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3 years ago

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The siren on an ambulance is emitting a sound whose frequency is 2450 Hz. The speed of sound is 343 m/s. If the ambulance is sta

tionary and you (the "observer") are sitting in a parked car, what are the wavelength and the frequency of the sound you hear?

2 answers:

lyudmila [28]3 years ago

5 0

Answer:

The wavelength is 0.14 m

Explanation:

Given that,

Frequency = 2450 Hz

Speed of sound = 343 m/s

We need to calculate the wavelength

Using formula of wavelength

$v= f\lambda$

Where, v = speed of sound

f = frequency

Put the value into the formula

$\lambda=\dfrac{v}{f}$

$\lambda=\dfrac{343}{2450}$

$\lambda=0.14\ m$

Hence, The wavelength is 0.14 m

Anna11 [10]3 years ago

5 0

Explanation:

It is given that,

Frequency of the siren, f = 2450 Hz

The speed of sound, v = 343 m/s

Here, both ambulance and the observer is stationary. The observed frequency is calculated using Doppler's effect as :

$f'=\dfrac{v+v_o}{v-v_s}\times f$

$v_o$ is the velocity of observer

$v_s$ is the velocity of source

v is the speed of sound wave

Here, $v_o=v_s=0$

So, f' = f

f' = 2450 Hz

Wavelength, $\lambda'=\dfrac{v}{f'}$

$\lambda'=\dfrac{343\ m/s}{2450\ Hz}$

$\lambda'=0.14\ m$

So, the frequency and wavelength of the observed sound is 2450 Hz and 0.14 meters. Hence, this is the required solution.

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Radar uses radio waves of a wavelength of 2.9 m . The time interval for one radiation pulse is 100 times larger than the time of

Mandarinka [93]

Answer:

145 m

Explanation:

Given:

Wavelength (λ) = 2.9 m

we know,

c = f × λ

where,

c = speed of light ; 3.0 x 10⁸ m/s

f = frequency

thus,

$f=\frac{c}{\lambda}$

substituting the values in the equation we get,

$f=\frac{3.0\times 10^8 m/s}{2.9m}$

f = 1.03 x 10⁸Hz

Now,

The time period (T) = $\frac{1}{f}$

or

T = $\frac{1}{1.03\times 10^8}$ = 9.6 x 10⁻⁹ seconds

thus,

the time interval of one pulse = 100T = 9.6 x 10⁻⁷ s

Time between pulses = (100T×10) = 9.6 x 10⁻⁶ s

Now,

For radar to detect the object the pulse must hit the object and come back to the detector.

Hence, the shortest distance will be half the distance travelled by the pulse back and forth.

Distance = speed × time = 3 x 10^8 m/s × 9.6 x 10⁻⁷ s) = 290 m {Back and forth}

Thus, the minimum distance to target = $\frac{290}{2}$ = 145 m

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2 years ago

Before colliding, the momentum of block A is -100 kg*m/, and block B is -150 kg*m/s. After, block A has a momentum -200 kg*m/s.

rjkz [21]

Answer:

Momentum of block B after collision = $-50\ kg\ ms^{-1}$

Explanation:

Given

Before collision:

Momentum of block A = $p_{A1}$ = $-100\ kg\ ms^{-1}$

Momentum of block B = $p_{B1}$ = $-150\ kg\ ms^{-1}$

After collision:

Momentum of block A = $p_{A2}$ = $-200\ kg\ ms^{-1}$

Applying law of conservation of momentum to find momentum of block B after collision $p_{B2}$ .

$p_{A1}+p_{B1}=p_{A2}+p_{B2}$

Plugging in the given values and simplifying.

$-100-150=-200+p_{B2}$

$-250=-200+p_{B2}$

Adding 200 to both sides.

$200-250=-200+p_{B2}+200$

$-50=p_{B2}$

∴ $p_{B2}=-50\ kg\ ms^{-1}$

Momentum of block B after collision = $-50\ kg\ ms^{-1}$

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2 years ago

A physics major is working to pay her college tuition by performing in a traveling carnival. She rides a motorcycle inside a hol

il63 [147K]

Answer:

v = 12.1 m/s

Explanation:

When at the top of the circle, there are two forces acting on the combined mass of the rider and the motorcycle.
These are the force of gravity (downward) and the normal force, which is directed from the surface away from it, perpendicular to the surface.
In this case, as the motorcycle runs in the interior of the circle, at the top point this force is completely vertical, and is also downward.
Since the motorcycle is moving in a vertical circle, there must be a force, keeping the object moving around a circle.
This force is the centripetal force, aims towards the center of the circle, and is just the net force aiming in this direction at any point.
At the top point, this force is just the sum of the normal force and the weight of the mass of the rider and the motorcycle combined, as follows (we take the direction towards the center as positive):

$F_{c} = N + m*g (1)$

Now, we know that the centripetal force is related with the tangential speed at this point and the radius of the circle as follows:

$F_{c} = m*\frac{v^{2}}{r} (2)$

Since the normal force takes any value as needed to make (1) equal to (2), if the speed diminishes, it will be needed less force to keep the equality valid.
In the limit, when the motorcyvle tires barely touch the surface, this normal force becomes zero.
In this condition, from (1) and (2), we can find the minimum possible value of the speed that still keeps the motorcycle touching the surface, as follows:
$v_{min} =\sqrt{r*g} =\sqrt{15.0m*9.8m/s2} = 12.1 m/s (3)$

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3 years ago

A soap bubble appears red (λ = 633nm) at the point on its front surface nearest to the viewer. Assuming n = 1.35, what is the sm

alexira [117]

Answer:

The smallest film thickness is 117 nm.

Explanation:

Light interference on thin films can be constructive or destructive. Constructive interference is dependent on the film thickness and the refractive index of the medium.

For the first interference (surface nearest to viewer), the minimum thickness can be expressed as:

$2t_{min} = \frac{wavelenth}{2n}$

where n is the refractive index of the bubble film.

Therefore,

$2t_{min} = \frac{633x10^{-9} }{(2)(1.35)}$

$2t_{min} =2.344x10^{-7}$

∴ $t_{min} =\frac{2.344x10^{-7} }{2}$

$t_{min} = 1.17x10^{-7} m = 117 nm.$

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3 years ago

An airplane is heading due south at a speed of 560 km/h . If a wind begins blowing from the southwest at a speed of 80.0 km/h (a

polet [3.4K]

Answer:

the magnitude of Vpg = 493.711 km/h

Explanation:

given data

speed Vpg = 560 km/h

speed Vwg = 80 km/h

solution

we get here magnitude of the plane velocity w.r.t. ground is

we know that the Vpg = Vpw + Vwg .....................1

writing the component of the velocity that is

Vpw = (0 km/h î - 560 km/h j )

Vwg = (80 cos 45 km/h î + 80 sin 45 km/h j)

adding these

Vpg = (0+80 cos 45 km/h ) î + ( -560 + 80 sin 45 km/h j)i

Vpg = (42.025 ) î (-491.92 km/h)j

now we take magnitude

the magnitude of Vpg = $\sqrt{(42.025^2+(-491.92)^2)} km/h$

the magnitude of Vpg = 493.711 km/h

5 0

2 years ago