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algol13
2 years ago
11

An elephant produces a 10Hz sound wave. Assuming the speed of sound in air is 344/s, determine the wavelength of this infrasonic

sound wave.​
Physics
1 answer:
user100 [1]2 years ago
4 0

Answer:

34.5 m

Explanation:

Given data : An elephant produces 10 Hz sound wave. Assuming the speed of sound in air is 345 m/s. To find : What is the wavelength of sound ? Answer : The wavelength of the sound is 34.5 m.

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What does the model below represent?
serious [3.7K]

Answer:

B. a chemical reaction

Explanation:

2CO + O2 → 2C02

8 0
3 years ago
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A kid pulls on a rope with 20 newtons of force. The block and tackle system pulls up a 160 newton box. What is the mechanical ad
Minchanka [31]
Answer:
The mechanical advantage of the system is 8

Explanation:
the mechanical advantage measures how much the system multiplies the input force to get the output.

In the given:
The input force (effort) is 20 Newton
The output force (load) is 160 Newton

This means that the mechanical advantage is:
mechanical advantage = load / effort = 160 / 20 = 8

Note that the mechanical advantage is unit-less (has no unit) since it is a ratio between two forces.

Hope this helps :)
3 0
3 years ago
You hear a sound in the distance. Suddenly the sound gets deeper, decreasing in pitch. Which can you assume about the sound wave
oksian1 [2.3K]

Answer:

A. The wavelengths of the new sound waves are longer

Explanation:

This is the Doppler effect which can be best illustraded  for the case of a siren of an ambulance approaching us having a greater frequency and getting lower in frequency and deeper as the ambulance passes  us.

Since the wavelength is inversely proportional to the frequency it follows the wavelengths are longer when the frequency decreases lowering its pitch and getting deeper.

8 0
3 years ago
Convert planks constant in cgs system
dezoksy [38]

in cgs system, plank's constant= h=6.626 x10⁻²⁶ erg s

Value of Plank's constant in SI system= 6.626 x10⁻³⁴ Js

now 1 Joule= 10⁷ ergs

so h= 6.626 x10⁻³⁴ Js (10⁷ ergs/1J)

h=6.626 x10⁻²⁷ erg s

7 0
3 years ago
A 20.0 kg crate sits at rest at the bottom of a 15.0-m-long ramp that is inclined at 34.0° above horizontal. A constant horizont
ankoles [38]

Answer:

987 joules, 3.01s

Explanation:

(A)

from the attached diagram

net force, Fnet, pulling the crate up the ramp is given by

Fnet = FcosФ - WsinФ - Fr

where FcosФ is the component of horizontal force 290N resolved parallel to the plane

WsinФ = mgsinФ = component of the weight of the crate resolved parallel to the plane

Fr = constant opposing frictional force

Fnet = 290cos34⁰ - 20 × 9.8 × sin34° - 65

Fnet = 240.421 - 109.602 - 65

Fnet = 65.82N

Work done on the crate up the ramp, W, is given by

W = Fnet × d (distance up the plane)

W = 65.819 × 15

W = 987.285 joules

W = 987 joules (to 3 significant Figures)

(B)

to calculate the time of travel up the ramp

we use the equation of motion

s = ut + \frac{1}{2}at^{2}

where s = distance up the plane, 15m

u = Initial velocity of the crate, which is 0 for a body that is initially at rest

a = acceleration up the plane, given by

a = \frac{Fnet}{m}

where m = mass of the crate, 20 kg

now, a = \frac{65.819}{20} \\a = 3.291\frac{m^{2} }{s}

from, s = ut + \frac{1}{2}at^{2}

15 = 0*t + \frac{1}{2}* 3.291 * t^{2}

15 = 0 + 1.645t^{2}

15 = 1.645t^{2}

t = \sqrt{\frac{15}{1.645} }

t = 3.019

t = 3.01s (to 3 sig fig)

7 0
3 years ago
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