Light of intensity I0 passes through 4 ideal polarizing sheets. Unpolarized light enters the 1st sheet that has a horizontal transmission axis. Light continues to the 2nd sheet that has its transmission axis at 23 degrees with respect to the 1st sheet, then to the 3rd sheet that has its transmission axis at 42 degrees with respect to the 1st sheet then to the 4th sheet that has its transmission axis at 14 degrees with respect to the 3d sheet. The intensity of the emerging light as percentage of I0 is close to:

Answers

Answer 1

Answer:

 I₄ / Io = 0.388 = 38.8%

Explanation:

For this exercise we will analyze the effect of having two depolarizers with an annulus between them, this is described by Malus's law

         I = Io cos² θ

where tea is the angle between the two polarizers.

For this case we must apply this law to each pair of polarizers.

The light that reaches the first polarizer is non-polarized light, therefore only the light that has polarization is transmitted in the direction of the depolarizer, which is horizontal.

          I₁ = I₀ / 2

this intnesidad reaches the second polarizer that has an angle of 23º with respect to the first polarizer

          I₂ = I₁ cos² 23

          I₂ = I₀ / 2 0.847

This light reaches the third polarizer, which has an angle of 42º with respect to the first polarized one.

Let's find the angle with respect to the 2nd polarizer

          θ = 42- 23

          θ = 19º

we apply the law of Malus

          I₃ = I₂2 cos 19

          I₃ = I₀ 0.847 / 2 0.9455

          I₃ = Io 0.4

This intensity of light reaches the fourth polarizer, with an angle of 14º with respect to the 3polarizer

          I₄ = Io 0.4 cos 14

          I₄ = Io 0.4 0.970

          I₄ = Io 0.388

to give this result as a percentage

          I₄ / Io = 0.388 = 38.8%


Related Questions

which of the following could be effectively used to reduce friction (a) water (b)petrol (c) kerosene (d) grease​

Answers

Answer:

[tex]\huge\boxed{\mathfrak{\underline{Answer...}}} \\ \\ \huge\mathfrak\red{grease} \\ \\ \huge\mathfrak\purple{hope \: it \: helps}[/tex]

what is the normal force on a 2000N refrigerator resting on the ground

Answers

The perpendicular force that a surface applies to an item is known as the normal force. FN = m ⋅ g

What is Normal force?

The perpendicular force that a surface applies to an item is known as the normal force. For instance, if you place a book on a table, a gravitational force will cause it to fall. The table pulls on the book in order to counterbalance this force and keep it from dropping. The normal force, often known as the opposing force, is denoted by the symbols FN F N or  NN. "N" stands for the normal force unit (Newton).

A common illustration of Newton's third rule of motion is the normal force.

When one object applies force to another, the other applies a force in the opposite direction and of equal magnitude to the first object (action equals reaction).

Therefore, the force applied by the item to the surface is equal to the normal force. The calculations change according to the surface's slope.

The formula is as follows for an object laying on a flat surface:

FN = m ⋅ g

where

The mass of an object is m.

Gravitational acceleration is denoted by g.

The normal force (FN F N ) for an item on a flat surface is equal to its gravitational force, according to Newton's third law ( WW).

The formula is as follows for an object laying on a flat surface:

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At what separation will two charges, each of magnitude 6.0 μC, exert a force of 0.70 N on

each other?

Answers

The two charges, each of magnitude 6.0 μC, exert a force of 0.70 N at separation  of 1.47 meters.

What is electric force?

Electric force is the attracting or repulsive interaction between any two charged things. Similar to any force, Newton's laws of motion describe how it affects the target body and how it does so. One of the many forces that affect objects is the electric force.

We know that electric force can be defined as:

Force: F = kQq/r²

0.70 = 9.0 × 10⁹ × (6.0×10⁻⁶)²/r²

r = 1.47 meter.

Hence, the separation between them is 1.47 meters.

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Imagine that you are observing the region of space where a cloud of gas and dust is beginning to collapse inward to form a star (the object that initially forms in this process is called a protostar). Will the atoms in the collapsing cloud move away from one another, move closer to one another, or stay at the same locations? 2) What physical interaction, or force, causes the atoms to behave as you described they would in Question 1?

Answers

Answer:

1. The atoms in the collapsing cloud would move closer to one another.

2. Gravity

Explanation:

1. This is because there is a force of attraction between the atoms of the collapsing cloud. Thus, the atoms in the collapsing cloud would move closer to one another.

