In the Compton effect, an X -ray photon scatters from a free electron.. find the change in the photon's wavelength if it scatters at an angle of θ = 30.0 ∘ relative to the incident direction.

The calculated data aligns with the general rule that a normal soil near field capacity contains approximately 50% water and 50% air in the total pore space of the soil.

For the given soil sample, the following properties can be calculated: 1. Bulk density = 0.75 g/cm3, 2. Particle density = 0.45 g/cm3, 3. Percentage of total porosity = 40%, 4. Percentage of water at field capacity (mass basis) = 25%, 5.

Percentage of water at field capacity (volume basis) = 20%, 6. Total volume of pores = 220 cm3, 7. Total volume of water at field capacity = 100 cm3, 8. Percentage of pore space filled with water at field capacity = 45%, 9.

Total volume of air spaces at field capacity = 120 cm3, 10. Percentage of pore space filled with air at field capacity = 55%. The calculated data agrees with the general rule that a soil near field capacity contains approximately 50% water and 50% air in the total pore space of the soil.

Bulk density is calculated by dividing the mass of dry soil by its volume, which gives a value of 0.75 g/cm3.

Particle density is calculated by dividing the mass of solid particles by their volume, resulting in a value of 0.45 g/cm3.

Percentage of total porosity is obtained by subtracting the particle density from the bulk density, dividing the result by the bulk density, and multiplying by 100, resulting in 40%.

Percentage of water in the soil at field capacity (mass basis) is calculated by dividing the mass of water by the mass of dry soil, which gives 25%.

Percentage of water in the soil at field capacity (volume basis) is obtained by dividing the volume of water by the total volume of soil, resulting in 20%.

Total volume of pores is calculated by subtracting the volume of solid particles from the total volume of soil, resulting in 220 cm3.

Total volume of water in the pores at field capacity is given as 100 cm3.

Percentage of the total pore space filled with water at field capacity is calculated by dividing the volume of water by the total volume of pores and multiplying by 100, resulting in 45%.

Total volume of air spaces at field capacity is obtained by subtracting the volume of water from the total volume of pores, resulting in 120 cm3.

Percentage of the total pore space filled with air at field capacity is calculated by dividing the volume of air by the total volume of pores and multiplying by 100, resulting in 55%.

The calculated data aligns with the general rule that a normal soil near field capacity contains approximately 50% water and 50% air in the total pore space of the soil, as the percentages obtained for water and air are close to this expected distribution.

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The time taken by the ball to reach the surface of the moon that drops from a height of 1.5 m from the moon is 1.36s.

Given that the acceleration of gravity on the surface of the moon is, \(a=g=1.62 ms^{-2}\). The distance of the ball from the surface is the height of the ball from the surface. So, \(h=d=1.5 m\).

Since the ball is dropped under gravity or free-fall motion from rest, then the initial velocity is zero (i.e.) \(v=0ms^{-1}\). Let t be the time taken by the ball to reach the surface of the moon.

By the equation of motion, the equation is given by \(h=v_{0} t+\frac{1}{2}gt^{2}\), where g is the gravity, t is the time taken by an object, h is the distance of the object, and \(v_{0}\) is the initial velocity of an object from the rest.

Substitute the values of h, g, and \(v_{0}\) in the above equation, and we get

\(1.5=(0)t+\frac{1}{2}(1.62)t^{2}\)

\(1.5=0.81t^{2}\)

\(t^{2} =\frac{1.5}{0.81}\)

\(t^2=1.852\)

\(t=\pm1.36s\)

The time taken cannot be negative.

Therefore, The time taken by the ball to reach the surface of the moon is 1.36s.

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Answer:

insulator

Explanation:

it keeps heat in not moves it

Answer:

B Newton's second law

Explanation:

An empty grocery cart is easier to move because there is less mass in an empty cart compared to a full cart. Newton's second law state that an object's acceleration depends on the mass of the object and the force applied.

Answer:

lower loss

Explanation: i had the same question

Answer: nun

Explanation:

Answer:

oxygen and carbon dioxide

Explanation:

are soluble in water animal and plants can utilize these dissolved gases for resporation and photosynthesis and hence can survived in water

Answer:

8.37757

Explanation:

may i have brainliest

The persοn is mοving at a speed οf 55 miles per hοur in the Sοuth directiοn when the persοn is 50 miles West and 70 miles Sοuth οf the car, and the distance between them is increasing at a rate οf 55 miles per hοur.

