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3 years ago

the speed of sound in the air is about 340m/s. what is the wavelength of a sound wave with a frequency of 400Hz?

Physics

1 answer:

olga2289 [7]3 years ago

5 0

Wavelength = (speed) / (frequency)

Wavelength = (340 m/s) / (400 /s)

<em>Wavelength = 0.85 meter</em>

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A fireperson is 50 m from a burning building and directs a stream of water from a fire hose at an angle of 300 above the horizon

notsponge [240]

Answer:

We can think the water stream as a solid object that is fired.

The distance between the fireperson and the building is 50m. (i consider that the position of the fireperson is our position = 0)

The angle is 30 above the horizontal. (yo wrote 300, but this has no sense because 300° implies that he is pointing to the ground).

The initial speed of the stream is 40m/s.

First, using the fact that:

x = R*cos(θ)

y = R*sin(θ)

in this case R = 40m/s and θ = 30°

We can use the above relation to find the components of the velocity:

Vx = 40m/s*cos(30°) = 34.64m/s

Vy = 20m/s.

First step:

We want to find the time needed to the stream to hit the buildin.

The horizontal speed is 34.64m/s and the distance to the wall is 50m

So we want that:

34.64m/s*t = 50m

t = 50m/(34.64m/s) = 1.44 seconds.

Now we need to calculate the height of the stream at t = 1.44s

Second step:

The only force acting on the water is the gravitational one, so the acceleration of the stream is:

a(t) = -g.

g = -9.8m/s^2

For the speed, we integrate over time and we get:

v(t) = -g*t + v0

where v0 is the initial speed: v0 = 20m/s.

The velocity equation is:

v(t) = -g*t + 20m/s.

For the position, we integrate again over time:

p(t) = -(1/2)*g*t^2 + 20m/s*t + p0

p0 is the initial height of the stream, this data is not known.

Now, the height at the time t = 1.44s is

p(1.44s) = -5.9m/s^2*(1.44s)^2 + 20m/s*1.44s + po

= 16.57m + p0

So the height at wich the stream hits the building is 16.57 meters above the initial height of the fire hose.

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3 years ago

The graph at the right shows the force needed to pull a bow back as the string is pulled further and further.

Sindrei [870]

A. 9 J

In a force-distance graph, the work done is equal to the area under the curve in the graph.

In this case, we need to extrapolate the value of the force when the distance is x=30 cm. We can easily do that by noticing that there is a direct proportionality between the force and the distance:

$F=kx$

where k is the slope of the line. We can find k, for instance chosing the point at x=5 cm and F=10 N:

$k=\frac{F}{x}=\frac{10 N}{5 cm}=2 N/cm$

And now we can calculate the work by calculating the area under the curve until x=30 cm, F=60 N:

$W=\frac{1}{2} (height) (base)= \frac{1}{2}(60 N)(0.30 m)=9 J$

B. 24.5 m/s

The mass of the arrow is m=30 g=0.03 kg. The kinetic energy of the arrow when it is released is equal to the work done by pulling back the bow for 30 cm:

$W=K=\frac{1}{2}mv^2$

where m is the mass of the arrow and v is its speed. By re-arranging the formula and using W=9 J, we find the speed:

$v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2\cdot 9J}{0.03 kg}}=24.5 m/s$

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3 years ago