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4 years ago

7

In a potentiometer circuit a 1m long potentiometer PQ of resistance 10ohms is connected in series with a cell of e.m.f 9v with i

nternal resistance 3 ohms calculate the p.d across PQ

1 answer:

Marta_Voda [28]4 years ago

6 0

Answer:6

Explanation:muliple

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what happens to the water after it rains? a. precipitation b. runoff c. condensation user: what do organisms from the ocean use

Vedmedyk [2.9K]

The rain gets evaporated in to water vapor and is returned to the clouds where they go through condensation and then they poud down as rain or A.K.A, Precipitation.

4 0

3 years ago

A spherical conductor is inside a uniform electric field with a magnitude of 1000 N/C to the right. What is the magnitude and di

Lyrx [107]

Answer:

1.-E=1000N/C to the LEFT

2.-The electric field inside a conductor in electrostatic state is always zero (conductor proprieties).

3.-The voltmeter read 0V as differential voltage between two points from the conductor

Explanation:

1.The electric field inside the conductor must be zero (conductor proprieties). Then the charges create a electric field equal an opposite to the external electric field. In other words E=1000N/C to the LEFT

2. The electric field inside a conductor in electrostatic state is always zero. As shown in the figure the electric field induced by the charges in the sphere surface cancelled the EXTERN electric field.

3.If the Electric field inside the conductor is zero, that means that the Voltage in the hole conductor is constant (conductor proprieties). In other words the the voltmeter read 0v as differential voltage between two points from the conductor.

3 0

3 years ago

A 0.500-nm x-ray photon is deected through 134 in a Compton scattering event. At what angle (with respect to the incident beam)

natita [175]

Answer:

The angle of recoil electron with respect to incident beam of photon is 22.90°.

Explanation:

Compton Scattering is the process of scattering of X-rays by a charge particle like electron.

The angle of the recoiling electron with respect to the incident beam is determine by the relation :

$\cot\phi = (1+\frac{hf}{m_{e}c^{2} })\tan\frac{\theta }{2}$ ....(1)

Here ∅ is angle of recoil electron, θ is the scattered angle, h is Planck's constant, $m_{e}$ is mass of electron, c is speed of light and f is the frequency of the x-ray photon.

We know that, f = c/λ ......(2)

Here λ is wavelength of x-ray photon.

Rearrange equation (1) with the help of equation (1) in terms of λ .

$\cot\phi = (1+\frac{h}{m_{e}c\lambda })\tan\frac{\theta }{2}$

Substitute 6.6 x 10⁻³⁴ m² kg s⁻¹ for h, 9.1 x 10⁻³¹ kg for $m_{e}$ , 3 x 10⁸ m/s for c, 0.500 x 10⁻⁹ m for λ and 134° for θ in the above equation.

$\cot\phi = (1+\frac{6.6\times10^{-34} }{9.1\times10^{-31}\times3\times10^{8}\times0.5\times10^{-9} })\tan\frac{134 }{2}$

$\cot\phi=2.37$

$\phi$ = 22.90°

8 0

4 years ago

In one of the original Doppler experiments, a tuba was played at a frequency of 64.0 Hz on a moving flat train car, and a second

wolverine [178]

Answer:

$f_{beat} = 1.64\ Hz$

Explanation:

given,

frequency of tuba.f = 64 Hz

Speed of train approaching, v = 8.50 m/s

beat frequency = ?

using Doppler's effect formula

$f' = f(\dfrac{v}{v-v_s})$

v_s is the velocity of the source

v is the speed of sound, v = 340 m/s

now,

$f' = 64\times (\dfrac{340}{340 - 8.50})$

f' = 65.64 Hz

now, beat frequency is equal to

$f_{beat} = f' - f$

$f_{beat} = 65.64 - 64$

$f_{beat} = 1.64\ Hz$

hence, beat frequency is equal to 1.64 Hz

3 0

3 years ago

Please help with this :(

Angelina_Jolie [31]

Answer:

I'm sorry I don't have a answer but I like your pfp

8 0

3 years ago