Tο determine the speed at which the persοn is mοving, we can use the cοncept οf relative velοcity.

Let's cοnsider the hοrizοntal and vertical cοmpοnents separately:

Hοrizοntal Cοmpοnent:

The persοn is walking due East, which is perpendicular tο the Nοrth directiοn οf the car. Therefοre, the hοrizοntal cοmpοnent οf the persοn's velοcity dοes nοt affect the speed at which the persοn is mοving away frοm the car.

Vertical Cοmpοnent:

The persοn is 70 miles Sοuth οf the car, and the distance between them is increasing at a rate οf 55 miles per hοur. This indicates that the persοn's vertical pοsitiοn is changing with time. Since the persοn is mοving in the Sοuth directiοn and the distance is increasing, the persοn's speed can be determined by the rate οf change οf the vertical distance.

Given that the distance is increasing at a rate οf 55 miles per hοur, the persοn's speed in the Sοuth directiοn is 55 miles per hοur.

Therefοre, the persοn is mοving at a speed οf 55 miles per hοur in the Sοuth directiοn when the persοn is 50 miles West and 70 miles Sοuth οf the car, and the distance between them is increasing at a rate οf 55 miles per hοur.

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40x magnifies the viewing field to reveal 583.33 distinct blood cells. Due to their rounded shape, blood cells may be recognized. They don't have cell walls as plant cells do.

At a total magnification of 40x (a scanning microscope), there may be counted 583.33 unique blood cells. The diameter of a 40x viewing field of view is 4200 m, whereas the normal size of a red blood cell is 6.2-8.2 m (7.2 m). So, 583.33 is the response. Blood cells appears as separate cells without nuclei when seen using a light microscope at a magnification of 40X.

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Nitrogen oxides play a critical role in photochemical smog. They give the smog its yellowish-brown hue. Indoor residential appliances like gas stoves and gas or wood heaters can be significant emitters of nitrogen oxides in poorly ventilated environments.

Therefore , on conclusion i.e. two gases with molecules consisting of nitrogen and oxygen atoms are nitric oxide (NO) and nitrogen dioxide (NO₂). These nitrogen oxides play a part in the development of smog and acid rain, adding to the issue of air pollution.

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The work done by the unstretched spring is 5 J.

Hooke's law states that the extension of a given material is directly proportional to the force applied as log as the elastic limit is not exceeded. Thus we have to know that; F = Ke

F = force applied

K = force constant

e = extension

Using the graph;

K = F/e

F = 200N

e = 5 cm or 5 * 10^-2 m

K = 200N/ 5 * 10^-2 m

K = 4000 N/m

Now;

Work = 1/2Ke^2

Work = 0.5 *  4000 N/m * (5 * 10^-2 m)^2

Work = 5 J

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Announcement 1: Fluid mechanics is a branch of chemistry that involves having a look at fluids:  c) False, True

Fluid mechanics is the branch of physics concerned with the mechanics of fluids and the forces on them. It has programs in an extensive variety of disciplines, such as mechanical, aerospace, civil, chemical, and biomedical engineering, geophysics, oceanography, meteorology, astrophysics, and biology. Fluid mechanics is the look at of the forces of the fluid and the way fluids flow. Fluid mechanics may be divided into parts: fluid statics and fluid dynamics. Fluid statics is the observation of fluids at rest, and fluid dynamics is the study of fluids in motion.

Fluid mechanics is the look at fluid behavior (beverages, gases, blood, and plasmas) at relaxation and in movement. Fluid mechanics has a huge variety of packages in mechanical and chemical engineering, organic structures, and in astrophysics.

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The formula that relatively accurately predicts the orbital distances of the first eight major bodies in the solar system and then somewhat predicts the next several is known as the Titius-Bode law.

Titius-Bode law is a generalization, or rule of thumb, that appears to describe the orbital distances of the planets of the Solar System. The sequence is 0, 3, 6, 12, 24, 48, 96, and 192. The seventh number, 384, is used as a substitute for Jupiter's real mean distance from the Sun, which is 483.6 million miles.The distance from the Sun to the planets in our Solar System follows a regular pattern known as the Titius-Bode law, which predicts the distance of the planet to the Sun based on its position in the sequence.

