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1 year ago

8

Mrs. Gonzalez is about to give birth and Mr. Gonzalez is rushing her to the hospital at a speed of 30.0 m/s. Witnessing the spee

ding car, Officer O'Malley jumps in his police car and turns on the siren, whose frequency is 800. Hz. If the officer chases after the Gonzalez' car with a speed of 35.0 m/s, what frequency do the Gonzalezes hear as the officer approaches?

1 answer:

valina [46]1 year ago

3 0

Answer: The frequency = 1714.3Hz

Explanation: The solution can be achieved by using doppler effect formula.

Since the source is moving toward the observer, the velocity of the observer will be positive.

Please find the attached file for the solution

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A group of students prepare for a robotic competition and build a robot that can launch large spheres of mass M in the horizonta

Dvinal [7]

Nobody will do that for 5 points loll

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2 years ago

A cylindrical rod of steel (E = 207 GPa, 30 × 10 6 psi) having a yield strength of 310 MPa (45,000 psi) is to be subjected to a

Yanka [14]

Answer:

Diameter of the cylinder will be $d=2.998\times 10^4m$

Explanation:

We have given young's modulus of steel $E=207GPa=207\times 10^9Pa$

Change in length $\Delta l=0.38mm$

Length of rod $l=500mm$

Load F = 11100 KN

Strain is given by $strain=\frac{\Delta l}{l}=\frac{0.38}{500}=7.6\times 10^{-4}$

We know that young's modulus $E=\frac{stress}{strain}$

So $207\times 10^9=\frac{stress}{7.6\times 10^{-4}}$

$stress=1573.2\times 10^{-5}N/m^2$

We know that stress $=\frac{force}{artea }$

So $1573.2\times 10^{-5}=\frac{11100\times 1000}{area}$

$area=7.055\times 10^{8}m^2$

So $\frac{\pi }{4}d^2=7.055\times 10^{8}$

$d=2.998\times 10^4m$

6 0

2 years ago

Sunlight strikes a piece of crown glass at an angle of incidence of 38.0°. Calculate the difference in the angle of refraction b

zhuklara [117]

Answer:

Difference in the angle of refraction = 0.3°

41.04° is the minimum angle of incidence.

Explanation:

Angle of incidence = 38.0°

For yellow light :

Using Snell's law as:

$\frac {sin\theta_2}{sin\theta_1}=\frac {n_1}{n_2}$

Where,

Θ₁ is the angle of incidence

Θ₂ is the angle of refraction

n₁ is the refractive index for yellow light which is 1.523

n₂ is the refractive index of air which is 1

So,

$\frac {sin\theta_2}{sin{38.0}^0}=\frac {1.523}{1}$

${sin\theta_2}=0.9377$

Angle of refraction for yellow light = sin⁻¹ 0.9377 = 69.67°.

For green light :

Using Snell's law as:

$\frac {sin\theta_2}{sin\theta_1}=\frac {n_1}{n_2}$

Where,

Θ₁ is the angle of incidence

Θ₂ is the angle of refraction

n₁ is the refractive index for green light which is 1.526

n₂ is the refractive index of air which is 1

So,

$\frac {sin\theta_2}{sin{38.0}^0}=\frac {1.526}{1}$

${sin\theta_2}=0.9395$

Angle of refraction for green light = sin⁻¹ 0.9395 = 69.97°.

The difference in the angle of refraction = 69.97° - 69.67° = 0.3°

Calculation of the critical angle for the yellow light for the total internal reflection to occur :

The formula for the critical angle is:

${sin\theta_{critical}}=\frac {n_r}{n_i}$

Where,

${\theta_{critical}}$ is the critical angle

$n_r$ is the refractive index of the refractive medium.

$n_i$ is the refractive index of the incident medium.

n₁ is the refractive index for yellow light which is 1.523 (incident medium)

n₂ is the refractive index of air which is 1 (refractive medium)

Applying in the formula as:

${sin\theta_{critical}}=\frac {1}{1.523}$

The critical angle is = sin⁻¹ 0.6566 = 41.04°

5 0

2 years ago

What is the change in internal energy (in J) of a system that does 4.50 ✕ 105 J of work while 3.20 ✕ 106 J of heat transfer occu

Dmitrij [34]

Answer:

$-3.25\times 10^6 J$

Explanation:

We are given that

Work done by the system= $4.5\times 10^5$ J

Heat transfer into the system= $U_1=3.2\times 10^6$ J

Heat transfer to the environment= $U_2=6\times 10^6$ J

We have to find the change in internal energy

By first law of thermodynamics

$\Delta Q=\Delta U+w$

$\Delta Q=U_1-U_2=3.2\times 10^6-6\times 10^6=-2.8\times 10^6J$

Substitute the values then we get

$-2.8\times 10^6=\Delta U+4.5\times 10^5$

$\Delta U=-2.8\times 10^6-4.5\times 10^5=-28\times 10^5-4.5\times 10^5=-32.5\times 10^5=-3.25\times 10^6 J$

Hence, the change in internal energy = $-3.25\times 10^6 J$

7 0

2 years ago

You wad up a piece of paper and throw it into the wastebasket. How far will

vitfil [10]

The range of the piece of paper is C) 1.4 m

Explanation:

The motion of the piece of paper is the motion of a projectile, which consists of two separate motions:

- A uniform motion along the horizontal direction, with constant velocity

- A uniformly accelerated motion along the vertical direction, with constant acceleration (the acceleration of gravity, $g=9.8 m/s^2$ )

From the equation of motion, it is possible to find an expression for the range (the total horizontal distance covered) of a projectile, which is given by:

$d=\frac{u^2 sin 2\theta}{g}$

where

u is the initial velocity

$\theta$ is the angle of projection

g is the acceleration of gravity

For the piece of paper in this problem,

u = 4.3 m/s

$\theta=65^{\circ}$

Substituting,

$d=\frac{(4.3)^2 sin(2\cdot 65^{\circ})}{9.8}=1.45 m \sim 1.4 m$

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

6 0

2 years ago

Read 2 more answers