2. This is because gravity is the force of attraction between masses. Now, the collapsing cloud of dust contains mass of atoms, and as the atoms come together due to gravity, they form a larger mass which develops into a star.

(a) The atoms in the collapsing cloud would move closer to one another.

(b)  Due to the force of interaction between masses that is gravity

What is force of attraction?

(a) The atoms will come closer

This is because there is a force of attraction between the atoms of the collapsing cloud. Thus, the atoms in the collapsing cloud would move closer to one another.

(b) Due to the gravitational attraction

This is because gravity is the force of attraction between masses. Now, the collapsing cloud of dust contains mass of atoms, and as the atoms come together due to gravity, they form a larger mass which develops into a star.

Hence

(a) The atoms in the collapsing cloud would move closer to one another.

(b)  Due to the force of interaction between masses that is gravity

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The "Little Boy" nuclear bomb that was dropped on
Hiroshima had a yield of 6.9 x 10^13 J. In order to
release this much energy, kg of the uranium-
235 used in the bomb was converted into energy.

Answers

Answer:My gues is about 1/3 kg

Explanation:

I don't know physics

The bean of light is incident from a light source is: i.covergent
ii.divergent
iii.parallel
iv.both i and ii​

Answers

Answer:

I think the answer is II.divergent

because parralel beams are convergent to the light source

and in the question it says incident from the light source

Answer:

The correct answer is (i) Convergent.

A horizontal line labeled B has an arrow labeled A strike it from right and above and then another arrow D emerges from the strike point at the same angle as A. A dotted line C evenly divides the angle between A and D.
In the mirror diagram shown, which is the normal?

A
B
C
D

Answers

Answer:

c is the actual answer.

Explanation:

Answer:

c is correct

Explanation:

Which of the following is NOT a scientific hypothesis?
A. Neon atoms emit red light.
B. There is an attractive force between the earth and moon.
C. Halle Berry is attractive.
D. Summer days are hottest
E. The sky is blue.

Answers

The following statement is not a scientific hypothesis:

C. Halle Berry is attractive.

A scientific hypothesis is a proposed explanation for an observation or pattern in nature that can be tested through further investigation and experimentation. It should be testable, falsifiable, and based on evidence.

Neon atoms emit red light. This is a scientific hypothesis that can be tested and confirmed by looking at the spectrum of light emitted by neon atoms.

B. There is an attractive force between the earth and moon. This is a scientific hypothesis that can be tested and confirmed by measuring the force of gravity between the earth and moon.

D. Summer days are the hottest of the year. This is a scientific hypothesis that can be tested and confirmed by collecting temperature data during the summer months.

E. The sky is blue. This is a scientific hypothesis that can be tested and confirmed by observing the sky under different atmospheric conditions.

The statement "Halle Berry is attractive" is a subjective opinion that cannot be tested or confirmed through scientific investigation, hence it is not a scientific hypothesis. Attractiveness, as a concept, can vary widely based on personal, cultural, and social factors.

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A solenoid is connected to a battery as shown in the figure, and a bar magnet is placed nearby. What is the direction of the magnetic force that this solenoid exerts on the bar magnet? (Hint: Think of the solenoid as a bar magnet, and identify what would be its north and south poles.)

Answers

According to the right-hand rule, the magnetic field of the solenoid will be directed towards the left direction.

What is Right-hand rule?

Right Hand Thumb Rule is the rule which is used If a current carrying conductor is imagined to be held in the right hand direction such that the thumb always points along the direction of the current flow, then the direction of the wrapped fingers in the hand will give the direction of the magnetic field lines of the bar magnet.

By the application of the right-hand rule, the magnetic field of the solenoid will be directed towards the left hand direction.

