Describe how the sound from the radio reaches all parts of the room?? (2m)

While bats emit a wide variety of sounds, one type emits pulses of sound at a frequency between 39 kHz and 78 kHz. What is the r

sdas [7]

Answer:

The range of wavelengths of the sound is 7692.30 m and 3846.15 m

Explanation:

A bat emits pulses of sound at a frequency between 39 kHz and 78 kHz. It is required to find the range of wavelengths of this sound.

Bat uses ultrasonic waves. It moves with the speed of light.

If f = 39 kHz,

$\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{39\times 10^3}\\\\\lambda=7692.30\ m$

If f = 78 kHz,

$\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{78\times 10^3}\\\\\lambda=3846.15\ m$

So, the range of wavelengths of the sound is 7692.30 m and 3846.15 m.

6 0

3 years ago

What is the basic building block of all matter

Natalka [10]

I think that is atom.

7 0

3 years ago

Read 2 more answers

A 1.00 kg particle has the xy coordinates (-1.20 m, 0.500 m) and a 4.50 kg particle has the xy coordinates (0.600 m, -0.750 m).

just olya [345]

Answer:

a) The x coordinate of the third mass is -1.562 meters.

b) The y coordinate of the third mass is -0.944 meters.

Explanation:

The center of mass of a system of particles ( $\vec r_{cm}$ ), measured in meters, is defined by this weighted average:

$\vec r_{cm} = \frac{\Sigma_{i=1}^{n}\,m_{i}\cdot \vec r_{i}}{\Sigma_{i=1}^{n}\,m_{i}}$ (1)

Where:

$m_{i}$ - Mass of the i-th particle, measured in kilograms.

$\vec r_{i}$ - Location of the i-th particle with respect to origin, measured in meters.

If we know that $\vec r_{cm} = (-0.500\,m,-0.700\,m)$ , $m_{1} = 1\,kg$ , $\vec r_{1} = (-1.20\,m, 0.500\,m)$ , $m_{2} = 4.50\,kg$ , $\vec r_{2} = (0.600\,m, -0.750\,m)$ and $m_{3} = 4\,kg$ , then the coordinates of the third particle are:

$(-0.500\,m, -0.700\,m) = \frac{(1\,kg)\cdot (-1.20\,m,0.500\,m)+(4.50\,kg)\cdot (0.600\,m,-0.750\,m)+(4\,kg)\cdot \vec r_{3}}{1\,kg+4.50\,kg+4\,kg}$

$(-4.75\,kg\cdot m, -6.65\,kg\cdot m) = (-1.20\,kg\cdot m, 0.500\,kg\cdot m) + (2.7\,kg\cdot m, -3.375\,kg\cdot m) +(4\cdot x_{3},4\cdot y_{3})$

$(4\cdot x_{3}, 4\cdot y_{3}) = (-6.25\,kg\cdot m,-3.775\,kg\cdot m)$

$(x_{3},y_{3}) = (-1.562\,m,-0.944\,m)$

a) The x coordinate of the third mass is -1.562 meters.

b) The y coordinate of the third mass is -0.944 meters.

5 0

2 years ago

the distance a spring will stretch varies directly with how much weight is attached to the spring. If a spring stretches 11 inch

Paha777 [63]

Direct variation involves ration and proportions, so

you need to set up the proportion:

<span>11 / 75 = x / 65

Cross multiplying:

75x = 11*65

x = (11*65)/75

Solving, we get x = 9.533, </span>

<span>which rounds off to 9.5

Therefore, the spring will stretch up to 9.5 inches with 65 attached.

I hope my answer has come to your help. Thank you for posting your question here in Brainly.

</span>

4 0

3 years ago

In medieval warfare, one of the greatest technological advancement was the trebuchet. The trebuchet was used to sling rocks into

NNADVOKAT [17]

Answer:

2) a_y= -g 3) vₓ=constant v_y = v_{oy} - g t, 4) vₓ = v₀ₓ - ax t

5) changes the horizontal speed, should change range

7) changes the vertical speed change the maximum height

Explanation:

1) After reading your long writing, we are going to solve the exercise, in the attachment you can see the different vectors.

2) The acceleration vectors are vertical and directed downwards due to the attraction of the Earth (gravity force) this force is constant, on the x axis there is no acceleration

3) the velocity vectors on the x-axis are constant because there are no relationships and the y-axis changes value according to the expression

v_y = v_{oy} - gt

at the point of maximum height, vy = 0 is equal to the maximum height

4) For someone to change the horizontal acceleration we must assume a friction with the air, in this case they relate it would be in the opposite direction to the horizontal speed

In the graph it would be directed to the left, therefore the velocity would be

vₓ = v₀ₓ - ax t

5 and 6) If someone changes the horizontal speed, they should change the range of the shot for greater horizontal speed, the rock goes further.

the equations of motion are

x = v₀ₓ t

y = v_{oy} t - ½ g t²

7) If someone changes the vertical speed change the maximum height, but not the scope of the shot, for higher speed higher maximum height,

the equations of motion are the same.

4 0

3 years ago