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Ymorist [56]
2 years ago
11

Two swift canaries fly toward each other, each moving at 15.0 m/s relative to the ground, each warbling a note of frequency 1750

Hz. (a) What frequency note does each bird hear from the other one? (b) What wavelength will each canary measure for the note from the other one?
Physics
1 answer:
lesya692 [45]2 years ago
3 0

Answer:

Part a)

f = 1911.5 Hz

Part b)

\lambda = 0.186 m

Explanation:

Here the source and observer both are moving towards each other

so we know that the apparent frequency is given as

f' = f_0 (\frac{v + v_o}{v - v_s})

here we know that

f_0 = 1750 Hz

v_o = 15 m/s

v_s = 15 m/s

now we will have

f' = (1750)(\frac{340 + 15}{340 - 15})

f' = 1911.5 Hz

Part b)

Apparent wavelength is given by the formula

\lambda = \frac{v_{relative}}{f_{app}}

here we will have

\lambda = \frac{340 + 15}{1911.5}

\lambda = 0.186 m

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A proton, starting from rest, accelerates through a potential difference of 1.0 kV and then moves into a magnetic field of 0.040
adell [148]

Answer:

r = 0.114 m

Explanation:

To find the speed of the proton, from conservation of energy, we know that

KE = PE

Thus, we have;

(1/2)mv² = qV

Where;

V is potential difference = 1kv = 1000V

q is charge on proton which has a value of 1.6 x 10^(-19) C

m is mass of proton with a constant value of 1.67 x 10^(-27) kg

Let's make the velocity v the subject;

v =√(2qV/m)

v = √(2•1.6 x 10^(-19)•1000)/(1.67 x 10^(-27))

v = 4.377 x 10^(5) m/s

Now to calculate the radius of the circular motion of charge we know that;

F = mv²/r = qvB

Thus, mv²/r = qvB

Divide both sides by v;

mv/r = qB

Thus, r = mv/qB

Value of B from question is 0.04T

Thus,

r = (1.67 x 10^(-27) x 4.377 x 10^(5))/(1.6 x 10^(-19) x 0.04)

r = 0.114 m

r = 8.76 m

8 0
1 year ago
Find the centripetal force needed by a 1275 kg car to make a turn of radius 40.0 m at a speed of 25.0 km/h
natima [27]
<span>v = 25.0 km/</span><span>h = 25*5/18 m/s = 6.94 m/s

</span><span>centripetal force = mv²/r = 1275*6.94²/40 = 1537.18 N </span>

4 0
2 years ago
Read 2 more answers
A student attaches a block to a vertical spring so that the block-spring system will oscillate if the block-spring system is rel
vodka [1.7K]

Answer:

Time period of the motion will remain the same while the amplitude of the motion will change

Explanation:

As we know that time period of oscillation of spring block system is given as

T= 2\pi\sqrt{\frac{M}{k}}

now we know that

M = mass of the object

k = spring constant

So here we know that the time period is independent of the gravity

while the maximum displacement of the spring from its mean position will depends on the gravity as

mg = kx

x = \frac{mg}{k}

so we can say that

Time period of the motion will remain the same while the amplitude of the motion will change

4 0
2 years ago
A sound technician is testing the sound acoustics in a theatre for an upcoming music concert. He sets up speakers in different l
velikii [3]
Consertive alex dggghh
4 0
2 years ago
When two resistors are wired in series with a 12 V battery, the current through the battery is 0.33 A. When they are wired in pa
MA_775_DIABLO [31]

Answer:

If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

Explanation:

R₁ = Resistance of first resistor

R₂ = Resistance of second resistor

V = Voltage of battery = 12 V

I = Current = 0.33 A (series)

I = Current = 1.6 A (parallel)

In series

\text{Equivalent resistance}=R_{eq}=R_1+R_2\\\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{0.33}\\\Rightarrow R_1+R_2=36.36\\ Also\ R_1=36.36-R_2

In parallel

\text{Equivalent resistance}=\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}\\\Rightarrow {R_{eq}=\frac{R_1R_2}{R_1+R_2}

\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{1.6}\\\Rightarrow \frac{R_1R_2}{R_1+R_2}=7.5\\\Rightarrow \frac{R_1R_2}{36.36}=7.5\\\Rightarrow R_1R_2=272.72\\\Rightarrow(36.36-R_2)R_2=272.72\\\Rightarrow R_2^2-36.36R_2+272.72=0

Solving the above quadratic equation

\Rightarrow R_2=\frac{36.36\pm \sqrt{36.36^2-4\times 272.72}}{2}

\Rightarrow R_2=25.78\ or\ 10.57\\ If\ R_2=25.78\ then\ R_1=36.36-25.78=10.58\ \Omega\\ If\ R_2=10.57\ then\ R_1=36.36-10.57=25.79\Omega

∴ If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

6 0
2 years ago
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