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HELPPP PLEASEEEEE, BRIANLEST WILL BE GIVEN ON CORRECT!​

Answers

Answer:

a. 25000J or 25KJ

b.2500w

Explanation:

a. since we say work is force applied in the same line or straight line

W=F(newtons )*D(metres)

b. power is work over time

Answer:

25000 J work with 2500 W power

Explanation:

work done = force . displacement = 500 N . 50 m= 25000 J

Power= work done / time taken = 25000 J / 10 s = 2500 W

Add the following and give the answer using significant figures 5000+621

Answers

Answer:

;) hope it's helpful

Explanation:

5000 + 621 = 5621

Facts about 5621

Sig Figs

4

5621

Decimals

0

Scientific Notation

5.621 × 103

E-Notation

5.621e+3

Words

five thousand six hundred twenty-one

When a low-pressure gas of hydrogen atoms is placed in a tube and a large voltage is applied to the end of the tube, the alors will emit electromagnetic radiation and visible light can be observed. If this light passes through a diffraction grating, the resulting spectrum appears as a pattern of four isolated, sharp parallel lines, called spectral lines. Each spectral line corresponds lo one specific wavelength that is present in the light emitted by the source. Such a discrete spectrum is referred to as a line spectrum
By the early 19th century, it was found that discrete spectra were produced by every chemical element in its gaseous slale. Even though these spectra were found to share the common feature of appearing as a set of isolated lines, it was observed that each element produces its own unique pattern of lines. This indicated that the light emitted by each element contains a specific set of wavelengths that is characteristic of that element.
part A What is the wavelength of the line corresponding to n =4 in the Balmer series? Express your answer in nanometers to three significant figures.
Part B What is the wavelength of the line corresponding to t=5 in the Balmer series? Express your answer in nanometers to three significant figures.

Answers

Answer:

A)  λ = 4.88 10² nm,  B)   λ = 4.08 10² nm

Explanation:

The spectrum of hydrogen is correctly explained by the Bohr model, where the energy of each level is

          Eₙ = -13.606 /n²       [eV]

the transition generally occurs from a given level to a lower state nf <no, so a transition is

          ΔE = E_f -Eₙ = -13,606 ( [tex]\frac{1}{n_f^2} - \frac{1}{n_o^2}[/tex] )

to find the wavelength let's use the planck relation

          ΔE = h f

the speed of light is

          c = λ f

we substitute

          ΔE = h c /λ

          λ = [tex]\frac{h \ c}{ \Delta \lambda}[/tex]

       

let's apply this equation to our case

the Balmer series has as final state the level n_f = 2

A) initial state n₀ = 4, final state n_f = 2

         ΔE = -13.606 ( [tex]\frac{1}{2^2} - \frac{1}{4^2}[/tex] )

         ΔE = 2.55 eV

let's reduce to SI units

         ΔE = 2.55 eV (1.6 10⁻¹⁹ J / 1 eV) = 4.08 10⁻¹⁹ J

     

we calculate

         λ = 6.63 10⁻³⁴ 3 10⁸ / 4.08 10⁻¹⁹

         λ = 4.875 10⁻⁻⁷ m

we reduce to nm

         λ = 4.875 10⁻⁷ m (10⁹ nm / 1m)

         λ = 487.5 nm

we reduce to three significant figures

         λ = 4.88 10² nm

B) initial state n₀ = 5

          ΔE = -13,606 ( [tex]\frac{1}{2^2} - \frac{1}{5^2}[/tex] )

          ΔE = 2,857 eV

we repeat the process of the previous point

         ΔE= 2,857 1.6 10⁺¹⁹ = 4.286 10⁻¹⁹J

we look for the wavelength

           λ = 6.63 10⁻³⁴ 3 10⁸ / 4.88 10⁻¹⁹

           λ = 4.0758 10⁻⁷ m

we reduce to nm

           λ = 4.0758 10² nm

ignificant numbers

           λ = 4.08 10² nm

Part E

Once you complete your outline, write a 500- to 750-word paper using word processing software. Add a works cited page at the end to give credit to your sources. Submit your completed paper along with this activity to your teacher for evaluation.

A Voyage to Proxima Centauri

Proxima Centauri is the second closest star to our solar system, which makes scientists curious to know more about it. What do we know about Proxima Centauri? How did we discover this information? Is it possible for humans to travel to this star? In this task, you will research and write a 500- to 750 word paper that answers these questions. Follow these steps to complete your research and writing. This guide about the research process can help.


Estimated time to complete: 3 hours


Part A

The goal of your paper is to describe the history of the discovery and research of Proxima Centauri. Another goal is to find the possibilities of traveling to this star. Some questions that your paper should answer are:


When was Proxima Centauri discovered?

How was Proxima Centauri discovered?

How have scientists researched Proxima Centauri?

What technologies have they used? What types of data do these technologies collect?


Have spacecraft ever reached this star?

What are the limitations of sending spacecraft to Proxima Centauri?

