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4 years ago

15

A power transmission line is hung from metal towers with glass insulators having a resistance of 1.00×109 Ω . What current flows

through the insulator if the voltage is 200 kV? (Some high-voltage lines are DC.)

1 answer:

Hunter-Best [27]4 years ago

5 0

Answer:

The current flows through the insulator is 2 mA.

Explanation:

Given that,

Resistance $R=1.00\times10^{9}\ \Omega$

Voltage = 200 kV

We need to calculate the current

Using ohm's law

$V=IR$

$I=\dfrac{V}{R}$

Where, I = current

V = voltage

R = resistance

Put the value into the formula

$I=\dfrac{200\times10^{3}}{1.00\times10^{9}}$

$I=0.0002\ A$

$I=2\ mA$

Hence, The current flows through the insulator is 2 mA.

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Suppose a rocket ship in deep space moves with constant acceleration equal to 9.80 m/s2, which gives the illusion of normal grav

DochEvi [55]

Answer:

a) 3673469.39 seconds

b) 6.61×10¹⁴ m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity = 0.12×3×10⁸ m/s

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

Equation of motion

$v=u+at\\\Rightarrow 0.12\times 3\times 10^8=0+9.8t\\\Rightarrow t=\frac{0.12\times 3\times 10^8}{9.8}=3673469.39\ s$

Time taken to reach 12% of light speed is 3673469.39 seconds

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{(0.12\times 3\times 10^8)^2-0^2}{2\times 9.8}\\\Rightarrow s=6.61 \times 10^{14}\ m$

The distance it would have to travel is 6.61×10¹⁴ m

7 0

3 years ago

A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum

s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle= $q=-2.74\times 10^{-6} C$

Kinetic energy of particle= $K_E=6.65\times 10^{-10} J$

Initial time= $t_1=6.36 s$

Final potential difference= $V_2=0.351 V$

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

$qV=K.E$

Using the formula

$2.74\times 10^{-6}V_1=6.65\times 10^{-10} J$

$V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V$

Initial voltage= $V_1=2.43\times 10^{-4} V$

$\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2$

Using the formula

$\frac{V_1}{V_2}=(\frac{6.36}{t})^2$

$\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}$

$t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}$

$t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}$

$t=241.7 s$

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

6 0

3 years ago

A 15 kg bucket of water is suspended by a very light ropewrapped around a solid uniform cylinder 0.300 m in diamter withmass 12.

matrenka [14]

Answer:

Part a)

$T = 42 N$

Part b)

$v_f = 11.8 m/s$

Part c)

$t = 1.7 s$

Part d)

$F = 159.7 N$

Explanation:

Part a)

While bucket is falling downwards we have force equation of the bucket given as

$mg - T = ma$

for uniform cylinder we will have

$TR = I\alpha$

so we have

$T = \frac{1}{2}MR^2(\frac{a}{R^2})$

$T = \frac{1}{2}Ma$

now we have

$mg = (\frac{M}{2} + m)a$

$a = \frac{mg}{(\frac{M}{2} + m)}$

$a = \frac{15 \times 9.81}{(6 + 15)}$

$a = 7 m/s^2$

now we have

$T = \frac{12 \times 7}{2}$

$T = 42 N$

Part b)

speed of the bucket can be found using kinematics

so we have

$v_f^2 - v_i^2 = 2 a d$

$v_f^2 - 0 = 2(7)(10)$

$v_f = 11.8 m/s$

Part c)

now in order to find the time of fall we can use another equation

$v_f - v_i = at$

$11.8 - 0 = 7 t$

$t = 1.7 s$

Part d)

as we know that cylinder is at rest and not moving downwards

so here we can use force balance

$F = T + Mg$

$F = 42 + (12 \times 9.81)$

$F = 159.7 N$

5 0

3 years ago

Which type of wave interaction is shown in the diagram?

masya89 [10]

Answer: Constructive interference

Explanation: Just took the test

5 0

3 years ago

A photon with an energy E = 2.12 GeV creates a proton-antiproton pair in which the proton has a kinetic energy of 96.0 MeV. What

Oksanka [162]

Answer:

The kinetic energy of the anti proton is 147.4 MeV.

Explanation:

Given that,

Energy = 2.12 GeV

Kinetic energy = 96.0 MeV

We need to calculate the kinetic energy of the anti proton

Using formula of energy

$E_{photon}=m_{p}c^2+m_{np}c^2+K.E_{p}+K.E_{np}$

We know that,

$m_{p}c^2=m_{np}c^2$

So, $E_{photon}=2mc^2+K.E_{p}+K.E_{np}$

$K.E_{np}=E_{photon}-(2mc^2+K.E_{p})$

Put the value into the formula

$K.E_{np}=2.12\times10^{9}-2\times938.3\times10^{6}-96\times10^{6}$

$K.E_{np}=147.4\ MeV$

Hence, The kinetic energy of the anti proton is 147.4 MeV.

6 0

3 years ago