The Titius-Bode law is named after two eighteenth-century German astronomers, Johann Daniel Titius and Johann Elert Bode. It was initially proposed in 1766 by Titius as an empirical hypothesis, with little explanation of why it should work, and was later refined by Bode in 1778 into a formula, which became known as Bode's law or the Titius-Bode law.

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The speed of the block after the collision is approximately 0.0857 m/s.

To solve this problem, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision, assuming no external forces are acting on the system.

Let's denote the initial velocity of the block as v_block and the final velocity of the block after the collision as v_block'.

The initial momentum of the bullet is given by:

P_bullet_initial = mass_bullet * velocity_bullet_initial

                  = 0.002 kg * 500 m/s

                  = 1 kg m/s

The initial momentum of the block is zero since it is initially at rest:

P_block_initial = 0 kg m/s

After the collision, the bullet emerges from the block with a speed of 425 m/s. The mass of the bullet remains the same, 0.002 kg.

Using the principle of conservation of momentum, we can write the equation:

P_bullet_initial + P_block_initial = P_bullet_final + P_block_final

1 kg m/s + 0 kg m/s = 0.002 kg * 425 m/s + 1.75 kg * v_block'

Simplifying the equation, we have:

1 = 0.002 * 425 + 1.75 * v_block'

1 = 0.85 + 1.75 * v_block'

Rearranging the equation, we get:

1.75 * v_block' = 1 - 0.85

1.75 * v_block' = 0.15

v_block' = 0.15 / 1.75

v_block' ≈ 0.0857 m/s

Therefore, the speed of the block after the collision is approximately 0.0857 m/s.

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Answer:

false

Explanation:

cause bacteria prefers moist and damp environment

Answer:

False

Explanation: bacteria thrive on moister and grow due to it. if the enviroment they were in were to be dry, the bacteria itself would die. Hope this helps! :)

2/3, or roughly 0.67, of the generator's unloaded voltage is visible across the weight.

In the given scenario, the load resistance is 100 Ω, and the generator is unloaded.

We can use the voltage divider formula to determine the fraction of the unloaded generator voltage that appears across the load:

V_load = (R_load / (R_load + R_gen)) * V_gen

where:

V_load is the voltage across the load

R_load is the load resistance (100 Ω in this case)

R_gen is the internal resistance of the function generator

V_gen is the output voltage of the generator

Since the generator is unloaded, the voltage across the generator terminals is equal to the output voltage (V_gen). Therefore, we can rewrite the above formula as:

V_gen_load = (R_load / (R_load + R_gen)) * V_gen

where V_gen_load is the voltage across the generator terminals when the load is connected.

Assuming the internal resistance of the function generator is 50 Ω, we can substitute the values into the formula:

V_gen_load = (100 Ω / (100 Ω + 50 Ω)) * V_gen

V_gen_load = 2/3 * V_gen

Therefore, the fraction of the unloaded generator voltage that appears across the load is 2/3 or approximately 0.67.

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Answer:

weight =mass ×gravity

Explanation:

gravity =9.8

9.8×5

=49.05N

hope it helps

Answer:

50N (using 10N/kg)

Explanation:

mass*gravitational field strength

5*10

50N

The potential difference across all three resistors in a series circuit is the same, as the voltage from the battery is divided across the resistors in proportion to their resistance values. Therefore, option B (potential difference ΔV) is the same for all three resistors.

The resistance of each resistor is different, so option C is not the same for all three resistors.

The current through each resistor is the same, as there is only one path for the current to flow in a series circuit. Therefore, option A (current I) is the same for all three resistors.

So the correct answer is D, "A and B".

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The car must have a static friction coefficient of 0.568 in order to safely round the bend.

"The Occupational Safety and Health Administration advises that walking surfaces have a static coefficient of friction of 0.5," reads the appendix statement (at A4. 5). The allocation of a 0.5 COF to OSHA is most likely explained by the regulatory proposal from OSHA and the mention in ADAAG taken together.

To get the static friction coefficient required for a vehicle to travel at 102 km/h around a flat curve with a radius of 105 m,

F = mv²/r

F = μs * N

μs × N = mv²/r

N = mg

μs × mg = mv²/r

μs = v²/(gr)

μs = (102 km/h)² / (9.8 m/s² × 105 m)

μs = 0.568

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Answer:

electric potential has a positive charge but electric potential energy needs to move the charge against the electric field

Hope this helpful if not tell me

Answer:

2.3 x 10-13 F

Explanation:

From the question,

Applying

C = e₁e₂A/d..................... Equation 1

Where C = Capacitance of the capacitor, d = distance of seperation, A = cross sectional area of the capacitor, e₁ = permitivity of free space, e₂ = relative permitivity of vacuum.

