how did einstein's explanation of the photoelectric effect eventually affect bohr's model of the atom? (2 points)

The electric force between two or more charged objects depends on the charge and the distance between the charged objects

Coulomb's law states that the force of attraction or repulsion between two charged bodies is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. Coulomb's law shows the relationship between the force , charge and the distance between the bodies

therefore F= kq1q2/r²

where q1 is the charge of body 1 and q2 is the charge of body 2. r is the distance between them and k is called electrostatics constant which have a value of 9,×10⁹ Nm²/C²

Therefore what can affect the force between two charges are the product of the charge of the two bodies and the distance between the two charge.

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The problem states that the rate at which the intensity of a vertical beam of light decreases in a transparent medium is proportional to the thickness of the medium. It is given that in clear seawater, the intensity 1 meter below the surface is 25% of the initial intensity of the incident beam. The task is to find the intensity of the beam 6 meters below the surface.

Let's denote the initial intensity of the incident beam as I₀. According to the given information, the intensity 1 meter below the surface is 25% of I₀, which means it is 0.25I₀.

Since the rate of intensity decrease is proportional to the thickness of the medium, we can set up a proportion using the ratio of intensities:

I₁ / I₀ = \(e^{-kt}\)

where I₁ is the intensity at a depth of t meters, and k is the proportionality constant.

To find the intensity 6 meters below the surface, we substitute t = 6 into the equation:

I = I₀ *  \(e^{-6k}\)

We don't have the specific value of k or I₀, but we can still determine the ratio of intensities:

I / I₀ =  \(e^{-6k}\)

Using the given information, we can write the equation:

0.25I₀ / I₀ =  \(e^{-6k}\)

Simplifying the equation, we find:

0.25 = \(e^{-6k}\)

To solve for the intensity I, we substitute this value back into the equation:

I = I₀ * 0.25.

Therefore, the intensity of the beam 6 meters below the surface is 0.25 times the initial intensity I₀.

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Answer:

1) a substance made by mixing other substances together.

2) diverse in character or content.

3) of the same kind; alike.

4) a means of solving a problem or dealing with a difficult situation.

5) a homogeneous noncrystalline substance consisting of large molecules or ultramicroscopic particles of one substance dispersed through a second substance. Colloids include gels, sols, and emulsions; the particles do not settle, and cannot be separated out by ordinary filtering or centrifuging like those in a suspension.

6) the temporary prevention of something from continuing or being in force or effect.

7) having assets in excess of liabilities; able to pay one's debts.

8) the minor component in a solution, dissolved in the solvent.

9) the state or process that occurs when no more of something can be absorbed, combined with, or added.

REFER TO IMAGE ...

Answer:

Compounds are two or more elements chemically combined in definite proportions by weight.

Mixtures are two or more substances not chemically combined.

Explanation:

Hope this helped Mark BRAINLIEST!!!

Answer:

A mixture can be separated by physical means

Explanation:

The smallest possible shaft AE is having the diameter equal to 22nm or 22 nanometer, which can be calculated by using the formula for the torque transmitted by the shaft.

The smallest permissible diameter of shaft AE can be determined using the formula for torque transmitted by a solid shaft:

T = (π/16) × τ × d³

where T is the torque transmitted by the shaft, τ is the maximum shear stress, and d is the diameter of the shaft.

Given that:

T = 45 kW / (2π × 500 rpm) = 0.143 Nm,

τ = 50 MPa, and

that 30 kW is taken off at gear G and 15 kW is taken off at gear H, we can calculate the torque transmitted by gears G and H as follows:

\(T_G\) = 30 kW / (2π × 500 rpm)

= 0.095 Nm

\(T_H\) = 15 kW / (2π × 500 rpm) = 0.048 Nm

The torque transmitted by shaft AE can be calculated as follows:

\(T_{AE}\) = T + \(T_{G}\) + \(T_{H}\) = 0.143 + 0.095 + 0.048 = 0.286 Nm

Substituting these values into the formula for torque transmitted by a solid shaft and solving for d, we get:

d = [(16 × T_AE) / (π × τ)]¹/³

= [(16 × 0.286) / (π × 50)]¹⁾³ ≈ 22 mm

Therefore, the smallest permissible diameter of shaft AE is approximately 22 mm.

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Student rolls two balls off of a table and ball A has twice the mass of ball B. Both balls leave the table with the same starting velocity, then they will land the same distance away from the table.

Projectile motion is a form of motion experienced by an object that is projected in gravitational field and moves along a curved path under the action of gravity only.

