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3 years ago

If the amplitude of the waves coming from a light bulb decreased, which would you expect to happen?

2 answers:

4 0

The amplitude of the electromagnetic waves coming from
a light bulb is exactly an indication of the energy carried by
the waves.

If their amplitude decreases, then the waves are carrying less
energy. The brightness of the light decreases, AND the bulb
runs cooler, because it produces less heat.

elena55 [62]3 years ago

3 0

C) The brightness of the bulb would decrease. Hope I helped!

You might be interested in

the length of iron rod at 100 C is 300.36 cm and at 159 C is 300.54 cm.Calculate its length at 0 c and coefficient of linear exp

Ugo [173]

Answer:

The length at 0 °C is 300.05 cm

Coefficient of linear expansion of iron is 1.02×10¯⁵ C¯¹

Explanation:

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

Length (L₁) at 100 °C = 300.36 cm

Temperature 1 (θ₁) = 100 °C

Length (L₂) at 159 °C = 300.54 cm

Temperature 2 (θ₂) = 159 °C

Length (L₀) at 0 °C =?

Coefficient of linear expansion (α) =?

L₁ = L₀ (1 + θ₁α)

300.36 = L₀ (1 + 100α) ....(1)

L₂ = L₀ (1 + θ₂α)

300.54 = L₀ (1 + 159α) ..... (2)

Divide equation (2) by (1)

300.54 / 300.36 = L₀ (1 + 159α) / L₀ (1 + 100α)

1.0006 = (1 + 159α) / (1 + 100α)

Cross multiply

1.0006 (1 + 100α) = (1 + 159α)

1.0006 + 100.06α = 1 + 159α

Collect like terms

1.0006 – 1 = 159α – 100.06α

0.0006 = 58.94α

Divide both side by 58.94

α = 0.0006 / 58.94

α = 1.02×10¯⁵ C¯¹

Substitute the value of α into anything of the equation to obtain L₀. Here we shall use equation (2).

300.54 = L₀ (1 + 159α)

α = 1.02×10¯⁵ C¯¹

300.54 = L₀ (1 + 159 ×1.02×10¯⁵)

300.54 = L₀ (1 + 0.0016218)

300.54 = L₀ (1.0016218)

Divide both side by 1.0016

L₀ = 300.54 / 1.0016

L₀ = 300.05 cm

Summary:

The length at 0 °C is 300.05 cm

Coefficient of linear expansion of iron is 1.02×10¯⁵ C¯¹

6 0

3 years ago

For the material in the previous question that yields at 200 MPa, what is the maximum mass, in kg, that a cylindrical bar with d

Sedaia [141]

Answer:

The maximum mass the bar can support without yielding = 32408.26 kg

Explanation:

Yield stress of the material ( $\sigma$ ) = 200 M Pa

Diameter of the bar = 4.5 cm = 45 mm

We know that yield stress of the bar is given by the formula

Yield Stress = $\frac{Maximum load}{Area of the bar}$

⇒ $\sigma$ = $\frac{P_{max} }{A}$ ---------------- (1)

⇒ Area of the bar (A) = $\frac{\pi}{4}$ × $D^{2}$

⇒ A = $\frac{\pi}{4}$ × $45^{2}$

⇒ A = 1589.625 $mm^{2}$

Put all the values in equation (1) we get

⇒ $P_{max}$ = 200 × 1589.625

⇒ $P_{max}$ = 317925 N

In this bar the $P_{max}$ is equal to the weight of the bar.

⇒ $P_{max}$ = $M_{max}$ × g

Where $M_{max}$ is the maximum mass the bar can support.

⇒ $M_{max}$ = $\frac{P_{max} }{g}$

Put all the values in the above formula we get

⇒ $M_{max}$ = $\frac{317925}{9.81}$

⇒ $M_{max}$ = 32408.26 Kg

There fore the maximum mass the bar can support without yielding = 32408.26 kg

3 0

3 years ago

Drag the tiles to the correct boxes to complete the pairs.

Brut [27]

Answer:

5 percent = normal matter

68 percent = dark energy

27 percent = dark matter

Explanation:

3 0

2 years ago

A bar magnet is dropped toward a conducting ring lying on the floor. As the magnet falls toward the ring, does it move as a free

Maslowich

Consider that the bar magnet has a magnetic field that is acting around it, which will imply that there is a change in the magnetic flux through the loop whenever it moves towards the conducting loop. This could be described as an induction of the electromotive Force in the circuit from Faraday's law.

In turn by Lenz's law, said electromotive force opposes the change in the magnetic flux of the circuit. Therefore, there is a force that opposes the movement of the bar magnet through the conductor loop. Therefore, the bar magnet does not suffer free fall motion.

The bar magnet does not move as a freely falling object.

3 0

3 years ago

A nylon guitar string is fixed between two lab posts 2.00 m apart. The string has a linear mass density of μ=7.20 g/m\mu=7.20~\t

vladimir2022 [97]

Answer:

4.6 m

Explanation:

First of all, we can find the frequency of the wave in the string with the formula:

$f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}$