What is it called when the moon passes through the penumbra of Earth’s shadow?(1 point)

total lunar eclipse

total solar eclipse

partial lunar eclipse

partial solar eclipse

Answer:

the minimum wall thickness that will enhance the reflection of light is 146.9 nm

Explanation:

Given the data in the question;

At the first interface, a phase shift occurs as the incident light is in air that has less refractive index compare to the thin film of soap bubble.

At the second interface, no shift occurs,

condition for constructive interference;

t = ( m + 1/2) × λ/2n

where m = 0, 1, 2, 3 . . . . . .

now, the condition for the constructive interference;

t = mλ/2n

where t is the thickness of the soap bubble,  λ is the wavelength of light and n is the refractive index of soap bubble.

so the minimum thickness of the film which will enhance reflection of light will be;

t[tex]_{min[/tex] =  ( m + 1/2) × λ/2n

we substitute

t[tex]_{min[/tex] =  ( 0 + 1/2) × 711 /2(1.21)

t[tex]_{min[/tex] = 0.5 × 711/2.42

t[tex]_{min[/tex] = 0.5 × 293.80165

t[tex]_{min[/tex] = 146.9 nm

Therefore,  the minimum wall thickness that will enhance the reflection of light is 146.9 nm

Answer:

0.08 ms^-2

Explanation:

by using v= u + at

initial velocity is zero as it is starting from rest

4= 0 + a x 50

4/50 = a = 0.08 ms^-2

Answer:

(a) 8.85×10⁻³ minutes

(b) 4.24×10¹⁹ electrons

Explanation:

(a) Using,

Q = it............................. Equation 1

Where Q = quantity of charge, i = current, t = time.

Make t the subject of the equation

t = Q/i............................. Equation 2

Given: Q = 6.8×10⁰ C, i = 12.8 A

Substitute these values into equation 2

t = 6.8×10⁰/12.8

t = 8.85×10⁻³ minutes

(b) n = Q/(1.602×10⁻¹⁹)................. Equation 3

Where n = number of electrons.

Given: Q = 6.8×10⁰ C

Substitute into equation 2

n = 6.8×10⁰/1.602×10⁻¹⁹

n = 4.24×10¹⁹ electrons

(a) The time taken by the charge to flow from the drier will be [tex]\d8.85[/tex]×[tex]\d10^{-3}[/tex]minutes

(b) Amount of the electrons in the charge will be  [tex]\d4.24[/tex]×[tex]\d10^{19}[/tex] electrons

As we know Q = IT

Where Q = quantity of charge, i = current, T = time.

From the above equation

                    T= Q/I.

Given: Q = [tex]6.8\times\d10^{0}[/tex] C, i = 12.8 A

Substitute these values  

T=  [tex]6.8[/tex]×[tex]\d10^{0}[/tex] /12.8

T =  [tex]8.85[/tex]×[tex]\d10^{-3}[/tex] minutes

Now the number of the electrons present in the charge will be

n = Q/( [tex]1.602[/tex]×[tex]\d10^{-19}[/tex])

Where n = number of electrons.

Given: Q = [tex]6.8\times\d10^{0}[/tex] C

Substitute Value of Q  

n =  [tex]6.8\times\d10^{0}[/tex]/ [tex]1.602\times\d10^{-19}[/tex]

n = [tex]4.24\times\d10^{19}[/tex] electrons

Thus

(a)The time taken by the charge to flow from the drier will be [tex]\d8.85[/tex]×[tex]\d10^{-3}[/tex]minutes

(b)Amount of the electrons in the charge will be  [tex]\d4.24[/tex]×[tex]\d10^{19}[/tex] electrons

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Answer:

0.375

Explanation:

For incompressible flow, we know that;

ρ1•v1•A1 = ρ2•v2•A2

Where;

ρ1 = density of fluid at position A

v1 = speed of fluid at position A

A1 = area of tube

ρ2 = density of fluid at position B

v2 = speed of fluid at position B

A2 = area of tube

We want to find ratio of the density of the fluid at position A to the density of the fluid at position B.

