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2 years ago

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What frequency is received by a person watching an oncoming ambulance moving at 110 km/h and emitting a steady 800-Hz sound from

its siren? The speed of sound on this day is 345 m/s. Group of answer choices

1 answer:

Bas_tet [7]2 years ago

7 0

Answer:

check photo for solve

Explanation:

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Several large firecrackers are inserted into the holes of a bowling ball, and the 6.3 kg ball is then launched into the air with

Anni [7]

Answer

given,

mass of the ball = 6.3 kg

speed of the ball = 10.4 m/s

angle made with horizontal = 43°

m_a = 1.8 kg v_a = 2.2 m/s

m_b = 1.6 kg v_b = 1.8 m/s

mass of third particle = 6.3 - 1.8 - 1.6

= 2.9 kg

u cos θ = 10.4 x cos 43° = 7.61 m/s

by using conservation momentum along x-axis

6.3 x 7.61 = 1.8 × (-2.2) + 0 + 1.6 × V₃ₓ

V₃ₓ = 32.44 m/s (toward right)

by using conservation momentum along y-axis

0 = 0 + 1.6 x 1.8 + 1.6 × V₃y

V₃y = -1.8 m/s (indicate downward)

velocity of the third particle

$v = \sqrt{32.44^2 + (-1.8)^2}$

v = 32.49 m/s

$tan \theta = \dfrac{-1.8}{32.44}$

$\theta = tan^{-1}(\dfrac{-1.8}{32.44})$

θ = 3.176° (downward with horizontal)

4 0

2 years ago

A police radar gun uses X-band microwave radiation at a frequency of 12.2 GHz. Microwaves travel at the speed of light, or 3x108

quester [9]

Answer:

A police radar gun uses X-band microwave radiation at a frequency of 13.1 GHz. Microwaves travel at the speed of light, or 3x108 m/s. Since the frequency shift will be small for practical car speeds and difficult to detect, the shifted frequency is compared to the original frequency, and the resulting beat frequency is used to determine the speed of the car.

a.) If Michael is traveling at 29 m/s, what is the resulting beat frequency that the radar gun detects?

ANSWER: 2533 Hz

Explanation:

6 0

3 years ago

A small glass bead has been charged to +20 nC. A small metal ball bearing 1.0 cm above the bead feels a 0.018 N downward electri

Alla [95]

Answer:

$q=1\times10^{-8}C$

Explanation:

Let the charge on the ball bearing is q.

charge on glass bead, Q = 20 nC = 20 x 10^-9 C

Force between them, F = 0.018 N

Distance between them, d = 1 cm = 0.01 m

By use of Coulomb's law in electrostatics

$F=\frac{KQq}{d^{2}}$

By substituting the values

$0.018=\frac{9\times10^{9}\times20\times10^{-9}q}{0.01^{2}}$

$q=1\times10^{-8}C$

Thus, the charge on the ball bearing is $q=1\times10^{-8}C$

7 0

2 years ago

(a) What is the minimum width of a single slit (in multiples of λ ) that will produce a first minimum for a wavelength λ ? (b) W

Kitty [74]

Answer:

The minimum value of width for first minima is λ

The minimum value of width for 50 minima is 50λ

The minimum value of width for 1000 minima is 1000λ

Explanation:

Given that,

Wavelength = λ

For D to be small,

We need to calculate the minimum width

Using formula of minimum width

$D\sin\theta=n\lambda$

$D=\dfrac{n\lambda}{\sin\theta}$

Where, D = width of slit

$\lambda$ = wavelength

Put the value into the formula

$D=\dfrac{n\lambda}{\sin\theta}$

Here, $\sin\theta$ should be maximum.

So. maximum value of $\sin\theta$ is 1

Put the value into the formula

$D=\dfrac{1\times\lambda}{1}$

$D=\lambda$

(b). If the minimum number is 50

Then, the width is

$D=\dfrac{50\times\lambda}{1}$

$D=50\lambda$

(c). If the minimum number is 1000

Then, the width is

$D=\dfrac{1000\times\lambda}{1}$

$D=1000\lambda$

Hence, The minimum value of width for first minima is λ

The minimum value of width for 50 minima is 50λ

The minimum value of width for 1000 minima is 1000λ

4 0

3 years ago

A young woman with normal distant vision has a 10.0% ability to accommodate (that is, increase) the lens strength (a.k.a, lens p

Kipish [7]

Answer:

20.0 cm

Explanation:

Here is the complete question

The normal power for distant vision is 50.0 D. A young woman with normal distant vision has a 10.0% ability to accommodate (that is, increase) the power of her eyes. What is the closest object she can see clearly?

Solution

Now, the power of a lens, P = 1/f = 1/u + 1/v where f = focal length of lens, u = object distance from eye lens and v = image distance from eye lens.

Given that we require a 10 % increase in the power of the lens to accommodate the image she sees clearly, the new power P' = 50.0 D + 10/100 × 50 = 50.0 D + 5 D = 55.0 D.

Also, since the object is seen clearly, the distance from the eye lens to the retina equals the distance between the image and the eye lens. So, v = 2.00 cm = 0.02 m

Now, P' = 1/u + 1/v

1/u = P'- 1/v

1/u = 55.0 D - 1/0.02 m

1/u = 55.0 m⁻¹ - 1/0.02 m

1/u = 55.0 m⁻¹ - 50.0 m⁻¹

1/u = 5.0 m⁻¹

u = 1/5.0 m⁻¹

u = 0.2 m

u = 20 cm

So, at 55.0 dioptres, the closet object she can see is 20 cm from her eye.

8 0

2 years ago