What year was the RoboSapien toy robot released?<br> a) 2007<br> b) 2004<br> c) 2020<br> d) dunno

A police car with its 300-Hz siren is moving toward a warehouse at 30 m/s, intending to crash through the door. The sound bounce

Lady bird [3.3K]

Answer: The frequency heard will be f = 275.675Hz

Explanation: When an object emitting sound is moving, it occurs a phenomenon called Doppler shift or Doppler effect. What happens is that the sound gets higher when the moving object comes closer the observer and becomes lower after it passes, This change is due to the quantity of waves that passes through an area in an unit of time.

The formula to calculate the Doppler effect is as follows

f = ( $\frac{c}{c+Vs}$ ) · f₀

f is the observed frequency;

c is the speed of sound;

Vs is velocity of the source;

f₀ is the emitted frequency of source;

Substituting and calculating,

f = $\frac{340}{340+30}$ · 300

f = 275.675 Hz

Thus, the frequency heard by the police officer is 275.675Hz.

8 0

4 years ago

The spreading of waves behind an aperture is more for long wavelengths and less for short wavelengths.Less for long wavelengths

Allisa [31]

Answer:

Increase in wavelength of incident wave also increases the spread angle or spread of the interference pattern.

Explanation:

Solution:-

- The diffraction occurs when light bends in the same medium. The bending is the result of light waves "squeezing" through small openings or "curving" around sharp edges.

- Moreover, waves diffract best when the size of the diffraction opening (or grting or groove) corresponds to the size of the wavelength. Hence, light diffracts more through small openings than through larger openings.

- The formula for diffraction shows a direct relationship between the angle of diffraction (theta) and wavelength:

d sin (θ) = m λ

Where,

λ : Wavelength , θ : The spread angle , d : Slit opening or grating

- We can see that the wavelength λ and spread angle θ are related proportionally. So if we increase the wavelength of incident wave we also increase the spread angle or spread of the interference pattern.

5 0

4 years ago

Allen Aby sets up an Atwood Machine and wants to find the acceleration and the tension in the string. Please help him. Two block

Vitek1552 [10]

<u>Answers:</u>

In order to solve this problem we will use Newton’s second Law, which is mathematically expressed after some simplifications as:

This can be read as: The Net Force $F$ of an object is equal to its mass $m$ multiplied by its acceleration $a$ .

We will also need to <u>draw the Free Body Diagram of each block</u> in order to know the direction of the acceleration in this system and find the Tension $T$ of the string (<u>See figure attached). </u>

We already know<u> $m_{2}$ is greater than $m_{1}$ </u>, this means the weight of the block 2 $P_{2}$ is greater than the weight of the block 1 $P_{1}$ ; therefore <u>the acceleration of the system will be in the direction of $P_{2}$ </u>, as shown in the figure attached.

We also know by the information given in the problem that <u>the pulley does not have friction and has negligible mass</u>, and <u>the string is massless</u>.

This means that the tension will be the same along the string regardless of the difference of mass of the blocks.

Now that we have the conditions clear, let’s begin with the calculations:

1) Firstly, we have to find the weight of each block, in order to verify that block 2 is heavier than block 1.

This is done using equation (1), where the force of the weight $P$ is calculated using the <u>acceleration of gravity</u> $g=9.8\frac{m}{s^{2}}$ acting on the blocks:

<u>For block 1: </u>

$P_{1}=m_{1}g$ (3)

$P_{1}=1.5kg(9.8\frac{m}{s^{2}})$

<u>For block 2: </u>

$P_{2}=m_{2}g$ (5)

$P_{2}=2.4kg(9.8\frac{m}{s^{2}})$

Then, we are going to <u>find the acceleration $a$ of the whole system: </u>

$F_{r}=P_{1}+P_{2}$ (7)

Where the Resulting Force $F_{r}$ is equal to the sum of the weights $P_{1}$ and $P_{2}$ .

In the figure attached, note that $P_{1}$ is in opposite direction to the acceleration $a$ , this means it must <u>have a negative sing</u>; while $P_{2}$ is in the same direction of $a$ .

Here we only have to isolate $a$ from equation (8) and substitute the values according to the conditions of the system:

$-14.7N+23.52N=(1.5kg+2.4kg)a$

$8.82N=(3.9kg)a$

Then:

$a=\frac{8.82N }{3.9kg}$

<h2> $a=2.26\frac{m}{ s^{2}}$ </h2><h2>This is the acceleration of the system. </h2>

2) For the second part of the problem, we have to find the tension $T$ of the string.

We can choose either the Free Body Diagram of block A or block B to make the calculations, <u>the result will be the same</u>.

Let’s prove it:

For $m_{1}$

we see in the free body diagram that the <u>acceleration is in the same direction of the tension of the string</u>, so:

$F_{r}=T-P_{1}$ (9)

$T-P_{1}=m_{1}a$ (10)

$T-14.7N=(1.5kg)( 2.26\frac{m}{ s^{2}})$

Then;

<h2> $T=18.09N$ This is the tension of the string </h2><h2> </h2>

For $m_{2}$

we see in the free body diagram that the acceleration is in opposite direction of the tension of the string and must <u>have a negative sign,</u> so:

$F_{r}=T-P_{2}$ (9)