The double bond between two oxygen atoms (a molecule of oxygen air) has
two characteristics. What are they?
A. Four valence electrons are shared.
B. A metallic bond is formed.
C. Valence electrons are shared between oxygen atoms.
D. An ionic bond is formed.

Answer:

4

Explanation:

Answer:

 [tex]\frac{t_{air}}{t_{vaccum}}[/tex] = 1.00029

Explanation:

The refractive index is defined

         n = c / v

         v = c / n

the speed of light per se wave is constant, so we can use the relations of uniform motion

         v = x / t

         t = x / v

we substitute

         t = x n / c

let's calculate the time

vacuum

        t₁ = 1000 1/3 10⁸

        t₁ = 3.333333 10⁻⁶ s

air

        t₂ = 1000 1.00029 / 3 10⁸

        t2 = 3.3343 10⁻⁶ s

the relationship between these times is

       t₂ / t₁ = 3.3343 / 3.3333333

       t₂ / t₁ = 1.00029

Answer:

Option (2) is correct.

The energy density of an electromagnetic wave is equally divided between the magnetic and electric fields.

Explanation:

An electromagnetic waves are the waves which are produced when the oscillating electric and magnetic field are interact each other perpendicular to each other. The direction of propagation of electro magnetic waves is perpendicular to each electric and magnetic fields.

The energy associated with the electromagnetic waves is equally distributed in form of electric and magnetic fields.

So, the correct option is (2).

The energy density is equally distributed among the magnetic field and electric field. Hence, option (2) is correct.

The given problem is based on the concept and fundamentals of electromagnetic waves.  The waves created as a result of vibrations between an electric field and a magnetic field is known as Electromagnetic waves.

In other words, an electromagnetic waves are the waves which are produced when the oscillating electric and magnetic field are interact each other perpendicular to each other. The direction of propagation of electro magnetic waves is perpendicular to each electric and magnetic fields.

Also, the energy associated with the electromagnetic waves is equally distributed in form of electric and magnetic fields. So, the energy density of an electromagnetic wave is equally divided between the magnetic and electric fields.

Thus, we can conclude that the energy density is equally distributed among the magnetic field and electric field.

Learn more about the electromagnetic waves here:

https://brainly.com/question/25559554

Answer:

The viscosity is 1.30 x 10^-3 deca poise.

Explanation:

Volume per minute, V = 10^-5 m^3

Volume per second, V = 1.67 x 10^-7 m^3

density, d = 800 kg/m^3

radius, r = 0.05 cm

Length, L = 0.5 m

Height, h = 60 cm

Pressure, P = h d g = 0.6 x 800 x 9.8 = 4704 Pa

Use the formula  of rate of flow

[tex]V = \frac{\pi p r^4}{8\eta L}\\\\1.67\times 10^{-7}\times8\times \eta\times 0.5 = 3.14\times 4707\times (0.05\times 10^{-2})^4\\\\\eta = 1.38\times 10^{-3} deca poise[/tex]

Answer:

Their measured results are closer to the exact or true value. Hence, their measured value is considered to be more accurate.

Explanation:

Considering the situation described above, the accuracy of a measured value depicts how closely a measured value is to the accurate value.

Hence, since the students' measured values have very low percent differences, it shows the similarity of computations or estimates to the actual values, which in turn offers a smaller measurement error.

Therefore, their measured results are closer to the exact or true value, which implies that their measured value is considered to be more accurate.

Answer:

A) C_{eq} = 15 10⁻⁶  F,  B)   U₃ = 3 J,  U₄ = 0.5 J

Explanation:

In a complicated circuit, the method of solving them is to work the circuit in pairs, finding the equivalent capacitance to reduce the circuit to simpler forms.

In this case let's start by finding the equivalent capacitance.

