If 710-nm and 655-nm light passes through two slits 0.65 mm apart, how far apart are the second-order fringes for these two wavelengths on a screen 1.1 m away?

Answer:

Explanation:

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According to the National Geographic, "Earth’s magnetic field dominates a region called the magnetosphere, which wraps around the planet and its atmosphere. Solar wind, charged particles from the sun, presses the magnetosphere against the Earth on the side facing the sun and stretches it into a teardrop shape on the shadow side. The magnetosphere protects the Earth from most of the particles, but some leak through it and become trapped. When particles from the solar wind hit atoms of gas in the upper atmosphere around the geomagnetic poles, they produce light displays called auroras. These auroras appear over places like Alaska, Canada and Scandinavia, where they are sometimes called “Northern Lights.” The “Southern Lights” can be seen in Antarctica and New Zealand. The magnetic field is the area around a magnet that has magnetic force. All magnets have north and south poles."

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Hope this helps! <3

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Answer:

e:10 m   b: -5 m/s b.The object decreases its velocity.

Explanation:

Answer:

Explanation:

The momentum of an object can be found by using the formula

momentum = mass × velocity

From the question we have

momentum = 6 × 4

We have the final answer as

Hope this helps you

Answer:

The distance is 0.96m

Explanation:

Given

m1= 900kg

m2= 1600kg

Force F= 0.0001nN

G=6.67430*10^-11 Nm^2/kg^2

Required

The distance r

Step two:

the formula for the force is given as

F = Gm1m2/r2

make r subject of the formula

[tex]r= \sqrt{\frac{Gm1m2}{F} }[/tex]

[tex]r= \sqrt{\frac{6.67430*10^-11*900*1600}{0.0001} }\\\\r= 0.00009610992/0.0001`}\\\\r= 0.96m[/tex]

Answer:

The distance is 0.96m

Explanation:

Given

m1= 900kg

m2= 1600kg

Force F= 0.0001nN

G=6.67430*10^-11 Nm^2/kg^2

Required:

The distance r

Step two:

the formula for the force is given as

F = Gm1m2/r2

make r subject of the formula

[tex]r= \sqrt{\frac{Gm1m2}{F} }[/tex]

[tex]r= \sqrt{\frac{6.67430*10^-11*900*1600}{0.0001} }\\\\r= 0.00009610992/0.0001`}\\\\r= 0.96m[/tex]

Answer:

The distance between the compact car and pickup truck is 0.96048 m

Explanation:

The gravitational force is directly proportional to the product of the masses of the interacting object, it is also inversely proportional to the square of the distance between them.  This is shown in equation 1;

[tex]F =G \frac{m_{1} X m_{2} }{d^{2} }[/tex]............ 1

Where F is the gravitational force = 0.0001 N

G is the gravitational constant = 6.673 x [tex]10^{-11} Nm^{2} kg^{-2}[/tex]

[tex]m_{1}[/tex]  is the mass of the compact car = 900kg

[tex]m_{2}[/tex] is the mass of the pickup truck = 1600kg

d is the distance and its unknown ?

Let us make d the subject formula in equation 1

[tex]d = \sqrt{G\frac{m_{1} m_{2} }{F } }[/tex] .... 2

Substituting into equation 2 we have

[tex]d = \sqrt{\frac{6.673x10^{-11} x 900 x 1600}{0.0001N} }[/tex]

d = 0.96048m

Therefore the distance between the compact car and pickup truck is 0.96048 m

Answer:

1 +2=3 so thats it

Explanation:

Because i need points

Answer:

it’s c not d

Explanation:

took the test

Answer: D!!!

Explanation: jus got it wrong from the other answer.

Answer:

1.2 kg

__________________________________________________________

We are given:

Mass of the block = 2 kg

Coefficient of Static Friction = 0.6

__________________________________________________________

Friction Force on the Block:

Finding the Normal Force:

We know that the normal force will be equal and opposite to the weight of the 2 kg block

So, Normal Force = mg

replacing the variables with the given values

Normal Force = (2)(9.8)                    [Taking g = 9.8]

Normal Force = 19.6 N

Friction force on the Block:

We know that:

Coefficient of Static Friction =  Static Friction Force/Normal Force

replacing the variables

0.6 = Static Friction force / 19.6

Static Friction force = 0.6*19.6 N                 [Multiplying both sides by 19.6]

Static Friction force = 11.76 N

__________________________________________________________

Largest Mass that can Hang:

We know that the Static Friction force is 11.76 N, this means that a force of 11.76 N will be applied to keep the object at rest

So, if the weight of the second block is less than the static friction force, it will hang

Weight of the second block ≤ 11.76

We know that weight = mg

mg ≤ 11.76

m(9.8) ≤ 11.76                                                   [since g = 9.8]

m ≤ 1.2 kg                                                        [dividing both sides by 9.8]

From this, we can say that the maximum mass of the second block is 1.2 Kg

Answer:

RMS velocity, [tex]v_{rms}=5748.75\ m/s[/tex]

Explanation:

We need to find the RMS speed of helium atoms near the surface of the Sun at a temperature of about 5300 K.

