An object is projected horizontally with the velocity 4m/s from a height of 20 m. the horizontal displacement is (angle x) :

The wavelength of the light is 550 nm

For a double slit interference pattern with slit spacing, d we have

dsinθ = mλ where d = slit spacing = 0.22 mm = 0.22 × 10⁻³ m, m = number of maximum fringe = 2(from the picture) and λ = wavelength of light.

Thus sinθ = mλ/d

Also, tanθ = L/D where L = distance between central maximum and fringe = 2.0 cm/2 = 1.0 cm = 1 × 10⁻² m and D = distance between slit and screen = 2.0 m

Since θ is small, sinθ ≅ tanθ

So, mλ/d = L/D

Making λ subject of the formula, we have

λ = dL/mD

Substituting the values of the variables into the equation, we have

λ = dL/mD

λ = 0.22 × 10⁻³ m × 1 × 10⁻² m/(2 × 2.0 m)

λ = 0.22 × 10⁻⁵ m²/4.0 m

λ = 0.055 × 10⁻⁵ m

λ = 0.55 × 10⁻⁶ m

λ = 550 × 10⁻⁹ m

λ = 550 nm

So, the wavelength of the light is 550 nm

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Answer:

Semantic memory is a category of long-term memory that involves the recollection of ideas, concepts and facts commonly regarded as general knowledge. Examples of semantic memory include factual information such as grammar and algebra.

Answer:

9/8

Explanation:

Answer:

metamarket (plural metamarkets) A group of businesses that offer products that are related from a consumer's perspective but which have no institutional connections. quotations . A market that trades in the medium of exchange of a lower-level market, such as money, derivatives, or credit. quotations.

Explanation:

The best examples of Meta Marketing can be selling family planning ideas or the idea of prohibition. ... Let's say a car selling in a Meta market would be a website, that sells cars but you will also find car parts there, add-ons for cars, colours for cars, mechanic's reviews, etc.

We have that for the Question, it can be said that

From the question we are told

From,

[tex]tan\theta = \frac{h}{2}[/tex]

differentiate with respect to h

[tex]sec^2\theta * \frac{do}{dz} = \frac{1}{2} * \frac{dh}{dz}\\\\\frac{dh}{dz} = 2 sec^\theta * \frac{d\theta}{dz}\\\\\theta = \frac{\pi}{6} and \frac{d\theta}{dz} = 0.1rad/min\\\\\frac{dh}{dz} = 2sec^2 (\frac{\pi}{6}) * (0.1)\\\\= 0.266miles/min[/tex]

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[tex]\\ \sf\longmapsto Acceleration\longrightarrow Change\:in\: Velocity\:over\:time[/tex]

[tex]\\ \sf\longmapsto Speed\longrightarrow Change\:in\: Distance\:over\:Time[/tex]

[tex]\\ \sf\longmapsto Velocity\longrightarrow Change\:in\: Displacement\: over\:time[/tex]

[tex]\\ \sf\longmapsto ∆t\longrightarrow Time\: interval [/tex]

[tex]\\ \sf\longmapsto Magnitude\:only\longrightarrow Scaler[/tex]

[tex]\\ \sf\longmapsto ∆x=Change\:in\: position [/tex]

Answer:

Explanation:

Let's just have our reference frame travel along with the original un broken mass. This way the original velocity is not relevant.

Each half will have a mass of 770/2 = 385 kg

Each half will have the same magnitude of velocity (conservation of momentum) which will be 2.6 x 10³/2 = 1.30 x 10³ m/s

Now add back the reference frame velocity to get velocity relative to earth.

Section one will have velocity 6.90 x 10³ + 1.30 x 10³ = 8.2 x 10³ m/s

Section two will have velocity 6.90 x 10³ - 1.30 x 10³ = 5.6 x 10³ m/s

In the moving reference frame, each half will have kinetic energy which could only come from the explosion

KE = ½(385)(1.3 x 10³)² = 325,325,000 J

2(325,325,000) = 650,650,000 J released in the explosion.