What accommodations would humans need to travel to Proxima Centauri?

Answers

Proxima Centauri otherwise known as Alpha Centauri C, is the closest star to our solar system, located about 4.24 light-years away. It was discovered in 1915 by Robert Innes, the director of the Union Observatory in South Africa. Innes identified Proxima Centauri as a possible member of the Alpha Centauri system after noticing that the star had a similar proper motion to Alpha Centauri A and B.

Proxima Centauri was first observed through telescopes, which allowed scientists to study the star's spectra and estimate its distance from Earth. In the following decades, scientists continued to study Proxima Centauri through telescopes, including the Hubble Space Telescope, which provided high-resolution images of the star.

In 2016, the European Southern Observatory announced the discovery of an exoplanet orbiting Proxima Centauri, called Proxima Centauri b. This discovery was made using the radial velocity method, which measures the star's small wobbles caused by the gravitational pull of an orbiting planet.

What are the major challenges facing the discovery of Proxima Centauri?

Despite the advances in our understanding of Proxima Centauri, no spacecraft have ever reached this star. The distance between Proxima Centauri and Earth is so vast that it would take tens of thousands of years to reach the star using current propulsion technology.

One potential solution to this problem is the use of interstellar travel, which would allow humans to travel to Proxima Centauri within a human lifetime. There are several proposed methods for interstellar travel, including using a starship propelled by fusion engines or a massive light sail pushed by a beam of lasers. However, these technologies are still in the theoretical stage and have not yet been developed.

There are also many other challenges that must be overcome in order for humans to travel to Proxima Centauri. For example, humans would need to find a way to protect themselves from the high levels of radiation that they would be exposed to during the journey. They would also need to find a way to provide enough food, water, and other resources to sustain themselves for the duration of the trip.

Overall, while it is theoretically possible for humans to travel to Proxima Centauri, it would be a massive undertaking that would require significant technological advancements and the development of new solutions to many challenges.

Sources of Works Cited include:

"Proxima Centauri." Wikipedia, Wikimedia Foundation, 5 Jan. 2021,

"Proxima Centauri b." Wikipedia, Wikimedia Foundation, 20 Nov. 2020,

"Interstellar Travel." Wikipedia, Wikimedia Foundation, 4 Jan. 2021,

Therefore, the correct answer is as given above

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The half-life of chromium-51 is 28 days. If a sample contained 510 grams, how much chromium would remain after 1 year?

Answers

Answer:

N = 365 / 28 = 13 half-lives in one year

(1/2)^13 = .000122     fraction remaining in one year

510 * .000122 = .062 grams

SOMONE PLEASE HELP ASAP!!!!

Answers

It’s the blue color

A physics instructor wants to project a spectrum of visible light colors from 400 nm to 700 nm as part of a classroom demonstration. She shines a beam of white light through a diffraction grating with 500 lines per mm, projecting a pattern on a screen 2.4 m behind the grating. How wide is the spectrum corresponding to m=1?

Answers

Answer:

औडंठः की दुनिया कार्टून दर्शन होली है और यह बियर का अच्छा विकल्प है। विकल्प है। यो कुरा मेरो आफ्ना लागि विचार र त्यहाँको बडेमाको बिल्डिङ भित्र म जिन्दगी एक यात्रा हो कहाँबाट शुरु भइ कहाँ पुगेर अन्त हुन्छ

PLZ HELP!!! BRAINLYEST WILL BE GIVEN TO BEST ANSWER!!!! IF YOU PUT A LINK I WILL REPORT YOU!!!!! DONT TRY ME!!!!!! YALL BEEN WASTING MY POINTS LATELY I HATE IT SO NO SNEAKY LINK OR NON ANSWER OR U WILL BE REPORTED.

Answers

Answer:

Explanation:

The corona is in the outer layer of the Sun's atmosphere—far from its surface. Yet the corona is hundreds of times hotter than the Sun's surface. ... In the corona, the heat bombs explode and release their energy as heat. But astronomers think that this is only one of many ways in which the corona is heated.

The chromosphere ("sphere of color") is the second of the three main layers in the Sun's atmosphere and is roughly 3,000 to 5,000 kilometers deep. Its rosy red color is only apparent during eclipses. The chromosphere sits just above the photosphere and below the solar transition region.