A) Let's solve the part where C1 and C3 are. These two capacitors are in serious

         [tex]\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_3}[/tex]            (you has an mistake in the formula)

         [tex]\frac{1}{C_{eq1}} = (\frac{1}{30} + \frac{1}{15}) \ 10^{6}[/tex]

         [tex]\frac{1}{C_{eq1}}[/tex] = 0.1   10⁶

         [tex]C_{eq1}[/tex] = 10 10⁻⁶ F

capacitors C₂, C₄ and C₅ are in series

          [tex]\frac{1}{C_{eq2}} = \frac{1}{C_2} + \frac{1}{C_4} + \frac{1}{C_5}[/tex]

          [tex]\frac{1}{C_{eq2} } = (\frac{1}{15} + \frac{1}{30} + \frac{1}{10} ) \ 10^6[/tex]

          [tex]\frac{1}{C_{eq2} }[/tex] = 0.2 10⁶

          [tex]C_{eq2}[/tex] = 5 10⁻⁶ F

the two equivalent capacitors are in parallel therefore

          C_{eq} = C_{eq1} + C_{eq2}

          C_{eq} = (10 + 5) 10⁻⁶

          C_{eq} = 15 10⁻⁶  F

B) the energy stored in C₃

The charge on the parallel voltage is constant

is the sum of the charge on each branch

         Q = C_{eq} V

         Q = 15 10⁻⁶ 6

         Q = 90 10⁻⁶ C

the charge on each branch is

         Q₁ = Ceq1 V

         Q₁ = 10 10⁻⁶ 6

          Q₁ = 60 10⁻⁶ C

         Q₂ = C_{eq2} V

         Q₂ = 5 10⁻⁶ 6

         Q₂ = 30 10⁻⁶ C

now let's analyze the load on each branch

Branch C₁ and C₃

In series combination the charge is constant    Q = Q₁ = Q₃

          U₃ = [tex]\frac{Q^2}{2 C_3}[/tex]

          U₃ =[tex]\frac{ 60 \ 10^{-6}}{2 \ 10 \ 10^{-6}}[/tex]

          U₃ = 3 J

In Branch C₂, C₄, C₅

since the capacitors are in series the charge is constant Q = Q₂ = Q₄ = Q₅

          U₄ = [tex]\frac{30 \ 10^{-6}}{ 2 \ 30 \ 10^{-6}}[/tex]

          U₄ = 0.5 J

Explanation:

Given that,

Two resistors 4.5 Ω and 2.3 Ω .

Potential difference = 30 V

When they are in series, the current through each resistor remains the same. First find the equivalent resistance.

R' = 4.5 + 2.3

= 6.8 Ω

Current,

[tex]I=\dfrac{V}{R'}\\\\I=\dfrac{30}{6.8}\\\\=4.41\ A[/tex]

So, the current through both lightbulb is the same i.e. 4.41 A.

When they are in parallel, the current divides.

Current flowing in 4.5 resistor,

[tex]I_1=\dfrac{V}{R_1}\\\\=\dfrac{30}{4.5}\\\\I_1=6.7\ A[/tex]

Current flowing in 2.3 ohm resistor,

[tex]I_2=\dfrac{V}{R_2}\\\\=\dfrac{30}{2.3}\\\\I_2=13.04[/tex]

In parallel combination, are brighter than bulbs in series.

Answer:

I think it is of science is it true na i knew it bro dont take tension

Answer:

A. 16 J

B. 16 J

C. 8 N

Explanation:

A. Determination of the kinetic energy.

Mass (m) = 0.5 Kg

Velocity (v) =. 8 m/s

Kinetic energy (KE) =?

KE = ½mv²

KE = ½ × 0.5 × 8²

KE = ½ × 0.5 × 64

KE = 0.5 × 32

KE = 16 J

B. Determination of the Workdone by the force.

Kinetic energy (KE) = 16 J

Workdone =.?

Workdone and kinetic energy has the same unit of measurement. Thus,

Workdone = kinetic energy

Workdone = 16 J

C. Determination of the force.

Workdone (Wd) = 16 J

Displacement (s) = 2 m

Force (F) =?