The formula for RMS speed of a gas is given by :

[tex]v_{rms}=\sqrt{\dfrac{3RT}{m}}[/tex]

Where

R is ideal gas constant, R = 8.314 J /mol K

T = 5300 K

m is molar mass of Helium, [tex]m = 4\times 10^{-3}\ Kg/mol[/tex]

Substituting all the values in above formula :

[tex]v_{rms}=\sqrt{\dfrac{3\times 8.314\times 5300}{4\times 10^{-3}}}\\\\=5748.75\ m/s[/tex]

So, the RMS speed Helium atoms 5748.75 m/s.

Answer:

16 meters

Explanation:

Use the formula that relates frequency velocity and wavelength:

velocity = wave-length x frequency

in our case:

12 m/s = wave-length * 0.75 Hz

wave-length= 12/0.75  m = 16 meters

Answer:

5.79

Explanation:

Vf=Vi+at

0=8.33+(-9.8)sin8.44t

t=8.33/(9.8sin8.44)=0.83/sin 8.44

=5.79

彼が好きな男のような話は妻と関係があったのでそう

Explanation:これが難しい場合はコメントで教えてくださいが何でも

lol okay

from: a random person :)

Answer: Iron

Explanation: Gradpoint

Answer:

The drill's angular displacement during that time interval is 24.17 rad.

Explanation:

Given;

initial angular velocity of the electric drill, [tex]\omega _i[/tex] = 5.21 rad/s

angular acceleration of the electric drill, α = 0.311 rad/s²

time of motion of the electric drill, t = 4.13 s

The angular displacement of the electric drill at the given time interval is calculated as;

[tex]\theta = \omega _i t \ + \ \frac{1}{2}\alpha t^2\\\\\theta = (5.21 \ \times \ 4.13) \ + \ \frac{1}{2}(0.311)(4.13)^2\\\\\theta = (21.5173 ) \ + \ (2.6524)\\\\\theta =24.17 \ rad[/tex]

Therefore, the drill's angular displacement during that time interval is 24.17 rad.

Answer:

25

Explanation:

Momentum is conserved, so the total momentum before collision is equal to the total momentum after collision. Take the right direction to be positive. Then

(3.00 kg) (2.09 m/s) + (2.22 kg) (-3.92 m/s) = (3.00 kg) (-1.71 m/s) + (2.22 kg) v

where v is the velocity of the 2.22 kg block after collision. Solve for v :

6.27 kg•m/s - 8.70 kg•m/s = -5.13 kg•m/s + (2.22 kg) v

(2.22 kg) v = 2.70 kg•m/s

v = (2.70 kg•m/s) / (2.22 kg)

v ≈ 1.22 m/s

i.e. a velocity of about 1.22 m/s to the right.

my bad I was in a herie last time can you please answer my question , I am going to give you the 5 points back for this question and extra 30 points ,

Explanation:

first its going to say 10 points for my question but after that I well make answer a small question and give you 30points. like whats your favorite color . stay tuned .

Answer:

The maximum height reached by the ride after it was pulled to the ground is 51.6 m.

The given parameters;

The distance half-way between the bands;

[tex]\frac{d}{2} = \frac{31}{2} = 15.5 \ m[/tex]

The maximum length of the band when stretched is calculated as;

[tex]c^2 = 15.5^2 + 76^2\\\\c^2 = 6016.25\\\\c = \sqrt{6016.25} \\\\c = 77.57 \ m[/tex]

The extension of the elastic band;

x = 77.57 m - 41 m

x = 36.37 m

The elastic potential energy stored in the band;

[tex]E = \frac{1}{2} kx^2\\\\E = \frac{1}{2} \times 310 \times (36.57)^2\\\\E = 207,291.56 \ J[/tex]

The elastic potential energy of the elastic band will be converted into kinetic energy of the ride and the speed of the ride is calculated as;

[tex]E = \frac{1}{2} mv^2\\\\207,291.56 = \frac{1}{2} \times 410 \times v^2\\\\v^2 = \frac{207,291.56}{(0.5\times 410)} \\\\v^2 = 1011.178\\\\v = \sqrt{1011.178} \\\\v = 31.8 \ m/s[/tex]

The maximum height reached by the ride is calculated as;

[tex]P.E = K.E\\\\mgh = \frac{1}{2} mv^2\\\\h = \frac{v^2}{2g} \\\\h = \frac{(31.8)^2}{(2\times 9.8)} \\\\h = 51.6 \ m[/tex]

Thus, the maximum height reached by the ride is 51.6 m.

Learn more here: https://brainly.com/question/156316

Answer:98

Explanation:hope this helps!

Answer:

Fx = 467.65N

Fy = 270N

Explanation:

Given

Force = 540N

angle of inclination = 40 degree

Horizontal component Fx = Fcos 30

Fx = 540cos30

Fx = 540(0.8660)

Fx = 467.65N

Hence the horizontal component is 467.65N

Vertical component Fy = Fsin 30

Fy = 540sin30

Fy = 540(0.5)

Fy = 270N

Hence the vertical component is 270N

Split up the forces into components acting parallel to and perpendicular to the slope. See the attached picture for the reference axes.