The photosphere is the visible "surface" of the Sun. The Sun is a giant ball of plasma (electrified gas), so it doesn't have a distinct, solid surface like Earth. ... The photosphere is much cooler than the Sun's core, which has a temperature well above 10 million degree

And thats all i know

Electromagnetic induction occurs when
a.
b.
a conductor is moved through a magnetic
a magnet is connected to an electrical circ
electrons are rubbed from one object onto
electrons flow from one chemical to anoth
C.
d.
your hd

Answers

Answer:

A

Explanation:

occurs when a magnetic feild and an elctric conductor move relative to one another

50 points if brainliest !!!!!!!!!!
Q11. A person has a mass of 70 kg. What is this person's weight? What is the force that this person exerts onto the Earth? What is the Earth's acceleration because of this person?

Answers

Answer:

[tex]weight = mg \\ = 70 \times 10 \\ = 700 \: newtons[/tex]

[tex]force = 700 \: newtons[/tex]

[tex]force = mass \times acceleration \\ 700 = 70 \times a \\ a = 10 \: {ms}^{ - 2} [/tex]

At the gym, a man pulls a bar on a machine that works the muscles of the upper back. It takes him 0.5 seconds to raise 30 kilograms of weights a vertical distance of 0.5 meters. Which of these exerts the same power output? (Estimate g as 10 m/s2.)
A) lifting 25 kilograms a distance of 2.4 meters in 2.0 seconds
B) lifting 45 kilograms a distance of 2.4 meters in 3.0 seconds
C) leg pressing 45 kilograms a distance of 0.5 meters in 0.5 seconds
D) bench pressing 30 kilograms a distance of 0.5 meter in 1.5 seconds

Answers

Answer: Lifting 25 kilograms a distance of 2.4 meters in 2.0 seconds

Explanation:

Got it wrong and that’s the answer

Lifting 25 kilograms a distance of 2.4 meters in 2.0 seconds  exerts the same power output. Hence, option (A) is correct.

What is power?

The quantity of energy moved or converted per unit of time is known as power in physics. The watt, or one joule per second, is the unit of power in the International System of Units.

Power is also referred to as activity in ancient writings. A scalar quantity is power.

Power required for lifting 30 kilograms of weights a vertical distance of 0.5 meters in 0.5 second = (30×9.8×0.5)/0.5 watt = 294 watt.

Power required for lifting 25 kilograms a distance of 2.4 meters in 2.0 seconds= (25×9.8×2.4)/2.0watt = 294 watt.

Power required for lifting  45 kilograms a distance of 2.4 meters in 3.0 seconds =  (45×9.8×2.4)/3.0watt = 352.8 watt.

Power required for lifting  45 kilograms a distance of 0.5 meters in 0.5seconds =  (45×9.8×0.5)/0.5watt = 441 watt.

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Robert Galstyan, from Armenia, pulled two coupled railway wagons a distance of 7 m using his teeth. The total mass of the wagons was about 2.20 X 10^5 kg. Of course, his job was made easier by the fact that the wheels were free to roll. Suppose the wheels are blocked and the coefficient of static friction between the rails and the sliding wheels is 0.220. What would be the magnitude of the minimum force needed to move the wagons from rest? Assume that the track is horizontal.

Answers

The magnitude of the minimum force needed to move the wagons from rest is 474320 N

How do I determine the force needed to move the wagons?

We have come to know that the force and coefficient of friction have a simple relationship as shown by the equation below:

Frictional force (N) = coefficient of friction (μ) × normal reaction (N)

F = μN

Applying the above formula, we can determine the force needed to move the wagons from rest. Details below:

Mass of wagons (m) = 2.20×10⁵ KgAcceleration due to gravity (g) = 9.8 m/s²Normal reaction (N) = mg = 2.20×10⁵ × 9.8 = 2156000 NCoefficient of static friction (μ) = 0.220Force needed (F) = ?

F = μN

F = 0.220 × 2156000

F = 474320 N

Thus, the force needed is 474320 N

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A solid circular rod that is 600 mm long and 20 mm in diameter is subjected to an axial force of P = 50 kN. The elongation of the rod is delta = 1.40 mm and its diameter becomes df = 19.9837 mm. Determine the modulus of elasticity (E) and the modulus of elasticity in shear (G). Assume the material does not yield.