Wd = F × s

16 = F × 2

Divide both side by 2

F = 16 / 2

F = 8 N

Answer:

杰恩斯克克斯克奇沃伊斯克克斯克什德布德克什恩克恩德恩克恩茨克杰兹姆克斯恩斯姆斯姆德恩德姆德武伊乔奥斯克斯杰德布德赫德夫赫富伊什杰吉迪赫德赫夫赫德

Answer:

Scientists learn more about the natural world through investigation by identifying a problem, researching related information, designing and conducting an investigation, analyzing the results, evaluating the conclusion, and communicating the findings. Engineers follow similar steps when creating new products or solutions through technological design. The four stages of technological design include identifying a need, designing and implementing a solution, and evaluating the solution.

Answer:

(a) The area vector is 0.00503 m² at 30⁰ from the magnetic field

(b) The induced emf is 12.58 V

Explanation:

Given;

angle between the magnetic field and the plane of the circular coil, = 30⁰

number of turns of the coil, N = 1000

radius of the coil, r = 4 cm = 0.04 m

change in the magnetic field with time, dB/dt = 5 T/s

(a) The area vector is calculated as;

A = πr²

A = π x (0.04)²

A = 0.00503 m²

The area vector is 0.00503 m² at 30⁰ from the magnetic field.

(b) The induced emf is calculated as;

[tex]emf = N\frac{\Delta \phi}{\Delta t} \\\\where;\\\\\phi = BAcos \theta\\\\emf = N\times \frac{dB}{dt} \times Acos (\theta)\\\\where;\\\\\theta \ is \ the \ angle \ between \ a \ perpendicular \ vector \ to \ the \ area\\ and\ the \ magnetic\ field\\\\\theta = 90 - 30 = 60^0\\\\emf = N\times \frac{dB}{dt} \times Acos (\theta)\\\\emf = (1000) \times (5 )\times (0.00503) \times cos (60)\\\\emf = 12.58 \ V[/tex]

Answer: [tex]377.3\ kPa[/tex]

Explanation:

Given

Water column height [tex]h=25\ cm[/tex]

After oil is poured, the total height becomes [tex]h'=40\ cm[/tex]

Pressure at the bottom will be the sum due to the water and oil column

Suppose the density of the oil is [tex]\rho=900\ kg/m^3[/tex]

Pressure at the bottom

[tex]\Rightarrow P=10^3\times g\times 25+900\times g\times 15\\\Rightarrow P=100g[250+135]\\\Rightarrow P=3773\times 100\ Pa\\\Rightarrow P=377.3\ kPa[/tex]

Answer:

B. according to Newton's Third Law of Motion, the force of the Moon on the Earth and the force of the Earth on the Moon are equal in magnitude and opposite in direction

Answer:

Explanation:

The fish will have horizontal velocity of 2.6 m/s which is also the velocity of eagle. Additionally , he will have vertical velocity due to fall under gravity .

v² = u² + 2 g H .

v² = 0  + 2 x 9.8 x 4.7 m

= 92.12

v = 9.6 m /s

The fish's final velocity will have two components

vertical component = 9.6 m/s downwards

Horizontal component = 2.6 m /s  .

Resultant velocity = √ ( 9.6² + 2.6² )

= √ ( 92.16 + 6.76 )

= 9.9 m /s

Answer:

The speed of fish at the time of hitting the surface is 9.95 m/s.

Explanation:

Horizontal speed, u = 2.6 m/s

height, h = 4.7 m

Let the vertical velocity at the time of hitting to water is v.

Use third equation of motion

[tex]v^2 = u^2 - 2 gh \\\\v^2 = 0 + 2 \times 9.8\times 4.7\\\\v = 9.6 m/s[/tex]

The net velocity with which the fish strikes to the water is

[tex]v' = \sqrt{9.6^2 + 2.6^2 }\\\\v' = 9.95 m/s[/tex]

Answer:

C

Explanation:

Similarity

Answer:

A teacher is more important than a famer.

