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3 years ago

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The wavelength of an electromagnetic waves is __________ also known as the period. the number of waves that pass a given point i

n space. the negative of the frequency. the spatial distance between two successive troughs. all of these are true.

1 answer:

Andreas93 [3]3 years ago

6 0

The sentence can be completed as follows:

The wavelength of an electromagnetic waves is the spatial distance between two successive troughs.

Note that the wavelength of a wave can also measured as the spatial distance between two successive crests of the wave. Also note that the second part of the sentence ("also known as the period") is not true, because period is another thing (in fact, the period is the time interval between two successive troughs).

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Two forces Upper FSubscript Upper A Baseline Overscript right-arrow EndScripts and Upper FSubscript Upper B Baseline Overscript

Pavel [41]

Answer:

Part a)

$F_A = 4.59 N$

Part B)

$F_B = 1.28 N$

Explanation:

As we know that when both the forces are acting on the object in same direction then we will have

$F_A + F_B = ma$

as we know that

$a = 0.554 m/s^2$

m = 10.6 kg

now we will have

$F_A + F_B = 10.6(0.554)$

$F_A + F_B = 5.87 N$

Now two forces are in opposite direction then we have

$F_A - F_B = 10.6(0.313)$

$F_A - F_B = 3.32 N$

Part A)

Now we will have from above two equation

$F_A = 4.59 N$

Part B)

Similarly for other force we have

$F_B = 1.28 N$

5 0

2 years ago

Two protons are released from rest when they are 0.720 nm apart. For related problem-solving tips and strategies, you may want t

Gnesinka [82]

Answer:

a) Speed of the electrons at maximum speed = (1.384 × 10⁴) m/s

The maximum speed occurs at the point where all of the initial potential energy is converted into kinetic energy.

b) Maximum acceleration of the protons = (2.660 × 10¹⁷) m/s²

The maximum acceleration occurs at the minimum distance apart for the two protons.

Explanation:

The maximum speed occurs when all the potential energy of the protons has been converted to kinetic energy.

The potential energy between the two protons at the instant of release is given by

U = (kq₁q₂/r)

k = Coulomb' s constant = (8.988 × 10⁹) Nm²/C²

q₁ = q₂ = charge on a proton = q = (1.602 × 10⁻¹⁹) C

r = separation between the two protons = 0.72 nm = (7.2 × 10⁻¹⁰) m

U = (kq²/r) = [(8.988 × 10⁹) × (1.602 × 10⁻¹⁹)²] ÷ (7.2 × 10⁻¹⁰) = (3.204 × 10⁻¹⁹) N/m or Joules

At the maximum speeds, the two protons will not possess any potential Energy, only kinetic energy.

The sum of kinetic and potential energies is always constant for the system

(Initial Kinetic Energy) + (Initial Potential Energy) = (Kinetic Energy at maximum speed) + (Potential Energy at maximum speed)

Initial Kinetic Energy of the system = 0 J (Since both protons were intially at rest)

Initial Potential Energy = (3.204 × 10⁻¹⁹) J

Kinetic Energy at maximum speed = Sum of the kinetic energies of the protons at this point = (½mv²) + (½mv²) = (mv²) J (Since theu are both protons, they have the same mass and the same speed at maximum speed)

Potential Energy at maximum speed = 0 J

0 + (3.204 × 10⁻¹⁹) = mv² + 0

mv² = (3.204 × 10⁻¹⁹)

m = mass of a proton = (1.673 × 10⁻²⁷) kg

v = speed of each of the protons at maximum speed = ?

v = √[(3.204 × 10⁻¹⁹) ÷ m]

v = √[(3.204 × 10⁻¹⁹) ÷ (1.673 × 10⁻²⁷)]

v = √(1.915 × 10⁸) = 13,838.8 m/s = (1.384 × 10⁴) m/s

b) Since the two protons repel each other and force of repulsion reduces as the dI stance between the protons increases, the maximum acceleration occurs at the minimum distance apart for the two protons.

