using the image provided, which moon phase is next?

Answer:

377 nm

Explanation:

Number of lines per meter is, [tex]N &=530 \times 1000 \\ &=530000 \text { lines } / \mathrm{m} \end{aligned}[/tex]

Grating element is, [tex]d=\frac{1}{N}[/tex]

[tex]=1.8868 \times 10^{-6} \mathrm{~m}[tex]

Order is, n=5

Condition for maximum intensity is, [tex]d \sin \theta=n \lambda[/tex]

 [tex]\lambda &=\frac{1.8868 \times 10^{-6}}{5(\sin 90)} \\ &=0.377 \times 10^{-6} \mathrm{~m} \\ &=377 \mathrm{~nm}[/tex]

Answer:

To protect us.

Explanation:

For ex. your dunking on a basketball hoop if that wasn't strong you would fall on your back and get injured.

Answer:

= 1200m/s or 1.2 x [tex]10^{3}[/tex] m/s

Explanation:

Answer:

Explanation:

b) light

Answer: how the government and teachers can address or help students experiencing stress and anxiety.

Explanation:

Answer:

1. F = 3400 N = 3.4 KN

2. [tex]K.E_f=92832.6\ J = 92.83\ KJ[/tex]

3. v = 14.9 m/s

Explanation:

1.

First, we will calculate the acceleration of Pam by using the third equation of motion:

[tex]2as = v_f^2-v_i^2[/tex]

where,

a = acceleration = ?

s = distance = 27.3 m

vf = final speed = 62 m/s

vi = initial speed = 0 m/s

Therefore,

[tex]2a(27.3\ m) = (62\ m/s)^2-(0\ m/s)^2\\\\a = 70.4\ m/s^2[/tex]

Now, we will calculate the force by using Newton's Second Law of Motion:

F = ma

F = (48.3 kg)(70.4 m/s²)

F = 3400 N = 3.4 KN

2.

Final kinetic energy is given as:

[tex]K.E_f = \frac{1}{2}mv_f^2\\\\K.E_f = \frac{1}{2} (48.3\ kg)(62\ m/s)^2[/tex]

[tex]K.E_f=92832.6\ J = 92.83\ KJ[/tex]

3.

According to the law of conservation of energy:

[tex]Potential\ Energy\ at\ top = Kinetic\ Energy\ at\ bottom\\mgh = \frac{1}{2}mv_2 \\\\v = \sqrt{2gh}[/tex]

where,

v = speed at bottom = ?

g = acceleration due to gravity = 9.81 m/s²

h = height at top = 11.3 m

Therefore,

[tex]v = \sqrt{(2)(9.81\ m/s^2)(11.3\ m)}[/tex]

v = 14.9 m/s

Answer:

Explanation:

comfortable distance is 25 cm .

He must be using convex lens . In that case rays coming from object placed at 25 cm appears to be coming from 59.1 cm due to converging nature of convex lens.

object distance u = -25 cm

image distance v = -59.1 cm

Lens formula

1 / v - 1 /u = 1 /f

-1 / 59.1 + 1 / 25 = 1/f

- .0169 + .04 = 1 / f

.0231 = 1 / f

f = 43.3 m

Answer:

  α = 0.357 ras / s²

Explanation:

This is a rotational kinematics exercise, it tells us that it performs 10 squats in 30 s, for which it performs one squat at t = 3 s, also indicates that the angle of the squat is θ = 92º

            θ = θ₀ + w₀ t + ½ α t²

the athlete starts from rest, whereby w₀ = 0 and the initial angle in the vertical position is zero (θ₀=0)

            θ = ½ α t²

            α = 2 θ /t²

let's reduce the magnitudes to the SI system

            θ = 92º (π rad /180º) = 0.511π rad

let's calculate

            α = 2 0.5111π /3²

            α = 0.1136π rad / s²

            α = 0.357 ras / s²

Answer:

5*10^2

Explanation:

A p e x

Answer:

d = 1.19 x 10⁻⁴ m = 0.119 mm

Explanation:

This problem can be solved by using Young's double-slit experiment formula:

[tex]Y = \frac{\lambda L}{d}[/tex]

where,

Y = fringe spacing = 32 mm = 0.032 m

L = slit to screen distance = 6 m

λ = wavelength of light = 633 nm = 6.33 x 10⁻⁷ m

d = slit width = ?