Explanation:

A teacher is more important than a famer because the knowledge of farming is gotten through the teacher. Thus, without a teacher; whether formal or informal, there cannot be farming, let alone farmers.

Answer:

a.work done by man A is zero

D is best option

Explanation:

D none of them because as we know that to do work some distance should be cover with some load as both picture they are covering some distance with carrying some load so A and B option are absolutely wrong and remaining C they are not doing same amount of work because distance cover by them and load carrying by them is different so how can work be same so D is best option none of A B C is correct answer

Answer:

we all are the human being we all dont no the all of 5he answer dont take tension beacause other one will give your answer

Answer:

pha của dao động là hàm bậc nhất của thời gian.

Answer:

c). It wouldn't change.

Explanation:

[tex]{ \bf{F = \frac{ \ \gamma _{o}NI }{l} }}[/tex]

Answer:

Current = 0.000294 A

Explanation:

Below is the given values:

Given the charge = 9.00 C

Time = 8.50 h

Use the below formula to find the current:

Current = Q / t

Now plug the values:

Current = 9 / (8.5 x 3600)

Current = 0.000294 A

Answer:

The correct answer is - low pitch

Explanation:

Now for the case it is mentioned that the tube closed on one end frequency is:

f = v/2l

Where,

l = length of the tube

v = velocity of longitudinal wave of gas filled in the tube

if frequency increases then pitch will be increase as well as pitch depends on frequency.

Now increase with the temperature the density of the gas decreases and velocity v is inversely proportional to density of gas so velocity increases. So if there is an increase in frequency so pitch also increases.

As the temperature inside the house is at 750 F more than outsideat 450 Fso pitch is more inside and the pitch is low outside.

This question is incomplete, the complete question is;

A 70-turn coil has a diameter of 11 cm. Find the magnitude of the emf induced in the coil (in V) if it is placed in a spatially uniform magnetic field of magnitude 0.70 T so that the face of the coil makes the following angles with the magnetic field, and the magnetic field is reduced to zero uniformly in 0.2 s. a) θ = 30° V b) θ = 60° V c) θ = 90° V

Answer:

the magnitude of the emf induced in the coil are;

a)- For θ = 30°, e = 1.16 V

b)- For θ = 60°, e = 2.01 V

c)- For θ = 90°, e = 2.33 V

Explanation:

Given the data in the question;

number of turns N = 70

Diameter of coil D = 11 cm

Radius r = D/2 = 11/2 = 5.5 cm = 0.055 m

magnitude of magnetic ΔB = 0.7T

Δt = 0.2 seconds

Now,

a)

For θ = 30°,

Angle of with area of vector θ' = 90° - 30° = 60°

so

emf  e = N( Δ∅ / Δt ) = N( ΔBAcosθ  / Δt )

hence

e =  NAcosθ'(ΔB / Δt )

where A is area ( πr² )

so we substitute

e =  70 × πr² × cos(60°) × ( 0.7 / 0.2 )

e =  70 × π(0.055)² × cos(60°) × ( 0.7 / 0.2 )

e = 1.16 V

b)

For θ = 60°,

Angle of with area of vector θ' = 90° - 60° = 30°

so

e =  NAcosθ'(ΔB / Δt )

we substitute

e =  70 × πr² × cos(30°) × ( 0.7 / 0.2 )

e =  70 × π(0.055)² × cos(30°) × ( 0.7 / 0.2 )

e = 2.01 V

c)

For θ = 90°,

Angle of with area of vector θ' = 90° - 90° = 0°

so

e =  NAcosθ'(ΔB / Δt )

we substitute

e =  70 × πr² × cos(0°) × ( 0.7 / 0.2 )

e =  70 × π(0.055)² × cos(30°) × ( 0.7 / 0.2 )

e = 2.33 V

Therefore, the magnitude of the emf induced in the coil are;

a)- For θ = 30°, e = 1.16 V

b)- For θ = 60°, e = 2.01 V

c)- For θ = 90°, e = 2.33 V

Answer: [tex]1.79\ rad/s^2[/tex]

