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Viktor [21]
3 years ago
12

What is the speed of sound at 33 °C (m/s)? For a frequency of 5 kHz, how large do you expect the wavelength to be (m)?

Physics
1 answer:
Vedmedyk [2.9K]3 years ago
7 0

Answer:

-The speed of sound at 33°C is 362.8 m/s.

-The wavelength at a frequency at 5 kHz is 0.07256 m .

Explanation:

let v = 343 m/s be the speed of sound.

let T be the temperature.

then the speed of sound V, at 33°C is given by:

V = v + 0.6×T

   = 343 + 0.6×33

   = 362.8 m/s

Therefore, the speed of sound at 33°C is 362.8 m/s.

the wavelength at a frequency of f = 5kHz = 5000 Hz is given by:

λ = V/f

  = (362.8)/(5000)

  = 0.07256 m

Therefore, the wavelength at a frequency at 5 kHz is 0.07256 m .

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A runner exerts a net force of 335 n to accelerate at a rate of 2.5 m/s what is the runners mass
Olegator [25]
F=ma
M=F/a
M=335/2.5
M=134 kg
8 0
3 years ago
Positive charge Q is distributed uniformly along the x-axis fromx=0 to x=a. A positive point charge q is located on the positive
dybincka [34]

Answer:

a. b- x= y

dx = -dy

b. F = \frac{-kQqi}{r (a+r)}

c.  F = \frac{-kQqi}{r^{2} }

Explanation:

a. x components:

dE = \frac{kdq}{(a+r-x^{2}) } \\

     = \frac{kQdx}{(a(a+r-x)^2}

Integrating and solving gives:

b- x= y

dx = -dy

b. the force is given by the equation derived from (a.):

F = \frac{-kQqi}{r (a+r)}

c. Given that r>>a, the expression becomes:

F = \frac{-kQqi}{r^{2} }

Explanation:

When the size of the charge distribution is less than the distance to the deviation point of the charge then the charge distribution would produce the same effect such as a linear charge.

6 0
3 years ago
A boat travels at 30 mph for a huge, solid cliff that is about 3,000 meters away. When the horn on the boat makes a toot, you ca
AleksandrR [38]

Answer:The frequency of the echo is slightly decreased

Explanation:

Given

speed of boat =30\ mph

cliff is 3000\ m away

when boat is still , suppose t is the time taken by the echo to reach observer on the boat

But as soon as boat starts moving  the distance between cliff and boat decreasing and time for echo to reach observer also decreases

and we know time \propto \frac{1}{frequency}

therefore frequency of the echo slightly decreased.

5 0
3 years ago
Assume the following values: d1 = 0.880 m , d2 = 1.11 m , d3 = 0.560 m , d4 = 2.08 m , F1 = 510 N , F2 = 306 N , F3 = 501 N , F4
dsp73

Answer:

= 2630.6 N.m

Explanation:

(FR)x = ΣFx = -F4 = -407 N

(FR)y = ΣFy =-F1-F2 -F3 = -510 - 306 - 501 = -1317 N

(MR)B =ΣM + Σ(±Fd)

= MA + F1(d1 +d2) + F2d2 - F4d3

= 1504 + 510(0.880+1.11) +306(1.11) - 407(0.560)

= 2630.64 N.m (counterclockwise)

6 0
3 years ago
A spring is compressed 0.035 m inside a dart gun. (K=500 N/m). The spring has elastic energy. Calculate it.
NARA [144]

The elastic potential energy of the spring is 0.31 J

Explanation:

The elastic potential energy of a spring is given by

E=\frac{1}{2}kx^2

where

k is the spring constant

x is the compression/stretching of the spring

For the spring in this problem, we have:

k = 500 N/m (spring constant)

x = 0.035 m (compression)

Substituting, we find the elastic potential energy:

E=\frac{1}{2}(500)(0.035)^2=0.31 J

Learn more about potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

6 0
3 years ago
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