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1 year ago

What is Doppler effect

1 answer:

5 0

Doppler effect is the shift in frequency as the object gets near an object/person or far away from an object or person.

When the frequency of a police car is heard from far away, the frequency is low and long wave lengths
When the frequency of a police car is heard from near, the frequency is high and short-wave lengths

Hope it helps!

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HELP ME PLEASE!!!!!!!!!

Stolb23 [73]

As per the given Figure attached here we know that both charges q1 and q2 will apply same force on charge q3 and hence the resultant force due to both charges will be along Y axis vertically upwards

So here we know that

$F = \frac{kq_1q_3}{d_{13}^2}$

now from the above equation

$F = \frac{(9\times 10^9)(2\times 10^{-6})(4 \times 10^{-6})}{0.5^2}$

$F = 0.288 N$

so both of the charges will apply 0.288 N force on q3 charge along the line joining them

now the net force due to vector sum is given by

$F_{net} = 2Fcos\theta$

here we know that angle is

$\theta = 37 degree$

now we have

$F_{net} = 2\times 0.288 cos37$

$F_{net} = 0.46 N$

so net force on q3 is 0.46 N vertically upwards along +Y axis

6 0

3 years ago

Water flows through a cast steel pipe (k = 50 W m.K, ε = 0.8) with an outer diameter of 104mm and 2 mm wall thickness. Calculate

masha68 [24]

Answer:

The heat loss per unit length is $\frac{Q}{L} = 2981 W/m$

Explanation:

From the question we are told that

The outer diameter of the pipe is $d = 104mm = \frac{104}{1000} = 0.104 m$

The thickness is $D = 2mm = \frac{2}{1000} = 0.002m$

The temperature of water is $T = 90^oC = 90 + 273 = 363K$

The outside air temperature is $T_a = -10^oC = -10 +273 = 263K$

The water side heat transfer coefficient is $z_1 = 300 W/ m^2 \cdot K$

The heat transfer coefficient is $z_2 = 20 W/m^2 \cdot K$

The heat lost per unit length is mathematically represented as

$\frac{Q}{L} = \frac{2 \pi (T - Ta)}{ \frac{ln [\frac{d}{D} ]}{z_1} + \frac{ln [\frac{d}{D} ]}{z_2}}$

Substituting values

$\frac{Q}{L} = \frac{2 * 3.142 (363 - 263)}{ \frac{ln [\frac{0.104}{0.002} ]}{300} + \frac{ln [\frac{0.104}{0.002} ]}{20}}$

$\frac{Q}{L} = \frac{628}{0.2107}$

$\frac{Q}{L} = 2981 W/m$

6 0

3 years ago

2. What is the smallest particle into which an

just olya [345]

Explanation:

atommmmmmmmmmmmmmmmmmmmm

7 0

3 years ago

Read 2 more answers

A diver named Jacques observes a bubble of air rising from the bottom of a lake (where the absolute pressure is 3.50 atm) to the

Studentka2010 [4]

Answer:

A) the ratio of volumes of the bubble is Vs/Vb= 3.74

B) would not be safe since Vs/Vb== 3.5 and there is no lung capacity to store such amount of volume, and thus would generate an unsafe pressure over the lungs. would be better to release air accordingly to maintain safe conditions.

Explanation:

assuming the gas of the bubble behaves as ideal gas

P * V = n * R * T

where P= absolute pressure, V= volume occupied by the gas, n = number of moles , R = ideal gas constant , T = absolute temperature

if we assume that the mass of the bubble remains constant ( that is, it does not capture other bubbles during ascension of disaggregate into smaller ones and there is no mass transfer into the bubble due to diffusion)

inicial state) Pb * Vb = n * R * Tb

final state) Ps * Vs = n * R * Ts

dividing both equations

(Ps/Pb)(Vs/Vb) = Ts/Tb

therefore

Vs/Vb= (Ts/Tb) (Pb/Ps)

since Tb = 4°C = 277 K and Ts= 23°C = 296 K

Vs/Vb= (Ts/Tb) (Pb/Ps) = (296K/277K)*(3.5 atm/1 atm) = 3.74

B) if the T remains constant Ts=Tb and thus

Vs/Vb= (Ts/Tb) (Pb/Ps)= 1* (Pb/Ps) = 3.5 atm/1 atm = 3.5

it would not be safe since there is no lung capacity to store such amount of volume, and thus would generate an unsafe pressure over the lungs. would be better to release air accordingly to maintain safe conditions.

8 0

2 years ago

What is equivalent to 21/12

Harrizon [31]

21/12 is equal to 1 3/4 hope this helps!

7 0

3 years ago

Read 2 more answers

What is Doppler effect​

What is Doppler effect