Explanation:

Given

Initial angular speed is [tex]\omega_1=26.2\ rad/s[/tex]

Final angular speed is [tex]\omega_2=36.5\ rad/s[/tex]

Time period [tex]t=5.75\ s[/tex]

Magnitude of the fan's acceleration is given by

[tex]\Rightarrow \alpha=\dfrac{\omega_2-\omega_1}{t}[/tex]

Insert the values

[tex]\Rightarrow \alpha=\dfrac{36.5-26.2}{5.75}\\\\\Rightarrow \alpha=\dfrac{10.3}{5.75}\\\\\Rightarrow \alpha=1.79\ rad/s^2[/tex]

Thus, fan angular acceleration is [tex]1.79\ rad/s^2[/tex]

Answer:

The angular acceleration is given by 1.8 rad/s^2.

Explanation:

initial angular speed, wo = 26.2 rad/s

final angular velocity, w = 36.5 rad/s

time, t = 5.75 seconds

The first equation of motion is

[tex]w = wo + \alpha t\\\\36.5 = 26.2 + 5.75\alpha\\\\\alpha = 1.8 rad/s^2[/tex]

Answer:

b

Explanation:

Calculate force of each car:

P1 = 900 kg x 12m/s = 10,800

P2= 600 x 24 = 14,400

Degree of travel = arctan(14,300/10800)

Degree of travel = 53.1 N of E

Answer:

The correct solution is "122.2211".

Explanation:

Given:

deceleration,

a = 22 ft/sec²

Initial velocity,

[tex]V_i=50 \ m/h[/tex]

Now,

[tex]V_i=50 \ m/h\times 5280 \ ft/m\times hr/3600 \ s[/tex]

    [tex]=73.333 \ ft/sec[/tex]

Now,

Final velocity,

[tex]V_f=0[/tex]

Initial velocity,

[tex]V_{initial} = 73.333 \ ft/sec[/tex]

hence,

⇒ [tex]V_f^2=V_i^2+2aD[/tex]

By putting the values, we get

      [tex]0=(73.333)^2+2\times( -22) D[/tex]

  [tex]44D=(73.333)^2[/tex]

      [tex]D=\frac{(73.333)^2}{44}[/tex]

          [tex]=122.2211[/tex]

Answer:

no parallel is the correct answer

105 km/h ≈ 29.2 m/s

60.0 km/h ≈ 16.7 m/s

Before the collision the test car has an acceleration a of

a = (29.2 m/s - 0) / (8.43 s) ≈ 3.46 m/s²

During the collision, the car is slowed to about 16.7 m/s, so that its (average) acceleration is

a = (16.7 m/s - 29.2 m/s) / (0.635 s) ≈ -19.7 m/s²

i.e. with magnitude about 19.7 m/s².

Overall, the car has an average acceleration of

a = (16.7 m/s - 0) / (8.43 s + 0.635 s) ≈ 1.84 m/s²

Answer:

T₁ = 232.5 ºC

Explanation:

For this exercise let's start by finding the value of the resistance for the two currents, using Ohm's law

           V = i R

            R = V / i

i₀ = 1.28 A

            R₀ = 120 / 1.28

            R₀ = 93.75 ohm

i₁ = 1.229 A

             R₁ = 120 / 1.229

             R₁ = 97.64 or

Resistance in a metal is linear with temperature

            ΔR = α R₀ ΔT

where the coefficient of thermal expansion for Nichrome is α=0.0002 C⁻¹

            ΔT = [tex]\frac{\Delta R}{\alpha R_o}[/tex]

            ΔT = [tex]\frac{97.64 \ -93.75}{ 0.00020 \ 93.75}[/tex]

            ΔT = 2,075 10² C

            ΔT = T₁-T₀ = 2,075 10²

            T₁ = T₀ + 207.5

             T₁ = 25+ 207.5

             T₁ = 232.5 ºC