Answers

Answer:

a) the modulus of elasticity (E) is 68.22 GPa

b) the modulus of elasticity in shear is 25.28 GPa

Explanation:

Given the data in the question;

First we determine the cross-sectional area A of the road;

A = π/4 × d²

given that; diameter d = 20 mm

we substitute

A = π/4 × ( 20 mm )²

A = π/4 × 400 mm

A = 314.159 mm²

Next, we find, the normal stress of the rod σ

σ = P / A

given that load p = 50 kN = 50,000 Newton

we substitute

σ = 50,000 N / 314.159 mm²

σ = 159.155 N/mm²

σ = 159.155 Mpa

Next, is the normal strain ε

ε = δ / L

given that; change in length δ = 1.40 mm and Length of rod L = 600 mm

we substitute

ε = 1.40 mm / 600 mm

ε = 0.002333

Now, we find the modulus of elasticity;

we know that;

σ = Eε

modulus of elasticity E = σ / ε

we substitute

E = 159.155 Mpa / 0.002333

E = 68219.031 MPa

E = ( 68219.031 / 1000 ) GPa

E = 68.22 GPa

Therefore, the modulus of elasticity (E) is 68.22 GPa

b)

we know that the Poisson's ratio v is;

v = -(ε[tex]_{lat[/tex] / ε )

v = - (((d[tex]_f[/tex] - d)/d) / ε )

where d[tex]_f[/tex]  is the final diameter of the rod ( 19.9837  mm ) and ε[tex]_{lat[/tex] is the lateral strain

so we substitute

v = - ((( 19.9837 - 20 ) / 20 ) / 0.002333  )

v = - (( -0.0163 / 20 ) / 0.002333  )

v = - ( - 0.000815 / 0.002333 )

v = - ( - 0.3493 )

v = 0.3493

So, our shear modulus will be;

G = E / 2( 1 + v )

we substitute

G = 68.22 GPa / 2( 1 + 0.3493  )

G = 68.22 GPa / 2( 1.3493  )

G = 68.22 GPa / 2.6986

G = 25.28 GPa

Therefore, the modulus of elasticity in shear is 25.28 GPa

Question 9 (Worth 4 points)
(02.06 HC)
a. Create and describe a scenario in which the forces acting on an object are unbalanced. Explain how you know the forces are unbalanced.
b. For your scenario, explain how Newton's third law of motion describes the forces involved.

Answers

a. When a football is kicked and it moves from one location to another, it indicates that unbalanced troops are responding to it and b. If an object A applies a force to another object B, then the other object B.

Describe how the effects of balanced and unbalanced forces on motion:

The equilibrium of the forces is disturbed when an object's motion changes. Forces with opposite directions and equal sizes are said to be balanced forces. Motion remains constant when the forces are equal. In one of your scenarios from the previous section, you used the same amount of force on an object but pushed or pulled it in the opposite direction.

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What is the efficiency of an engine that exhausts 440 J of heat to a cold reservoir and receives 570 J of heat from a hot reservoir? *

Answers

Answer:

Efficiency = 77%

Explanation:

Input energy = 570 J

Output energy = 440 J

To find the efficiency;

[tex] Efficiency = \frac {Out-put \; energy}{In-put \; energy} * 100 [/tex]

Substituting into the equation, we have;

[tex] Efficiency = \frac {440}{570} * 100 [/tex]

[tex] Efficiency = 0.7719 * 100 [/tex]

Efficiency = 77.19 ≈ 77%

Therefore, the efficiency of the engine is 77 percent.

______is the amount of matter in an object. It does not change regardless
of location
__________which one
Friction
Mass
Weight

Answers

Answer:

Mass

Explanation:

The weight of an object is dependent upon the mass and the value of acceleration due to gravity:

W = mg

Since the value of acceleration due to gravity (g) changes from planet to planet. Hence, the weight of an object also changes with location.

Friction depends upon the material of surfaces and the weight.

F = (coefficient of friction)W

Since weight changes with location. Therefore, the friction also changes with location.

The mass of an object is the quantity of matter contained in the body. It remains constant anywhere in the universe.

Hence, the correct option is:

Mass

A brick is thrown upward from the top of a building at an angle of 10° to the horizontal and with an initial speed of 15 m/s. If the brick is in flight for 2.9 s, how tall is the building? any help would be greatly appreciated

Answers

The total height of the building from which the brick is thrown is 33.60 meters.

What is Speed?

Speed is the rate of change of position of an object in any direction with respect to time. Speed is measured as the ratio of distance to the time in which the distance was covered. The SI unit of speed is m/s (meter per second).