The box stays on the surface of the plane, so that the net force acting perpendicular to it is 0, and the only acceleration is applied in the parallel direction.

Let m be the mass of the box, θ the angle the plane makes with the ground, and a the acceleration of the box. By Newton's second law, we have

• net parallel force

∑ Force (//) = W (//) - F = m a

(that is, the net force in the parallel direction is the sum of the parallel component of the weight W and the friction F which acts in the negative direction)

• net perpendicular force

∑ Force (⟂) = W (⟂) + N = 0

Notice that

W (//) = W sin(θ) … … … which is positive since it points down the plane

W (⟂) = -W cos(θ) … … … which is negative since it points opposite the normal force N

So the equations become

W sin(θ) - F = m a

-W cos(θ) + N = 0

Solving for a gives

a = (W sin(θ) - F ) / m

which is good enough if you know the magnitude of the friction force.

If you don't, you can write F in terms of the coefficient of kinetic friction between the box and plane, µ, as

F = µ N

so that

a = (W sin(θ) - µ N ) / m

and the normal force itself has a magnitude of

N = W cos(θ)

so that

a = (W sin(θ) - µ W cos(θ) ) / m

The weight W has magnitude m g, where g is the magnitude of the acceleration due to gravity, so

a = (m g sin(θ) - µ m g cos(θ) ) / m

a = g (sin(θ) - µ cos(θ))

Answer:

its D: A frozen lake melts under the sun.

Explanation:

Radiation is the transfer of heat energy through space by electromagnetic radiation. Most of the electromagnetic radiation that comes to the earth from the sun is invisible. Only a small portion comes as visible light. Light is made of waves of different frequencies.

Answer:

133.8 N

Explanation:

Recall that the acceleration of gravity in Neptune is estimated as 11.15 m/s^2

Therefore, the weight of the dog on this planet would be:

Weight = mass x acceleration of gravity = 12 kg x 11.15 m/s^2 = 133.8 N

Answer:

Plants the chloroplasts

Answer:

the forces are the same

Explanation:

Answer

I Think Its 150

Answer:

F = 11226.02 N

Explanation:

The roller coaster  car and passengers have a combined mass of 1620kg.

They descend the first hill at  an angle of 45.0 degrees to the horizontal.

We need to find force is the rollercoaster pulled  down the hill.

We firstly find the rectangular component of the downward. The force acting in the downward direction is mgsinθ such that,

F = mgsinθ

= 1620 × 9.8 × sin(45)

= 11226.02 N

So, the roller coaster is pulled down the hill with a force of 11226.02 N.

Answer:

f = 1 Hz

Explanation:

From the attached figure, we find that the time period of the wave is 1 second.

It is a longitudinal wave. It travel in the form of compression and rarefaction. The relation between time period T and frequency f is given by :

[tex]f=\dfrac{1}{T}\\\\\text{As T = 1 s}\\\\f=1\ s^{-1}\\\\=1\ Hz[/tex]

Hence, the frequency of this wave is 1 Hz.

Answer:

A) 2.650 km

Explanation:

The relationship between acceleration of gravity and gravitational constant is:

[tex]g = \frac{Gm}{R^2}[/tex] ---- (1)

Where

[tex]R = 6,400 km[/tex] -- Radius of the earth.

From the question, we understand that the gravitational field of the rocket is 50% of its original value.

This means that:

[tex]g_{rocket} = 50\% * g[/tex]

[tex]g_{rocket} = 0.50 * g[/tex]

[tex]g_{rocket} = 0.5g[/tex]

For the rocket, we have:

[tex]g_{rocket} = \frac{Gm}{r^2}[/tex]

Where r represent the distance between the rocket and the center of the earth.

Substitute 0.5g for g rocket

[tex]0.5g = \frac{Gm}{r^2}[/tex] --- (2)

Divide (1) by (2)

[tex]\frac{g}{0.5g} = \frac{Gm}{R^2}/\frac{Gm}{r^2}[/tex]

[tex]\frac{g}{0.5g} = \frac{Gm}{R^2}*\frac{r^2}{Gm}[/tex]

[tex]\frac{1}{0.5} = \frac{1}{R^2}*\frac{r^2}{1}[/tex]

[tex]2 = \frac{r^2}{R^2}[/tex]

Take square root of both sides

[tex]\sqrt 2 = \frac{r}{R}[/tex]

Make r the subject

[tex]r = R * \sqrt 2[/tex]

Substitute [tex]R = 6,400 km[/tex]

[tex]r = 6400km * \sqrt 2[/tex]

[tex]r = 6400km * 1.414[/tex]

[tex]r = 9 049.6\ km[/tex]

The distance (d) from the earth surface is calculated as thus;

[tex]d = r - R[/tex]

[tex]d = 9049.6\ km - 6400\ km[/tex]

[tex]d = 2649.6\ km[/tex]

[tex]d = 2650\ km[/tex] --- approximated

Answer:

x=-3/4

Explanation: