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3 years ago

How did Newton use creativity and logic in his approach to investigating light?

2 answers:

soldier1979 [14.2K]3 years ago

8 0

Isaac Newton was creative in his use of prisms to show how white light is actually made up of multiple colors. He used logic in the way he presented his arguments rhetorically in order to convince readers of the correctness of his conclusions.

Newton was not the first to experiment with passing light through prisms to determine how light works. French philosopher Rene Descartes had done prism experiments of his own. But Descartes had thought that passing through a prism actually modified the light in order to produce the color spectrum. Newton correctly understood that when light refracted through the prism, it revealed the range of colors that were naturally in the light. He then used a second prism, blocking all but one color, to show that a single color passing through a prism was not modified in color. He also showed--by positioning the second prism differently--how the multiple colors of light could be recombined into white light again.

Newton's 1672 paper on light refracting through prisms established his reputation as a scientist. He continued to study light throughout his scientific career, publishing a larger work in 1704 on <em>Opticks</em> (as they spelled optics then).

ryzh [129]3 years ago

3 0

He used the quantum mechanic theory to approach light by describing relativity and perspective.

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If a company like Waste Management collects all of the paper, bottles, plastic, ete in a single bin, how

Tju [1.3M]

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yes they can be recycled

Explanation:

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4 years ago

For a plane mirror,how is the object s related to the distance s?

dlinn [17]

Light is a very complex phenomenon, but in many situations its behavior can be understood with a simple model based on rays and wave fronts. A ray is a thin beam of light that travels in a straight line. A wave front is the line (not necessarily straight) or surface connecting all the light that left a source at the same time. For a source like the Sun, rays radiate out in all directions; the wave fronts are spheres centered on the Sun. If the source is a long way away, the wave fronts can be treated as parallel lines.

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4 years ago

Calculate the energy and power required for a student to bike their way to college from home. The student bikes from home for 0.

gregori [183]

Answer:

Check the explanation

Explanation:

To solve the problem, we need to analyze all forces acting on a bicycle individually. In question, student bikes on flat terrain so gravity force doesn't affect the road load. This is the case of uniform acceleration and deceleration so need to calculate average velocity to find Air resistance.

Kindly check the attached images below to see the step by step explanation to the question above.

8 0

4 years ago

A small rubber wheel is used to drive a large pottery wheel. The two wheels are mounted so that their circular edges touch. The

Goshia [24]

Answer:

$0.629\ \text{rad/s}^2$ counterclockwise

$9.98\ \text{s}$

Explanation:

$r_1$ = Small drive wheel radius = 2.2 cm

$\alpha_1$ = Angular acceleration of the small drive wheel = $8\ \text{rad/s}^2$

$r_2$ = Radius of pottery wheel = 28 cm

$\alpha_2$ = Angular acceleration of pottery wheel

As the linear acceleration of the system is conserved we have

$r_1\alpha_1=r_2\alpha_2\\\Rightarrow \alpha_2=\dfrac{r_1\alpha_1}{r_2}\\\Rightarrow \alpha_2=\dfrac{2.2\times 8}{28}\\\Rightarrow \alpha_2=0.629\ \text{rad/s}^2$

The angular acceleration of the pottery wheel is $0.629\ \text{rad/s}^2$ .

The rubber drive wheel is rotating in clockwise direction so the pottery wheel will rotate counterclockwise.

$\omega_i$ = Initial angular velocity = 0

$\omega_f$ = Final angular velocity = $60\ \text{rpm}\times \dfrac{2\pi}{60}=6.28\ \text{rad/s}$

t = Time taken

From the kinematic equations of linear motion we have

$\omega_f=\omega_i+\alpha_2t\\\Rightarrow t=\dfrac{\omega_f-\omega_i}{\alpha_2}\\\Rightarrow t=\dfrac{6.28-0}{0.629}\\\Rightarrow t=9.98\ \text{s}$

The time it takes the pottery wheel to reach the required speed is $9.98\ \text{s}$

4 0

3 years ago

Six artificial satellites complete one circular orbit around a space station in the same amount of time. Each satellite has mass

oee [108]

Answer:

The ranking of the net force acting on different satellite from largest to smallest is ${F_E} > {F_F} > {F_A} = {F_B} = {F_D} > {F_C}$

Explanation:

In order to get a good understanding of this solution we need to understand that the main concepts used to solve this problem are centripetal force and velocity of satellite.

Initially, use the expression of the velocity of satellite and find out its dependence on the radius of orbit. Use the dependency in the centripetal force expression.

Finally, we find out the velocity of the six satellites and use that expression to find out the force experienced by the satellite. Find out the force in terms of mass (m) and radius of orbit (L) and at last compare the values of force experienced by six satellites.

Fundamentals

The centripetal force is necessary for the satellite to remain in an orbit. The centripetal force is the force that is directed towards the center of the curvature of the curved path. When a body moves in a circular path then the centripetal force acts on the body.

The expression of the centripetal force experienced by the satellite is given as follows:

${F_{\rm{c}}} = \frac{{m{v^2}}}{L}$

Here, m is the mass of satellite, v is the velocity, and L is the radius of orbit.

The velocity of the satellite with which the satellite is orbiting in circular path is given as follows:

$v = \frac{{2\pi L}}{T}$

Here, T is the time taken by the satellite.

The velocity of the satellite with which the satellite is orbiting in circular path is given as follows;

$v = \frac{{2\pi L}}{T}$

Since, all the satellites complete the circular orbit in the same amount of time. The factor of $\frac{{2\pi }}{T}$ is not affected the velocity value for the six satellites. Therefore, we can write the expression of v given as follows:

Substitute $v = \frac{{2\pi L}}{T}$ in the force expression ${F_{\rm{c}}} = \frac{{m{v^2}}}{L}$ as follows:

$\begin{array}{c}\\{F_c} = \frac{{m{{\left( {\frac{{2\pi L}}{T}} \right)}^2}}}{L}\\\\ = \frac{{4{\pi ^2}}}{{{T^2}}}mL\\\end{array}$

Since, all the satellites complete the circular orbit in the same amount of time. The factor of $\frac{{4{\pi ^2}}}{{{T^2}}}$ not affect the force value for six satellites.Therefore, we can write the expression of ${F_c}$ given as follows:

${F_c} = kmL$

Here, k refers to constant value and equal to $\frac{{4{\pi ^2}}}{{{T^2}}}$

${F_A} = k{m_A}{L_A}$

Substitute 200 kg for ${m_A}$ and 5000 m for LA in the expression ${F_A} = k{m_A}{L_A}$

$\begin{array}{c}\\{F_A} = k\left( {200{\rm{ kg}}} \right)\left( {5000{\rm{ m}}} \right)\\\\ = {10^6}k{\rm{ N}}\\\end{array}$

The force acting on satellite B from their rocket is given as follows: ${F_B} = k{m_B}{L_B}$

Substitute 400 kg for ${m_B}$ and 2500 m for in the expression ${F_B} = k{m_B}{L_B}$

$\begin{array}{c}\\{F_B} = k\left( {400{\rm{ kg}}} \right)\left( {2500{\rm{ m}}} \right)\\\\ = {10^6}k{\rm{ N}}\\\end{array}$

The force acting on satellite C from their rocket is given as follows: ${F_C} = k{m_C}{L_C}$

Substitute 100 kg for ${m_C}$ and 2500 m for in the above expression ${F_C} = k{m_C}{L_C}$

$\begin{array}{c}\\{F_C} = k\left( {100{\rm{ kg}}} \right)\left( {2500{\rm{ m}}} \right)\\\\ = 0.25 \times {10^6}k{\rm{ N}}\\\end{array}$

The force acting on satellite D from their rocket is given as follows: ${F_D} = k{m_D}{L_D}$

Substitute 100 kg for ${m_D}$ and 10000 m for ${L_D}$ in the expression ${F_D} = k{m_D}{L_D}$

$\begin{array}{c}\\{F_D} = k\left( {100{\rm{ kg}}} \right)\left( {10000{\rm{ m}}} \right)\\\\ = {10^6}k{\rm{ N}}\\\end{array}$

The force acting on satellite E from their rocket is given as follows: ${F_E} = k{m_E}{L_E}$

Substitute 800 kg for ${m_E}$ and 5000 m for ${L_E}$ in the expression ${F_E} = k{m_E}{L_E}$

$\begin{array}{c}\\{F_E} = k\left( {800{\rm{ kg}}} \right)\left( {5000{\rm{ m}}} \right)\\\\ = 4.0 \times {10^6}k{\rm{ N}}\\\end{array}$

The force acting on satellite F from their rocket is given as follows: ${F_F} = k{m_F}{L_F}$

Substitute 300 kg for ${m_F}$ and 7500 m for ${L_F}$ in the expression ${F_F} = k{m_F}{L_F}$

$\begin{array}{c}\\{F_F} = k\left( {300{\rm{ kg}}} \right)\left( {7500{\rm{ m}}} \right)\\\\ = 2.25 \times {10^6}k{\rm{ N}}\\\end{array}$

The value of forces obtained for the six-different satellite are as follows.

$\begin{array}{l}\\{F_A} = {10^6}k{\rm{ N}}\\\\{F_B} = {10^6}k{\rm{ N}}\\\\{F_C} = 0.25 \times {10^6}k{\rm{ N}}\\\\{F_D} = {10^6}k{\rm{ N}}\\\\{F_E} = 4.0 \times {10^6}k{\rm{ N}}\\\\{F_F} = 2.25 \times {10^6}k{\rm{ N}}\\\end{array}$

The ranking of the net force acting on different satellite from largest to smallest is ${F_E} > {F_F} > {F_A} = {F_B} = {F_D} > {F_C}$

7 0

4 years ago