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3 years ago

“What is the relationship between wavelength and frequency?”

Physics

1 answer:

Kisachek [45]3 years ago

7 0

Answer:

They are inversely proportional. If the frequency goes up, the wavelength will go down

Explanation:

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An infinite line of charge with linear density λ1 = 6 μC/m is positioned along the axis of a thick insulating shell of inner rad

Anna11 [10]

Answer: λ2= 2.34 * 10^-6 C/m

Explanation: In order to calculate the value of the linear charge density of the insulating shell we have to multiply ρ* Volume of the hollow cylinder, so

Volume of cylinder:2*π*b*L *(b-a) where (b-a) is the thickness, then

λ2=Q/L = 634 *10^-6 C/m^3* 2*π*0.042 m*(0.042-0.26)== 2.34 μ C/m

5 0

3 years ago

At what point does the external energy enter the system?

Phoenix [80]

The correct answer as the first one above !

8 0

3 years ago

Beats are the difference in (a) Frequency (b) Amplitude (c) Intensity (d) None

sasho [114]

Answer:

Beats are the difference in frequency.

(a) is correct option.

Explanation:

Beat :

Beat is the difference of the frequency of two waves.

The difference in frequency is equal to the number of beat per second.

Amplitude :

Amplitude of the wave is the maximum displacement.

Frequency :

Frequency is the number oscillations of wave in per second.

Intensity :

Intensity is the power per unit area.

Hence, Beats are the difference in frequency.

7 0

3 years ago

A reconnaissance plane flies 560 km away from its base at 602 m/s, then flies back to its base at 903 m/s.

IrinaVladis [17]

Answer:

Approximately $722\; \rm m\cdot s^{-1}$ .

Explanation:

The average speed of a vehicle is calculated as:

$\displaystyle \text{average speed} = \frac{\text{total distance}}{\text{total time}}$ .

In this question, the total distance is $2 \times 560\; \rm km = 1120\; \rm km$ .

The unit of the speeds in this question is meters per second, while the unit of distance is kilometers. Convert the unit of distance to meters:

$560 \; \rm km = 560 \times 10^{3} \; \rm m = 5.6 \times 10^{5}\; \rm m$ .

$1120 \; \rm km = 1120 \times 10^{3} \; \rm m = 1.12 \times 10^{6}\; \rm m$ .

Time required for the first part of this trip:

$\displaystyle \frac{5.60 \times 10^{5}\; \rm m}{602\; \rm m\cdot s^{-1}} \approx 930\; \rm s$ .

Time required for the second part of this trip:

$\displaystyle \frac{5.60 \times 10^{5}\; \rm m}{903\; \rm m\cdot s^{-1}} \approx 620\; \rm s$ .

The time required for the entire trip would be approximately $930 + 620 = 1550\; \rm s$ .

Calculate the average speed of this plane:

$\begin{aligned} \text{average speed} &= \frac{\text{total distance}}{\text{total time}} \\ &\approx \frac{1.12\times 10^{6}\; \rm m}{1550\; \rm s} \approx 722\; \rm m \cdot s^{-1}\end{aligned}$ .

6 0

2 years ago

A 12 kg object speeds up from an initial velocity of 10 m:s-1

amid [387]

Momentum = m • v

Original momentum = m • 10 m/s north

Final momentum = m • 15 m/s north

Change = m • (15 - 10) m/s north

Change = m • +5 m/s north

Change = +60 kg-m/s north

5 0

3 years ago