Projectile motion can be said as the motion of an object thrown or projected into the air and subject to only the acceleration of gravity. The object is known as a projectile and its path is called trajectory.

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(D) The temperature and medium through which the sound is traveling would most impact the speed of the sound waves coming from the ringing bell ?

The motion of the inferior angle of the scapula in the superior and lateral direction is called upward rotation.

The scapula or the shoulder blade has about six different types of motion it undergoes.

The six ways of movement of the scapula are:

When the inferior angle of the scapula moves in the superior and lateral direction, the motion is called upward rotation.

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Answer:

Electrical Resistance is a measure of the opposition to current flow in an electrical circuit

Types: variable resistance and set resistance

Explanation:

Answer:

4.395kg

Explanation:

1kg=1000g

Therefore, 4395g/1000g=4.395kg

If we know the frequency or wavelength of a wave, we can use this equation to calculate its speed or vice versa.

The correct relationship between the frequency of the incident radiation (ν), its wavelength (λ), and speed (c) is:

A. ν = λ/c

This equation is known as the wave equation and describes the relationship between the frequency, wavelength, and speed of a wave. It states that the frequency of a wave is inversely proportional to its wavelength and directly proportional to its speed. The speed of light (c) is a constant in a vacuum and its value is approximately 3.0 x 10^8 m/s.

Therefore, if we know the frequency or wavelength of a wave, we can use this equation to calculate its speed or vice versa.

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Answer:

A. The range of A and B are equal.

Explanation:

Let u be the initial speed of both the projectile.

The gravitational force acting in the projectile is in the downward direction, sp the speed of the projectile in the horizontal direction remains constant and equals to the initial horizontal speed.

For projectile, a projectile having initial velocity u at an angle \theta with the horizontal direction,

The speed in the horizontal direction \(= u\cos\theta\)

and the speed in the vertical direction is \(= u\sin\theta\) upward.

For A:

The speed in the horizontal direction \(= u\cos75^{\circ}\)

and the speed in the vertical direction is \(= u\sin75^{\circ}\) upward.

For B:

The speed in the horizontal direction \(= u\cos15^{\circ}\)

and the speed in the vertical direction is \(= u\sin15^{\circ}\) upward.

Let \(t_A\) and \(t_B\) are the time of flight for projectile A and B respectively.

As the range is the horizontal distance traveled by the projectile, so

The range for the projectile A \(= u\cos75^{\circ}\times t_A\cdots(i)\)

The range for the projectile B = \(u\cos15^{\circ}\times t_B\cdots(ii)\)

At the highest point, the vertical velocity is 0.

Bu using the equation of motion \(v^2=u^2 +2a s.\)

Here, the final velocity v=0, the initial velocity \(u = u \sin \theta\) , h= vertical distance up to the highest point, and \(a= -g\) (as per sign convention).

So, \(s= \frac{u^2\sin^2 \theta}{2g}\)

For projectile A: The maximum height attained.

\(s_A= \frac{u^2\sin^2 75^{\circ}}{2g}\)

For projectile B: The maximum height attained.

\(s_B= \frac{u^2\sin^2 15^{\circ}}{2g}\)

As \(\sin^2 75^{\circ} > \sin^2 15^{\circ}\), the height of A is attained by A is more than the heigHt attained by B.

Now, the times required to reach the highest point from the ground and again form the highest point to the ground are the same.

So, the total time of flight = 2 x (Time to reach the highest point)

In a similar way, by using the equation of motion v=u+at,

The time to reach the highest point \(=\frac {u\sin\theta}{g}\)

where g is the acceleration due to gravity.

So, the total time of flight

\(= 2 \times \frac {u\sin\theta}{g}\)

The total time of flight for A

\(=2 \times \frac {u\sin75^{\circ}}{g}\)

The total time of flight for A

\(=2 \times \frac {u\sin15^{\circ}}{g}\)

Now, from equations (i) and (ii),

The range for the projectile A =

\(u\cos75^{\circ}\times \frac {2u\sin75^{\circ}}{g}=\frac {u^2 \sin 150^{\circ}}{g}= \frac {u^2 \sin 30^{\circ}}{g}\)

The range for the projectile B =

\(u\cos15^{\circ}\times \frac {2u\sin15^{\circ}}{g}=\frac {u^2 \sin 30^{\circ}}{g}.\)

Both the projectile have the same range.

Hence, option (A) is correct.

Explanation:

0.75

According to the formula, P=IV

150 = I×200

I'm pretty sure this poor guy could never have been accepted for astronaut training, much less a mission to the Moon.  

He weighs something like 335 pounds when he's ON THE MOON !