Force of repulsion acting on each proton is given through Coulomb's law as

F = (kq₁q₂/r²)

And the force acting on each proton is obtainable using Newton's law that

F = ma

So, the acceleration of each proton at any time is obtainable through a relation of these 2 formulas.

ma = (kq₁q₂/r²)

a = (kq₁q₂/r²m)

k = Coulomb' s constant = (8.988 × 10⁹) Nm²/C²

q₁ = q₂ = charge on a proton = q = (1.602 × 10⁻¹⁹) C

r = separation between the two protons = 0.72 nm = (7.2 × 10⁻¹⁰) m

m = mass of a proton = (1.673 × 10⁻²⁷) kg

a = [(8.988 × 10⁹) × (1.602 × 10⁻¹⁹)²] ÷ [(7.2 × 10⁻¹⁰)² × (1.673 × 10⁻²⁷)]

a = (2.660 × 10¹⁷) m/s²

Hope this Helps!!!

5 0

2 years ago

A car traveling at 28 m/s starts to decelerate steadily. It comes to a complete stop in 13 seconds. What is its acceleration?

Maru [420]

Acceleration = (change in speed) / (time for the change)

change in speed = (speed at the end) minus (speed at the beginning)

change in speed = (zero) minus (28 m/s) = -28 m/s

Acceleration = (-28 m/s) / (13 sec)

Acceleration = -2.15 m/s²

5 0

3 years ago

Consider a 20 cm thick granite wall with a thermal conductivity of 2.79 W/m·K. The temperature of the left surface is held const

kozerog [31]

Answer:

The right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²

Explanation:

Thickness of the wall is L= 20cm = 0.2m

Thermal conductivity of the wall is K = 2.79 W/m·K

Temperature at the left side surface is T₁ = 50°C

Temperature of the air is T = 22°C

Convection heat transfer coefficient is h = 15 W/m2·K

Heat conduction process through wall is equal to the heat convection process so

$Q_{conduction} = Q_{convection}$

Expression for the heat conduction process is

$Q_{conduction} = \frac{K(T_1 - T)}{L}$

Expression for the heat convection process is

$Q_{convection} = h(T_2 - T)$

Substitute the expressions of conduction and convection in equation above

$Q_{conduction} = Q_{convection}$

$\frac{K(T_1 - T_2)}{L} = h(T_2 - T)$

Substitute the values in above equation

$\frac{2.79(50- T_2)}{0.2} = 15(T_2 - 22)\\\\T_2 = 35.5^\circC$

Now heat flux through the wall can be calculated as

$q_{flux} = Q_{conduction} \\\\q_{flux} = \frac{K(T_1 - T_2)}{L}\\\\q_{flux} = \frac{2.79(50 - 35.5)}{0.2}\\\\q_{flux} = 202.3W/m^2$

Thus, the right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²

6 0

2 years ago

A small branch is wedged under a 200 kg rock and rests on a smaller object. The smaller object is 2.0 m from the large rock and

Alexxandr [17]

Answer:

a

$F =326.7 \ N$

b

$M = 6$

Explanation:

From the question we are told that

The mass of the rock is $m_r = 200 \ kg$

The length of the small object from the rock is $d = 2 \ m$

The length of the small object from the branch $l = 12 \ m$

An image representing this lever set-up is shown on the first uploaded image

Here the small object acts as a fulcrum

The force exerted by the weight of the rock is mathematically evaluated as

$W = m_r * g$

substituting values

$W = 200 * 9.8$

$W = 1960 \ N$

So at equilibrium the sum of the moment about the fulcrum is mathematically represented as

$\sum M_f = F * cos \theta * l - W cos\theta * d = 0$

Here $\theta$ is very small so $cos\theta * l = l$

and $cos\theta * d = d$

Hence

$F * l - W * d = 0$

=> $F = \frac{W * d}{l}$

substituting values

$F = \frac{1960 * 2}{12}$

$F =326.7 \ N$

The mechanical advantage is mathematically evaluated as

$M = \frac{W}{F}$

substituting values

$M = \frac{1960}{326.7}$

$M = 6$

6 0

3 years ago