Therefore,

[tex]0.032\ m = \frac{(6.33\ x\ 10^{-7}\ m)(6\ m)}{d}\\\\d = \frac{(6.33\ x\ 10^{-7}\ m)(6\ m)}{0.032\ m}[/tex]

d = 1.19 x 10⁻⁴ m = 0.119 mm

Answer:

The person's velocity is zero.

Explanation:

Answer:

If we use the equation for the transformation of velocities for moving frames:

v' = (v - u) / (1 - u * v / c^2) where we measure the speed of v' approaching from the left where v is in a frame moving at -u towards v'

v' = (.6 c - (-.6 c)) / (1 - (-.6 c) * .6 c / c^2) = 1.2 c / (1 + .6 * .6)

or v' = 1.2 c / (1 + .36) = .88 c

v is approaching from the left at .6 c in the reference frame and the other frame approaches from the right at -.6 c with speed u  (-.6 c) and we measure the speed of v as seen in the frame moving to the left

Answer:

Height of the door = 2m = 2000 cm

1 in = 2.54 cm

So 1 cm = 1/2.54 in

2000 cm = 200000/ 254

=

787.401574803

So no.c is correct

The door is  78.7 inch tall. Hence, option (A) is correct.

Any arbitrarily selected and widely used reference standard for length measurement is referred to as a unit of length. The metric system, which is adopted by every nation on earth, is the most widely utilized in modern times.

The American customary units are also in use in the United States. In the UK and several other nations, British Imperial units are still used sometimes. There are SI units and non-SI units in the metric system.

Given that: the height of the door is = 2 meter

= 2*100 centimeter

= 200 centimeter.

There are 2.54 centimeter in 1 inch.

Hence,  the height of the door is = 2 meter  = 200 centimeter

= (200/2.54) inch

= 78.7 inch.

The door is 78.7 inch tall.

Learn more about length here:

https://brainly.com/question/17139363

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So they give us this

V=IR

V= 1.8

I=0.4

R=?

So we insert the thing that we know.

1.8=0.4*R

We need to leave our unknown value alone. So if our value of 0.4 is multiplying the unknown value it passes to the other side dividing.

So we have this.

Lastly we solve.

R=4.5ohms

The formula to find R is V=IR

V/I=R

So the resistance will be the Voltage divided by the Current

The sum of the resistance = 2 ohms x 80 lights = 160 ohms.

Current = Total voltage / total resistance:

Current = 120V / 160 ohms

Current = 0.75 Amps

Complete question is;

A small car of mass m and a large car of mass 2m drive around a highway curve of radius R. Both cars travel at the same speed (v). The

centripetal acceleration (Grad) of the large car is the centripetal acceleration of the small car. How does the Force of the small car FS compare to the force of the large car FL as they round the curve.

four times

twice

half

equal to

Answer:

Half

Explanation:

Formula for centripetal force is given as;

F = mv²/R

Where;

v is velocity

R is radius

Now, centripetal acceleration is given by;

a = v²/R

Since they both travel with the same velocity V and radius remains the same, we can say that;

F = ma

For the small car;

FS = ma

For the big car;

FL = 2ma

This means the force of the small car is half of that of the Large car

Thus;

FS = ½FL

Answer:

Explanation:

Rate of volume flow at two points will be same at two points .

A₁ V₁ = A₂V₂

A₁ and A₂ are area of cross section at two points and V₁ and V₂ are velocities .

A₁ = 18.2 x 3.55 = 64.61 m²

V₁ = 2 .55 x 10⁻² m/s

A₂ = 16.3 x d = 16.3 d  m²

d is depth at second point .

V₂ = 11.6 x 10⁻² m/s

64.61 m² x 2 .55 x 10⁻² m/s = 16.3 d  m² x 11.6 x 10⁻² m/s

d = .87 m

so canal is .87 m deep.

Answer:

T = 1.12 10⁻⁷ s

Explanation:

This exercise must be solved in parts. Let's start looking for the electric field in the axis of the ring.