Rounding to the three significant figures of the problem numerals

E = 6.50 x 10⁸ J or 650 MJ released

I think the answer is d.right?

Answer: Varies

Explanation: It became more relied upon, technology is responsible for this.

Answer:

it would remain the same speed

Explanation: the rock isnt going down a hill or anything so therefore if gravity isnt pulling it down a slope then it would remain the same pace

Answer:

ω = 3.1 rad/s

θ = 36° from vertical

Explanation:

I will ASSUME that the bob and string is acting as a pendulum.

Please understand that the string will break when the bob is at the lowest point of the swing where the vectors of gravity and centripetal acceleration align. It will NOT break at the angle of maximum inclination measured from vertical. This angle is only a component of the maximum potential energy that gets converted to maximum kinetic energy at the lowest point of the swing.

At the bottom of the swing, the string must support the weight of the bob plus supply the required centripetal acceleration.

F = mg + mω²R

F/m = g + ω²R

F/m - g = ω²R

ω = √((F/m - g)/R)

ω = √((3/0.220 - 9.8)/0.40)

ω = 3.09691...

ω = 3.1 rad/s

Potential energy will convert to kinetic energy

       mgh = ½mv²

             h = v²/2g

R - Rcosθ = v²/2g

R(1 - cosθ) = v²/2g

   1 - cosθ = v²/2gR

        cosθ = 1 - v²/2gR

        cosθ = 1 - (Rω)²/2gR

        cosθ = 1 - Rω²/2g

        cosθ = 1 - 0.40(3.1²)/(2(9.8))

        cosθ = 0.804267

              θ = 36.46045...

              θ = 36°

Answer:

Explanation:

Answer:

14.625 years

Explanation:

Because acceleration/deceleration are negligent we can simplify this problem into 2 basic parts:

First: how long before the astronaut arrives? Since the speed of the craft is 0.8 times the speed of light and the distance to travel is 6.5 light-years, the equation to answer this part is:

0.8t=6.5

solve for t by dividing by 0.8

t=6.5/0.8 or 8.125

so, she will arrive in 8.125 years.

Then, once she sends her message, it will travel towards earth at 1 times the speed of light or:

1.0t=6.5 which is just t=6.5

so, her message will arrive back to Earth (8.125+6.5) 14.625 years after takeoff.

The time elapsed in earth will be 14.625 years.

To understand this we need to find how much time she took to travel 6.5 light years and then how much time the signal took to reach the earth.

 0.8t = 6.5 ( because 1 light year = distance travelled by light in a year)

      t = 6.5/0.8

      t = 8.125 years

So, she travelled for 8.125 year.

1.0 t = 6.5 ( because radio signal travells with the same speed of light)

     t  = 6.5/1.0

     t  = 6.5 years

So, 8.125 + 6.5 = 14.625 years

So, the time elapsed on earth between the launch and the arrivel of the first radio signal is 14.625 years.

#SPJ2

Answer:

1m² = 10000cm²

1m² = 1000000mm²

1m² = 1 × 10^-6

Explanation:

When coverting from square meter to square centimetre, multiply the area value by 10.

When coverting from square meter to millimetre, multiply the area value by 10⁶ ( 1 million )

When converting from meter square to kilometre square, divide the area value by 10^-6 ( 0.000001)

Answer:  (B) 1 month

Answer: In ridge push, the mantle wells upward because of the convection and elevates the edges of spreading oceanic plates. Because these plates are higher at the spreading center, they are forced downhill due to gravity and eventually flatten out to the ocean floor.

the answer I need points for my math test

Answer:

The new gravitational pull would be three times as strong as the first which would be 0.000000003 or 3 × 10^-9N.

Explanation:

Hope this helps. . . <3

[tex]\mu = 0.306[/tex]

Explanation:

Assume that the direction down the incline is the +x-direction. We can apply Newton's 2nd law along the x-axis as

[tex]x:\:\:\:\:\:mg\sin17° - f_s = ma[/tex]

[tex]\Rightarrow mg\sin17° - \mu N = 0[/tex]

where m is the mass of the bear and g is the acceleration due to gravity and acceleration a is zero since the bear is moving with a constant velocity. Along the y-axis, we can write Newton's 2nd law as

[tex]y:\:\:\:\:\:N - mg\cos17° = 0 \Rightarrow N = mg\cos17°[/tex]

Combining these two equations together, we get

[tex]mg\sin17° = \mu(mg\cos17°)[/tex]

Solving for [tex]\mu, [/tex]

[tex]\mu = \dfrac{\sin17°}{\cos17°} = \tan17° = 0.306[/tex]

Answer: Moses

Explanation: The Israelites are angry at Aaron and Moses, because they lead them into this desert where there is no food, or water. It was told that after the Israelites left Egypt they would wander the desert for forty years. The Israelites complain that they would have had plenty of food if they had just stayed enslaved in Egypt. They are not grateful that God has liberated them from slavery.

Explanation:

formula: Vi = Vf - (at)

Vi: intial velocity

Vf: final velocity

a: acceleration

t: time

fill in formula with the numbers you are given

Vi= 41.6m - ((9.81 m/s^2)(1.89s))

parenthesis first according to pemdas

Vi= 41.6m - 18.54m/s

Answer: 23.06m/s

disclaimer: I havent done physics in awhile so I have no idea if this is right. just an attempt to help steer you in the right direction hopefully. good luck

Measuring the distance from  [tex]I_1 \ to \ I_1'[/tex] and dividing by 2 has advantage of achieving a more accurate result compared to measurement taking directly.

Measuring the distance from [tex]I_1 \ to \ I_1'[/tex] is equivalent to the double of the distance.

This method of measurement gives a more accurate result because it minimizes error that may result from taking readings form the intervals. It ensures more accurate result compared to the measurement obtained when [tex]OI_1[/tex] is measured directly.

[tex]OI_1 = \frac{I_1 + I_1'}{2}[/tex]

Thus, we can conclude that measuring the distance from  [tex]I_1 \ to \ I_1'[/tex] and dividing by 2 has advantage of achieving a more accurate result compared to measurement taking directly.

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Given the information supplied in the question, the height of the building is  196 m.

The work done on you is a work done in a gravitational field. Recall that the work done in a gravitational field is given by mgh where;

m = mass of the body

g = acceleration due to gravity

h = height

Given that;

Work done = 1.15x10^5 J

Acceleration due to gravity = 9.8 ms-2

Mass =  60.0 kg

Substituting values;

h = W/mg

h =  1.15x10^5 J/ 60.0 kg ×  9.8 ms-2

h = 196 m

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The projectile launch equations allow to find the results for the questions about the movement of the ball are:

Given parameters

To find

Projectile launching is an application of kinematics, where on the x-axis there is no acceleration and on the y-axis is the gravity acceleration.

The range is the distance traveled for the same departure height, see attached.

.

          R =[tex]\frac{v_o^2 \ sin 2\theta}{g}[/tex]  

         [tex]v_o^2 = \frac{ g R}{sin 2 \theta }[/tex]  

Let's calculate.

          v₀² = [tex]\frac{9.8 \ 31.5}{sin \ (2 \ 40)}[/tex]9.8 31.5 / sin (2 40.0)

          [tex]v_o = \sqrt{313.46}[/tex]o = ra 313.46

          v₀ = 17.7 m / s

At the point of maximum height the vertical speed is zero.

          v² = v₀² - 2 g y

          0 = v₀² - 2g y

          y = [tex]\frac{v_o^2}{2g}[/tex]  

Let's calculate.

         y = [tex]\frac{17.7^2}{2 \ 9.8}[/tex]  

         y = 16 m

In conclusion, using the projectile launch equations we can find the results for the questions about the movement of the ball are:

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Answer:

you have patience the distance.

Explanation:

the train leaves at 6.30.

Answer:

Explanation:

acceleration a = F/m = 44000/33000 = 1.3m/s²

F = ma = 70(1.3) = 93 N

Answer:

yes its okay i think so it is the correct answer

Answer:

83.8851 mi/h