Question 2 of 10

Urgent!!!!!! 20 points!!!!!! how are the speed, wavelength and frequency of a wave related

zheka24 [161]

Answer:

The higher the frequency, the shorter the wavelength. Because all light waves move through a vacuum at the same speed, the number of wave crests passing by a given point in one second depends on the wavelength. Speed shows how long it takes for wavelengths to travel.

6 0

3 years ago

An isotope undergoes radioactive decay by emitting radiation that has a –1 charge. What other characteristic does the radiation

weeeeeb [17]

Answer: The radiation emitted will have negligible mass number.

Explanation:

Radioactive decay is defined as the process in which an unstable nuclei breaks down into stable nuclei via various methods.

An isotope undergoes a radioactive decay to attain stability.

There are three types of decay process, but the process in which the emitted radiation carries a charge of -1 is beta decay.

Beta decay is defined as the decay process in which a neutron gets converted to a proton and an electron. In this decay process, beta particle is emitted. The emitted particle carries a charge of -1 units and has a mass of 0 units. The released beta particle is also known as electron.

$_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta$

Hence, the radiation emitted will have negligible mass number.

3 0

3 years ago

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Which of the following statements is true?

kupik [55]

Answer:

2. The hydrogen atom has quantized energy levels.

Explanation:

The Bohr model of the atom states that the structure of the atom is quantized, that is, that electrons can only orbit the nucleus in specific orbits with a fixed radius. Therefore, the electron cannot be in energy levels that do not correspond to these quantized levels.

4 0

4 years ago

You are working as a summer intern for the Illinois Department of Natural Resources (DNR) and are assigned to some initial work

hammer [34]

Answer:

1) F = 71.6 N , 2) F = 120.2 N , 3) P_m= 68.600 Pa, 4) V = 2.4210-5 m³

Explanation:

This is a problem of fluid mechanics, to find the force we must use its definition

P = F / A

F = P A

The area of the circular pipe is

A = π r² = π d²/4

The pressure is given by the expression

P = P_atm + ρ g h

1) the force on the outer side is

P = P_atm

we substitute in the expression of force

F = P_atm π d² / 4

let's calculate

F = 1,013 10⁵ π 0.03²/4

F = 7.16 10¹ N

F = 71.6 N

2) the force on the inner side

the pressure

P = P_atm + ρ g h

P = 1,015 10⁵ + 1,000 9.8 7

P = 1,701 10⁵ Pa

F = 1,701 10⁵ π 0.03² / 4

F = 1,202 10²

F = 120.2 N

3) manometric pressure

Pm = ρ g h

P m = 1000 9.8 7

P_m= 68.600 Pa

4) In this part they ask for the volume that comes out in time t= 3 h

to calculate this volume we can use the flow ratio

Q = A v

V t = A v

V = A v / t

sent is the velocity of the water that comes out, to calculate it we use the Bernoulli equation

we will use index 1 for the lake surface and ionice 2 apa the position of the plug

P₁ + ρ g v₁² + ρ g y₁ = P₂ + ρ g v₂² + ρ g y₂

As the lake has much more capacity than the pipeline, the velocity of the surface of the lake is peeling, in this case we approximate it steel

(P₁-P₂) + ρ G (y₁ -y₂) = ρ g v₂²

1000 9.8 v₂² = ρ g h + 1000 9.8 (7-0)

9800 v₂² = 1000 9.8 7 + 68600

v₂ = √ (137200)

v₂ = 370.4 m / s

t = 3 h (3600s / h) = 10800 s

we substitute in the volume equation

V = π d²/4 370.4 / 10800

V = 2.4210-5 m³

4 0

3 years ago

A 0.350-m-long cylindrical capacitor consists of a solid conducting core with a radius of 1.20 mm and an outer hollow conducting

Brut [27]

Answer: a. 6.52*EXP{-10}C/m. b. 2.28*EXP{-10}C. c. 38picoFarad.

d. 6.84*EXP{-10}joules

Explanation: We first calculate for the capacitance first. For a concetric cylinder with two radius R1 and R2 the capacitance C is given as

C= {2*pi*permitivity of

freespace*lenght}/In(R2/R1) from the given Question R2 is 2mm and R1 is 1.2mm, lenght is 0.35meter, permitivity of free space is 8.85*EXP {-12} and pi is 3.142.

Therefore Capacitance would be,

C = 2*3.142*8.85*EXP

{-12}*0.35/In(2/1.2)

C = 3.8*EXP {-11} Farad which is also

38*EXP {-12} Farad or 38picoFarad.

Next, we solve for our total charge Q. Charge, capacitance and voltage are related by

Q = C*V = 38*EXP {-12}*6

=2.28*EXP {-10} Coulombs

Next, we obtian charge per unit lenght, which is

Q/L = 2.28*EXP {-10}/0.35

= 6.52*EXP {-10} Coulombs/meter

Next, we obtain the energy stored in the capacitor from

Energy stored = 1/2*(C*V²)

=1/2*38*EXP {-12}*6²

=6.84*EXP {-10} Joules

Note: EXP means 10^

4 0

3 years ago

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