All the charge dq is at a distance r

           dE = k dq / r²

Due to the symmetry of the ring, the field perpendicular to the axis is canceled, leaving only the field in the direction of the axis, if we use trigonometry

            cos θ =[tex]\frac{dE_x}{dE}[/tex]

             dEₓ = dE cos θ

              cos θ = x / r

substituting

                dEₓ = [tex]k \frac{dq}{r^2 } \ \frac{x}{r}[/tex]

                DEₓ = k dq x / r³

let's use the Pythagorean theorem to find the distance r

             r² = x² + a²

where a is the radius of the ring

we substitute

              dEₓ = [tex]k \frac{x}{(x^2 + a^2 ) ^{3/2} } \ dq[/tex]

we integrate

               ∫ dEₓ =k \frac{x}{(x^2 + a^2 ) ^{3/2} }  ∫ dq

               Eₓ = [tex]k \ Q \ \frac{x}{(x^2+a^2)^{3/2}}[/tex]

In the exercise indicate that the electron is very central to the center of the ring

                x << a

                Eₓ = [tex]k \ Q \frac{x}{a^3 \ ( 1 +(x/a)^2)^{3/2})}[/tex]

if we expand in a series

                  [tex](\ 1+ (x/a)^2 \ )^{-3/2} = 1 - \frac{3}{2} (\frac{x}{a} )^2[/tex]

we keep the first term if x<<a

                 Eₓ = [tex]\frac{ k Q}{a^3} \ x[/tex]

the force is

                 F = q E

                 F = [tex]- \frac{kQ }{a^3} \ x[/tex]

this is a restoring force proportional to the displacement so the movement is simple harmonic,

                 F = m a

                 [tex]- \frac{keQ}{a^3} \x = m \frac{d^2 x}{dt^2 }[/tex]

                 [tex]\frac{d^2 x}{dt^2} = \frac{keQ}{ma^3} \ x[/tex]

the solution is of type

                  x = A cos (wt + Ф)

with angular velocity

                w² = [tex]\frac{keQ}{m a^3}[/tex]

angular velocity and period are related

                 w = 2π/ T

we substitute

               4π² / T² = \frac{keQ}{m a^3}

                T = 2π  [tex]\sqrt{\frac{m a^3 }{keQ} }[/tex]

let's calculate

                T = 2π [tex]\sqrt{ \frac{ 9.1 \ 10^{-31} \ 2.2^3 }{9 \ 10^9 \ 1.6 \ 10^{-19} \ 0.021 \ 10^{-3} } }[/tex]

                 T = 2π pi [tex]\sqrt{320.426 \ 10^{-18} }[/tex]

                 T = 2π  17.9 10⁻⁹ s

                 T = 1.12 10⁻⁷ s

Answer:

The air in the soil is similar in composition to that in the atmosphere with the exception of oxygen, carbon dioxide, and water vapor. In soil air as in the atmosphere, nitrogen gas (dinitrogen) comprises about 78%. In the atmosphere, oxygen comprises about 21% and carbon dioxide comprises about 0.036%.

hope this helps

have a good day :)

Explanation:

Answer:

That is equal to R1 + R2. If three or more unequal (or equal) resistors are connected in series then the equivalent resistance is: R1 + R2 + R3 +…, etc. One important point to remember about resistors in series networks to check that your maths is correct.

The work-energy theorem says that the total work done on the block is equal to the difference of its kinetic energies at points B and A. Then the total work done on the block is

[tex]W_{\rm total} = K_B - K_A = 4.0\,\mathrm J - 5.0\,\mathrm J = -1.0\,\mathrm J[/tex]

Friction acts on the block to oppose its motion, so it does negative work on the block, -4.5 J.

The only other force acting on the block as it moves is the force P. Let [tex]W_P[/tex] be the work done by the force P. Then the total work done on the block is

[tex]W_{\rm total} = W_P + W_{\rm friction} \iff -1.0\,\mathrm J = W_P - 4.5 \,\mathrm J \implies W_P = \boxed{3.5\,\mathrm J}[/tex]

Answer:

In which school you are???

Explanation:

Answer:

static friction=0.126