If he has enough fuel, and survives a trip back to Earth, he'll weigh around 9,080 Newtons here.  That's like 2,040 pounds ... durn close to a ton ! !

Answer:

Gravitational field strength:

Moon is 1.6N/kg

Earth is 9.8N/kg or 10N/kg

Moon:

Weight = mass x gravitational field

1500N = m x 1.6N/kg

÷1.6 both sides

937.5 = m

Mass of astronaut is 937.5kg

(Your mass doesn't change, no matter where you go, in space)

Earth:

Weight = mass x gravitational field

= 937.5kg x 9.8 N/kg

= 9187.5N

Or

Weight = mass x gravitational field

= 937.5kg x 10 N/kg

= 9375N

Thus, the astronaut's weight will be 9187.5N or 9375N on Earth.

Hope this helps!

Given:

The resistance is,

The capacitance is,

The applied voltage is,

The frequency of the AC is,

To find:

The capacitive reactance

Explanation:

The capacitive reactance is,

Hence, the capacitive reactance is 0.106 ohm.

To solve this problem, we need to first find the angular acceleration and then the final angular velocity of the flywheel. After that, we can determine the time it takes to complete the remaining 5 revolutions at constant angular velocity.

1. Determine the angular acceleration:
Since the flywheel makes one complete revolution during the 2 seconds of angular acceleration, it rotates through an angle of 2π radians (1 revolution = 2π radians). Using the equation θ = ω₀t + (1/2)αt², where θ is the angle in radians, ω₀ is the initial angular velocity (0 since it starts from rest), α is the angular acceleration, and t is the time (2 seconds), we can solve for α: 2π = 0(2) + (1/2)α(2)²
α = 2π/2² = π rad/s²
2. Determine the final angular velocity:
Using the equation ω = ω₀ + αt, we can find the final angular velocity ω:
ω = 0 + π(2) = 2π rad/s
3. Calculate the time to complete the remaining 5 revolutions:
Now that the flywheel has a constant angular velocity of 2π rad/s, we can calculate the time it takes to complete the remaining 5 revolutions. To do this, we need to find the angle θ for 5 revolutions (5 * 2π = 10π radians) and use the equation θ = ωt: 10π = (2π)t
t = 5 seconds
4. Determine the total time for 6 revolutions:
Finally, we add the initial 2 seconds of acceleration to the 5 seconds it takes to complete the remaining revolutions:
Total time = 2 + 5 = 7 seconds
So, it takes the flywheel 7 seconds to make a total of 6 full revolutions.

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A longitudinal wave makes up the second wave. The two waves will have traveled the same distance after 10 seconds since they are both longitudinal waves. Option a is Correct.

In the graphic below, an earthquake wave is depicted: An earthquake's second wave simultaneously emerges from the epicenter. They are in particular to blame for the terrible noises that follow powerful earthquakes. These waves move along the path of wave travel at a speed of around 5 miles per second, causing the crustal rock to shake.

These waves in a solid medium might either be longitudinal waves or transverse waves. P waves (for "principal" waves) denote longitudinal or compressional seismic waves through bulk material, whereas S waves denote transverse seismic waves ("secondary" ). Option a is Correct.

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Correct Question:

A wave produced by an earthquake appears in the diagram below:

A second wave leaves the center of an earthquake at the same time. The second wave is a longitudinal wave. After 10 seconds, will the two waves travel the same distance?

answer choices

a. Yes, because both waves are longitudinal waves.

b. Yes, because all waves travel at the same speed in similar mediums.

c. No, because only one type of wave can travel through a medium at a time.

d. No, because transverse waves travel at different speeds than longitudinal waves.

The speed of the moving train is approximately 33.5 m/s.

The beat frequency is given by the difference in frequency between the two horns, which is equal to the Doppler shift in frequency due to the motion of the moving train. Using the formula for the Doppler effect, we can solve for the speed of the train:

\(f_b = f_s\dfrac{(v + v_o)}{(v + v_s)}\)

where \(f_b\) is the beat frequency, \(f_s\) is the horn frequency, v is the speed of sound, \(v_o\) is the observer's speed, and \(v_s\) is the speed of the source.

We know that \(f_s\) = 410 Hz and \(f_b\) = 35 Hz. The speed of sound in air at standard temperature and pressure is approximately 343 m/s. Since the observer is stationary, \(v_o\) = 0.

Solving for \(v_s\), we get:

\(v_s = \dfrac{(f_s + f_b)}{f_s - 1} \times v\)

\(v_s\) = ((410 Hz + 35 Hz) / 410 Hz - 1) * 343 m/s

\(v_s\) = 33.5 m/s

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The maximum height attained by the arrow is 1109 ft.

As arrow rises to its maximum height, its velocity decreases from 256 ft/s to 0 ft/s at the rate of 32 ft/s each second. Use the following equation to determine the distance it rises.

vf^2 = vi^2 + 2 * a * d, vf = 0, a = -32

0 = 256^2 + 2 * -32 * d

64 * d = 65,536

d = 65,536 ÷ 64 = 1024 ft

Maximum height = 85 + 1024 = 1109 ft.

Distance is defined as an object's total movement without regard for direction. Distance can be defined as how much ground an object has covered regardless of its starting or ending point.

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Answer:

Noble gases

also known as group 0

Explanation:

First you have to warm up the ice to melting point ....melt it ....then heat the water to boiling point.....boil it to steam ...then heat the steam to 118 C

35 g = .035 kg

.035 kg (  2090  J  /( kg C) * (12C)  +  3.33 x 10^5 J/kg  +  4186 J/(kg C) * 100 C   +  2.26 x 10^6 J/kg   +  2010 J/(kg C) * 18 C ) =

  107550.1 J     round as appropriate

Explanation:

Density = mass ÷ volume

             = 12 g ÷ 6 cm³

             =  2 g/cm³

Answer:

B, A, C

Explanation:

The momentum is equal to the mass times the velocity, so for each car the momentum is equal to

Car A

Momentum = Mass x Velocity

Momentum = 1,500 kg x 10 m/s

Momentum = 15,000 kg m/s

Car B

Momentum = Mass x Velocity

Momentum = 1,500 kg x 25 m/s

Momentum = 37,500 kg m/s

Car C

Momentum = Mass x Velocity

Momentum = 1,000 kg x 10 m/s

Momentum = 10,000 kg m/s

Then, the decreasing momentum is B, A, C because

37,500 is greater than 15,000 and 15,000 is greater than 10,000

So, the answer:

B, A, C

The core of a highly evolved high mass star, depending on the type of stellar remnant, can be as small as the size of the Earth (for a white dwarf) or as compact as about 10 kilometers (for a neutron star).

The core of a highly evolved high mass star is typically a compact object known as a stellar remnant. There are two main types of stellar remnants that can form depending on the mass of the star: white dwarfs and neutron stars.

A white dwarf is the remnant of a star with a mass up to about 8 times that of the Sun. Its core is about the size of the Earth, which is much smaller compared to the original size of the star.

On the other hand, a neutron star is formed when a star with a mass greater than about 8 times that of the Sun undergoes a supernova explosion. The core of a neutron star is incredibly dense and compact, with a radius typically on the order of 10 kilometers (6.2 miles). Neutron stars are composed primarily of tightly packed neutrons and are extremely massive.

In summary, the core of a highly evolved high mass star, depending on the type of stellar remnant, can be as small as the size of the Earth (for a white dwarf) or as compact as about 10 kilometers (for a neutron star).

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Answer:

1. 7.105 kg    2. 69.629 N    3. 7.105 kg    4. 11.368 N

Explanation:

Firstly the main point is that " mass " is independentnof the the place wheather it is no earth or moon or some else place.

So mass is

   W = mg

    27 = m 3.8

     m =7.105 kg

1. Mass on Earth :

     m = 7.105 kg

2. Weight on Earth:

    W=mg

    W=7.105 × 9.8

   W = 69.629 N

3.Mass on Moon:

   m = 7.105 kg

4. Weight on Moon:

    W =mg

    W =7.015 ×1.6

    W= 11.368 N

Answer:

4 cm/s

Explanation:

Speed = distance over time

The conclusion that you might arrive at is that the speed of the car affects the gas mileage.

We know that an experiment is the only way that we can be able to establish cause and effect relationship. We know that the speed would affect the consumption of the gas. By varying the speed of the car, we can be able to obtain the effect that it has on the mileage.

Thus, the conclusion that you might arrive at is that the speed of the car affects the gas mileage.

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Answer:

Mass, m = 1561.23kg

Explanation:

Given the following data;

Weight = 15300N

We know that acceleration due to gravity is equal to 9.8m/s²

To find mass;

Weight can be defined as the force acting on a body or an object as a result of gravity.

Mathematically, the weight of an object is given by the formula;

\( Weight, W = mg\)

Where;

Substituting into the equation, we have;

\( 15300 = m*9.8\)

\( m = \frac {15300}{9.8}\)

Mass, m = 1561.23 Kg

Therefore, the mass of the car is 1561.23 kilograms.