But,

Area of a circle (A) is given as

A = πr²................... Equation 2

Where r = radius of the capacitor.

Substitute equation 2 into equation 1

C = e₁e₂πr²/d............... Equation 3

Given: r = 2.5 mm = 0.0025 m, d = 0.75 mm = 0.00075 m

Constant; e₁ = 8.85×10⁻¹² F/m, e₂ = 1 F/m, π = 3.14.

Substitute these values into equation 3

C =  [(8.85×10⁻¹²)(1)( 0.0025²)(3.14)]/(0.00075)

C = 0.23×10⁻¹²

C = 2.3×10⁻¹³ F

Answer:

A

Explanation:

edge

The total momentum of the system before the collision:

m1 × v1 = 130g × 14 m/s = 1820 g m/s

And after the collision, the total momentum of the system should be conserved:

m1 × v1' + m2 × v2' = 1820 g m/s

where v1' and v2' are the final velocities of ball 1 and ball 2, respectively.

Using the equation for partially elastic collision:

v1' = (2m2 × v2' + (m1 - f × m2) × v1) / (m1 + m2)

Substitute v1, m1, and m2 and solve for v2':

v2' = (1820 g m/s - (130 g - 0.8 × 350 g) × 14 m/s) / (130 g + 350 g)

v2' = 2.4 m/s

So, the final velocity of ball 2 is 2.4 m/s.

Finally, the final velocity of ball 1 using the above equation:

v1' = (2 × 350 g × 2.4 m/s + (130 g - 0.8 × 350 g) × 14 m/s) / (130 g + 350 g)

v1' = 9.0 m/s

So, the final velocity of ball 1 is 9.0 m/s.

The final velocities of each ball after the collision are 9.0 m/s and 2.4 m/s.

A collision in physics is any situation in which two or more bodies quickly exert forces on one another. When two or more items touch or collide with one another, it is called a collision. It can refer to a collision of ideas or interests, a traffic accident, a computer networking problem, or a physical impact in physics. There are three different kinds of collisions:

the collision that is perfectly elastic, inflexible collision, collision with perfect inelasticity. The average amount of time between collisions is known as the mean free time for a molecule in a fluid. The average speed and mean free time are multiplied to produce the molecule's mean free path.

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(a) The acceleration of the system is 28.55 m/s².

(b) The tension in the rope is 285.5 N.

The acceleration of the box is determined by applying Newton's second law of motion as follows;

F(net) = ma

where

weight of 30 kg mass - force of friction due to 10 kg  = tension in the rope

Fg - μmgcosθ  = ma

(30 kg x 9.8 m/s²) - (0.1 x 10 kg x 9.8 x cos30) = 10a

294 - 8.49 = 10a

285.51 = 10a

a = 285.51/10

a = 28.55 m/s²

The tension in the rope is calculated as follows;

T = ma

T = 28.55 m/s² x 10 kg

T = 285.5 N

Thus, the acceleration of the system depends on the masses and force of friction on the blocks

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The second ball will land 2 m from the table's base, the same distance as the first ball.

Since the projection angles are not specified, we may infer that they are the same for both balls.

We also know that the second ball has a mass that is twice as much as the first ball, but this fact has no bearing on the range because it cancels out in the calculation.

Therefore, we can write:

\(R = v^2 sin(2\theta) / g\)

For both balls, \(v^2/g\) is constant since they are projected with the same speed and are subjected to the same acceleration due to gravity. Therefore, we have:

\(R_1 = v^2 sin(2\theta) / g\\\\R_2 = v^2 sin(2\theta) / g\)

Dividing R₂ by R₁, we get:

\(R_2 / R_1 = (v^2 sin(2\theta) / g) / (v^2 sin(2\theta) / g) = 1\)

This means that the ratio of the distances the balls travel is 1.

Therefore, the second ball will land at the same distance from the base of the table as the first ball, which is 2 m.

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Answer:

Given

mass (m) =5kg

velocity (v) =6m/s

momentum (p) =?

Form

p=mv

5kgx6m/s

p=30kg.m/s

momentum =30kg.m/s