Thus;

ρ1/ρ2 = (v2•A2)/(v1•A1)

Now, the tube will have the same height.

But we are given;

diameter of A = 12.00 cm = 0.12 m

diameter of B = 6 cm = 0.06 m

Thus;

A1 = π(d²/4)h = πh(0.12²/4)

A2 = πh(0.06²/4)

We are also given;

v1 = 12 m/s

v2 = 18 m/s

Thus;

ρ1/ρ2 = (18 × πh(0.06²/4))/(12 × πh(0.12²/4))

πh/4 will cancel out to give;

ρ1/ρ2 = (18 × 0.06²)/(12 × 0.12²)

ρ1/ρ2 = 0.375

Answer:

Generally, the lowest overtone for a pipe open at one end and closed would be at  y / 4  where y represents lambda, the wavelength.

Since F (frequency) = c / y       Speed/wavelength

F2 / F1 = y1 / y2      because c is the same in both cases

F2 = y1/y2 * F1

F2 = 3 F1 = 750 /sec

Note that L = y1 / 4 = 3 y2 / 4 for these wavelengths to fit in the pipe

and y1 = 3 y2

The second harmonic will be three times the first harmonic. The answer is 750 Hz

Closed pipes have odd multiples of frequencies or harmonics. That is,

If  [tex]F_{0}[/tex] = fundamental frequency = first harmonic

[tex]F_{1}[/tex] = 3[tex]F_{0}[/tex] = second harmonic

[tex]F_{2}[/tex] = 5[tex]F_{0}[/tex] = third harmonic

[tex]F_{3}[/tex] = 7[tex]F_{0}[/tex] = fourth harmonic

Let assume that the first harmonic is 250 Hz, If you blow it much harder, second, third or fourth harmonic can be produced.

By using the formula above,

second harmonic will be 3 x 250 = 750Hz

Therefore, the frequency of the next harmonic heard if you blow much harder will be 750 Hz

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Answer:

oil heats faster

Explanation:

Answer:

work done= force × displacement

=800×5

=4000J

Explanation:

The amount of work done is the result of the magnitude of force applied and the displacement of the body due to the force applied. Therefore, work done is defined as the product of the applied force and the displacement of the body.

Answer:

the amplitude of the wave

the energy of the wave

the type of wave

the type of medium

Answer: The amount of light that enters the pupil is controlled by the Iris

Explanation:

Answer:

static force

Explanation:

mark me brainliest

Answer:

533.33 nm

Explanation:

Since dsinθ = mλ  for each slit, where m = order of slit and λ = wavelength of light. Let m' = 10 th order fringe of the first slit of wavelength of light, λ = 640 nm and m"= 12 th order fringe of the second slight of wavelength of light, λ'.

Since the fringes coincide,

m'λ = m"λ'

λ' = m'λ/m"

= 10 × 640 nm/12

= 6400 nm/12

= 533.33 nm

1) Does a 1 kg object weight 9.8 newtons on the moon? why?

No. 1kg of mass does not weigh 9.8N on the moon.

Weight = (mass) x (gravity).

Gravity is 9.8 m/s² on Earth, but gravity is only 1.62 m/s² on the moon.

2) How much does a 3-kg object weigh (on earth) in newtons?

Weight = (mass) x (gravity)

Gravity = 9.8 m/s² on Earth.

Weight = (3 kg) x (9.8 m/s² )

Weight = 29.4 N

3) How much does a 20-kg object weigh (on earth) in newton?

Weight = (mass) x (gravity)

Gravity = 9.8 m/s² on Earth.

Weight = (20 kg) x (9.8 m/s² )

Weight = 196 N

4) What must happen for the mass of an object to change?

When an object moves, its mass increases.  The faster it moves, the greater its mass gets.  But this is all part of Einstein's "Relativity".  The object has to move at a significant fraction of the speed of light before any change can be noticed or measured.  So as far as we are concerned, in everyday life, the mass of an object doesn't change, no matter where it is, or what you do to it.

5) What are 2 ways the weight of an object can change?