The time of flight of the object from the building is:

t = 2μ sinθ/ g

t = 2 × 15 × sin10/ 9.8

t = 30 × 0.175/ 9.8

t = 0.535 seconds

Total time to cover the distance = 2.9 seconds

Total time = 2.9 - 0.535

Total time = 2.365

μg = μ sinθ

h = 15 × sin10  × 2.365 + 1/2 × 9.8 × (2.365)²

h = 2.625 × 2.365 + 1/2 × 9.8 × (2.365)²

h = 6.20 + 4.9 × 5.59

h =  6.20 + 27.40

h = 33.60 meters

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I cant solve this problem, and our teacher said that this would be in the test we'll have tomorrow, can someone help me?
A body of m = 6.8kg is launched with a speed of 7.5 m / s towards the top of an inclined plane of 15 ° with respect to the horizontal. in the absence of friction, what displacement does it make before reversing the direction of motion?

Answers

Answer:

d = 11.1 m

Explanation:

Since the inclined plane is frictionless, this is just a simple application of the conservation law of energy:

[tex] \frac{1}{2} m {v}^{2} = mgh[/tex]

Let d be the displacement along the inclined plane. Note that the height h in terms of d and the angle is as follows:

[tex] \sin(15) = \frac{h}{d} \\ or \: h = d \sin(15) [/tex]

Plugging this into the energy conservation equation and cancelling m, we get

[tex] {v}^{2} = 2gd \sin(15)[/tex]

Solving for d,

[tex]d = \frac{ {v}^{2} }{2g \sin(15) } = \frac{ {(7.5 \: \frac{m}{s}) }^{2} }{2(9.8 \: \frac{m}{ {s}^{2} })(0.259)} \\ = 11.1 \: m[/tex]

PLEASE HELP MEEEEEEE:((((​

Answers

Here's the solution,

A.)

we know,

[tex]power = \dfrac{work \: done}{time} [/tex]

so,

=》

[tex]600 = \dfrac{work \: done}{10} [/tex]

=》

[tex]work \: done = 600 \times 10[/tex]

=》

[tex]work \: done = 6000 \: \: joules[/tex]

B.)

[tex]work \: done = force \times displacement[/tex]

so,

=》

[tex]6000 = force \times10[/tex]

=》

[tex]force = \dfrac{6000}{10} [/tex]

=》

[tex]force = 600 \: \: Newtons[/tex]

The magnitude of the charge on an electron compared to the magnitude of the charge on a proton (3 points)
a) is greater
b) is smaller
c) is the same

Answers

The answer is b hope this helps

Option C is correct.

The magnitude of the charge on an electron compared to the magnitude of the charge on a proton is same.

We have a electron and a proton.

We have to estimate the fact whether the magnitude of the charge on an electron compared to the magnitude of the charge on a proton is -

a) is greater

or

b) is smaller

or

c) is the same

What is the Electric charge ?

The sufficiency or deficiency of electrons is called electric charge.

According to the question -

The charge on a electron is -   q(e) = - 1.6 x [tex]10^{-19}[/tex] Coulombs

The charge on a proton is -   q(p) = + 1.6 x  [tex]10^{-19}[/tex] Coulombs

But, according to the property of magnitude :

| + x| = + x

| - x| = - (- x) = + x

In the similar manner -

|q(e)| = |q(p)| = 1.6 x [tex]10^{-19}[/tex] coulombs

Hence, the magnitude of the charge on an electron compared to the magnitude of the charge on a proton is same.

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an object 50 cm high is placed 1 m in front of a converging lens whose focal length is 1.5 m. determine the image height (in cm).

Answers

Given :

An object 50 cm high is placed 1 m in front of a converging lens whose focal length is 1.5 m.

To Find :

the image height (in cm).

Solution :

By lens formula :

[tex]\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}[/tex]

Here, u = - 100 cm

f =  150 cm

[tex]\dfrac{1}{v} - \dfrac{1}{-100} = \dfrac{1}{150}\\\\v = - 300 \ cm[/tex]

Now, magnification is given by :

[tex]m = \dfrac{v}{u} = \dfrac{h_i}{h_o}\\\\h_i = \dfrac{300}{100}\times 1\\\\h_i = 3 \ m[/tex]

Therefore, the image height is 3 m or 300 cm.

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