Because of the anode heel effect, where is the intensity of the x-ray beam the greatest?

Physics

1 answer:

Aneli [31]3 years ago

4 0

Answer:

Along the anode-cathode axis

Explanation:

The Anode Heel Effect, in an X-ray tube states that the direction of emission of X-rays by the anode along the cathode-anode axis causes a variation in the intensity of X-rays.

The geometry of the anode causes the X-ray emitted perpendicular to the cathode-anode axis less intense than the X-rays emitted towards the cathode along the axis.

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80 grams of iron at 100°C is dropped into 200 of water at 20°C contained in an iron vessel of mass 50 gram find the resulting te

vodka [1.7K]

Answer:

the resulting temperature is 23.37 ⁰C

Explanation:

Given;

mass of the iron, m₁ = 80 g = 0.08 kg

mass of the water, m₂ = 200 g = 0.2 kg

mass of the iron vessel, m₃ = 50 g = 0.05 kg

initial temperature of the iron, t₁ = 100 ⁰C

initial temperature of the water, t₂ = 20 ⁰C

specific heat capacity of iron, c₁ = 462 J/kg⁰C

specific heat capacity of water, c₂ = 4,200 J/kg⁰C

let the temperature of the resulting mixture = T

Apply the principle of conservation of energy;

heat lost by the hot iron = heat gained by the water

$m_1c_1 \Delta t_1 = m_2c_2\Delta t_2\\\\m_1c_1 (100 - T) = m_2c_2 (T- 20)\\\\0.08 \times 462 (100-T) = 0.2 \times 4,200 (T-20)\\\\36.96 (100-T) = 840 (T-20) \\\\100 - T = 22.72 (T-20)\\\\100-T = 22.72 T - 454.4 \\\\554.4 = 23.72T\\\\T = \frac{554.4}{23.72} \\\\T = 23.37 \ ^0C$

Therefore, the resulting temperature is 23.37 ⁰C

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3 years ago

A force of 10N propels a 350-g toy rocket upwards. The rocket initially starts from rest and the rocket motor burns for 0.1 s. W

dimaraw [331]

Impulse=force x time
=10N x 0.1s
=1.0Ns

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3 years ago

Read 2 more answers

A 15x10^-6c charge is placed at the origin and a 9x10^-6C charge is placed on the x-axis at x=1.00m. where, on the x-axis is the

Harlamova29_29 [7]

The Electric field is zero at a distance 2.492 cm from the origin.

Let A be point where the charge $15\times10^-6$ C is placed which is the origin.

Let B be the point where the charge $9\times 10^-6$ C is placed. Given that B is at a distance 1 cm from the origin.

Both the charges are positive. But charge at origin is greater than that of B. So we can conclude that the point on the x-axis where the electric field = 0 is after B on x - axis.

i.e., at distance 'x' from B.

Using Coulomb's law, $\frac{kQ_A^2}{d_A^2} = \frac{kQ_B^2}{d_B^2}$ ,