First, remember that the mass of an object doesn't change, no matter where it is, what you do to it, or what else is around it.

But its weight can change, because its weight depends on the strength of gravity in the place where the object is, and that gravity is the result of what else is around it in the neighborhood.  So the weight can change even though the mass doesn't.

The weight of an object changes if you take it to a place where gravity is stronger or weaker.

Let's say we have an object whose mass is 90.72 kilograms.  Like me !    

As long as I stay on earth, where gravity is 9.8 m/s² , I weigh 889 Newtons  (200 pounds).

. . . Fly me to the moon. Gravity = 1.62 m/s²  Weight = 147 Newtons (33 lbs)

. . . Drag me to Jupiter.  Gravity = 24.8 m/s²  Weight = 2,249 N (506 pounds)

My mass never changed, but my weight sure did.

Answer:

a)    σ = [tex]\frac{Q_1}{ a^2 - b^2}[/tex] ,  b)  σ = [tex]\frac{Q_2}{d^2}[/tex] , c) Q_ {total} = Q₁ + Q₂,  σ_ {net} = [tex]\frac{Q_1 + Q_2}{\pi \ a^2}[/tex]

Explanation:

a) The very useful concept of charge density is defined by

          σ = Q / A

In this case we have a circular disk

The are of a circle is

         A = π r²

in this case we have a hole in the center of radius r = b, so

         A_net = π r² - π r_ {hollow} ²

         A_ {net} = π (a² - b²)

whereby the density is

          σ = [tex]\frac{Q_1}{ a^2 - b^2}[/tex]

b) The density of the other disk is

          σ = Q₂ / A₂

          σ = [tex]\frac{Q_2}{d^2}[/tex]

c) The total waxed load is requested by the larger circle

           Q_ {total} = Q₁ + Q₂

the net charge density, in the whole system is

          σ = [tex]\frac{Q_{total} }{ A_{total} }[/tex]

the area  is

          A_{total} = π a²

since the other circle is inside, we are ignoring the space between the two circles

          σ_ {net} = [tex]\frac{Q_1 + Q_2}{\pi \ a^2}[/tex]

Answer:

athletic

Explanation:

because internet system has been down since we were in few days

Answer:

[tex]3.52\times 10^{25}\ \text{J}[/tex]

Explanation:

G = Gravitational constant = [tex]6.674\times 10^{-11}\ \text{Nm}^2/\text{kg}^2[/tex]

M = Mass of Sun = [tex]1.989\times 10^{30}\ \text{kg}[/tex]

m = Mass of Earth = [tex]5.972\times 10^{24}\ \text{kg}[/tex]

[tex]r_i[/tex] = Initial radius of orbit = [tex]1.5\times 10^{11}\ \text{m}[/tex]

[tex]r_f[/tex] = Final radius of orbit = [tex]((1.5\times 10^{11})+2\times 10^3)\ \text{m}[/tex]

Energy required is given by

[tex]E=\dfrac{1}{2}\Delta U\\\Rightarrow E=\dfrac{GMm}{2}(\dfrac{1}{r_i}-\dfrac{1}{r_f})\\\Rightarrow E=\dfrac{6.674\times 10^{-11}\times 1.989\times 10^{30}\times 5.972\times 10^{24}}{2}(\dfrac{1}{1.5\times 10^{11}}-\dfrac{1}{(1.5\times 10^{11})+2\times 10^3})\\\Rightarrow E=3.52\times 10^{25}\ \text{J}[/tex]

The energy required would be [tex]3.52\times 10^{25}\ \text{J}[/tex].

Answer:

a. 3.81

Explanation:

F = GMm/r^2

F = (6.67 x 10^-11 x 28 x 1) / 7^2

F = 3.81 x 10^-11 N

Answer:

B. in the direction of the force

Explanation:

Sana nakatulong

The answer is Motion

Explanation:

रात में घूमने वाला arthaarat निशाचर

Answer:

The answer should be D

Explanation:

Answer: if weight affects how fast they go?

Explanation:

Answer:

How can we change the speed of a toy car on a racetrack to describe the car’s motion?

