =1.4$

specific heat is given as $=\frac{\gamma R}{\gamma -1}$

gas constant =287 J⋅kg−1⋅K−1

$Cp = \frac{1.4*287}{1.4-1} = 1004.5 Jkg^{-1} k^{-1}$

specific heat at constant volume

$Cv = \frac{R}{\gamma -1} = \frac{287}{1.4-1} = 717.5 Jkg^{-1} k^{-1}$

change in internal energy $= Cv(T_2 -T_1)$

$\Delta U = 717.5 (800-295) = 3.62*10^5 J kg^{-1}$

change in enthalapy $\Delta H = Cp(T_2 -T_1)$

$\Delta H = 1004.5*(800-295) = 5.07*10^5 J kg^{-1}$

change in entropy

$\Delta S =Cp ln(\frac{T_2}{T_1}) -R*ln(\frac{P_2}{P_1})$

$\Delta S =1004.5 ln(\frac{800}{295}) -287*ln(\frac{8.74*10^5}{1.01*10^5})$

$\Delta S = 382.79 J kg^{-1} K^{-1}$

7 0

3 years ago

Which form of

Soloha48 [4]

I think it’s C b/c it works for me

3 0

3 years ago

A 1.5-kg ball is throw at 10 m/s. What is the balls momentum?

alexdok [17]

Answer : The momentum of ball is, 15 kg.m/s

Explanation :

Momentum : It is defined as the motion of a moving body. Or it is defined as the product of mass of velocity of an object.

Formula of momentum is:

where,

p = momentum = ?

m = mass = 1.5 kg

v = velocity = 10 m/s

Now put all the given values in the above formula, we get:

Therefore, the momentum of ball is 15 kg.m/s

3 0

3 years ago