Explanation:

thats the sample respond

Answer:

D. The object moves away from the origin at a speed of 3 m/s, stands still 6 m away from the origin for 3 seconds, then moves toward the origin at a speed of 2 m/s.

Explanation:

I just got it right lol

Answer:

It's plastic.

trust me it's plastic, i've rad it somewhere.

All of them have something that's not like the others.

-- Rubber is the only one on the list that has two repeated letters.

-- Plastic is the only one on the list thagt has no repeated letters.

-- Plastic is the only one on the list that has no 'r' in its name.

-- Copper is the only one on the list that is an element, not a compound.

-- Copper is the only good electrical conductor on the list.

-- Plastic is the only one on the list with more than six letters in its name.

-- Rubber is the only one on the list with no 'p' in its name.

-- Plastic is the only one on the list that doesn't end in "-er".

Answer:

141.18 ohms

Explanation:

From the question given above, the following data were obtained:

Voltage (V) = 12

Current (I) = 0.085 A

Resistance (R) =?

The resistance needed can be obtained as follow:

V = IR

12 = 0.085 × R

Divide both side by 0.085

R = 12 / 0.085

R = 141.18 ohms

Therefore, a resistor of resistance 141.18 ohms is needed.

Answer:

If you have two batteries and they have precisely the same voltage then placing one backwards will effectively cancel out the voltages and no current will flow. However, batteries aren't like that. The slightest difference in voltages mean that current will flow.

Explanation:

Answer:

a) space only    t = 1.28 s

b) space+ atmosphere   t_ {total} = 1.28000003 s

Explanation:

The speed of light in each material medium is constant, which is why we can use the uniform motion relations

           v= x / t

a) let's look for time when it only travels through space

          t = x / c

          t = 3.84 10⁸/3 10⁸

          t = 1.28 s

b) we look for time when it travels part in space and part in the atmosphere

space

as it indicates that the atmosphere has a thickness of e = 30 10³ m

           t₁ = (D-e) / c

           t₁ = (3.84 10⁸ - 30.0 10³) / 3 10⁸

           t₁ = 1.2799 s

atmosphere

we use the refractive index

           n = c / v

           v = c / n

we substitute in the equation of time

           t₂ = e n / c

           t₂ = 30 10³   1,000293 /3 10⁸

           t₂ = 1.000293 10⁻⁴ s

therefore the total travel time is

           t_ {total} = t₁ + t₂

           t_ {total} = 1.2799+ 1.000293 10⁻⁴

           t_ {total} = 1.28000003 s

we can see that the time increase due to the atmosphere is very small

Answer:

Regular reflection

Explanation:

Regular reflection occurs when light reflects off a very smooth surface and forms a clear image.

i hope this helps a bit.

According to the context, a sharp image is formed when light reflects from a regular reflection.

It is reflection without diffusion that obeys the laws of geometrical optics, as in mirrors.

This reflection of light happens when the angles that the two rays determine with the surface are equal.

Therefore, we can conclude that according to the context, a sharp image is formed when light reflects from a regular reflection.

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This question is incomplete, the complete question is;

Car B is rounding the curve with a constant speed of 54 km/h, and car A is approaching car B in the intersection with a constant speed of 72 km/h. The x-y axes are attached to car B. The distance separating the two cars at the instant depicted is 40 m. Determine: the angular velocity of Bxy rotating frame (ω).

Answer:

the angular velocity of Bxy rotating frame (ω) is 0.15 rad/s

Explanation:

Given the data in the question and image below and as illustrated in the second image;

distance S = 40 m

V[tex]_B[/tex] = 54 km/hr

V[tex]_A[/tex] = 72 km/hr

α = 100 m

now, angular velocity of Bxy will be;

ω[tex]_B[/tex] = V[tex]_B[/tex] / α

so, we substitute

ω[tex]_B[/tex] = ( 54 × 1000/3600) / 100

ω[tex]_B[/tex] = 15 / 100

ω[tex]_B[/tex] = 0.15 rad/s

Therefore, the angular velocity of Bxy rotating frame (ω) is 0.15 rad/s