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erma4kov [3.2K]
3 years ago
7

How is frequency related to the sound we hear?

Physics
1 answer:
Leviafan [203]3 years ago
8 0
Frequency is the vibration of noise and the vibration determines the pitch, which we depend on to be a pitch or frequency we can hear. If it's too high or too low our ears can't hear it 
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The eagle drops the trout a height of 6.1 m the fish travels 7.9 m horizontaly before hitting the water what is the velocity of
Semenov [28]

The velocity of the eagle is 7.0 m/s

Explanation:

The motion of the fish is a projectile motion, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction, where the horizontal velocity of the fish is equal to the horizontal velocity of the eagle

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction  

We start by analyzing the vertical motion, using the following suvat equation:

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

where:

s = 6.1 m is the vertical displacement of the fish

u = 0 is the initial vertical velocity of the fish

a=g=9.8 m/s^2 is the acceleration of gravity

t is the time

Solving for t, we find the time of flight of the fish:

t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(6.1)}{9.8}}=1.12 s

Now we know that during this time, the fish travels a horizontal distance of

x = 7.9 m

Therefore, the horizontal velocity of the fish is

v_x = \frac{x}{t}=\frac{7.9}{1.12}=7.0 m/s

And therefore, this is the initial velocity of the eagle.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

6 0
3 years ago
a golfer hits a golf ball with a velocity of 36.0 meters/second at an angle of 28.0. if the hang time of the golf ball is 33.4 s
Kobotan [32]
Solving this using the time, we know that range = horizontal velocity x time of flight 

since there are no horizontal forces acting on the ball, there are no horizontal accelerations and the initial horizontal velocity of 36 cos 28 will be constant throughout. If we use the correct time of flight given the launch parameters, we have 

range = 36 cos 28 x 3.44 s = 109.3 m

6 0
3 years ago
What is newton's second law of motion?
lord [1]

Answer:

Newton's second law of motion can be formally stated as follows: The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.

3 0
3 years ago
A gasoline tank has the shape of an inverted right circular cone with base radius 4 meters and height 5 meters. Gasoline is bein
RSB [31]

Answer:

h'=0.25m/s

Explanation:

In order to solve this problem, we need to start by drawing a diagram of the given situation. (See attached image).

So, the problem talks about an inverted circular cone with a given height and radius. The problem also tells us that water is being pumped into the tank at a rate of 8m^{3}/s. As you  may see, the problem is talking about a rate of volume over time. So we need to relate the volume, with the height of the cone with its radius. This relation is found on the volume of a cone formula:

V_{cone}=\frac{1}{3} \pi r^{2}h

notie the volume formula has two unknowns or variables, so we need to relate the radius with the height with an equation we can use to rewrite our volume formula in terms of either the radius or the height. Since in this case the problem wants us to find the rate of change over time of the height of the gasoline tank, we will need to rewrite our formula in terms of the height h.

If we take a look at a cross section of the cone, we can see that we can use similar triangles to find the equation we are looking for. When using similar triangles we get:

\frac {r}{h}=\frac{4}{5}

When solving for r, we get:

r=\frac{4}{5}h

so we can substitute this into our volume of a cone formula:

V_{cone}=\frac{1}{3} \pi (\frac{4}{5}h)^{2}h

which simplifies to:

V_{cone}=\frac{1}{3} \pi (\frac{16}{25}h^{2})h

V_{cone}=\frac{16}{75} \pi h^{3}

So now we can proceed and find the partial derivative over time of each of the sides of the equation, so we get:

\frac{dV}{dt}= \frac{16}{75} \pi (3)h^{2} \frac{dh}{dt}

Which simplifies to:

\frac{dV}{dt}= \frac{16}{25} \pi h^{2} \frac{dh}{dt}

So now I can solve the equation for dh/dt (the rate of height over time, the velocity at which height is increasing)

So we get:

\frac{dh}{dt}= \frac{(dV/dt)(25)}{16 \pi h^{2}}

Now we can substitute the provided values into our equation. So we get:

\frac{dh}{dt}= \frac{(8m^{3}/s)(25)}{16 \pi (4m)^{2}}

so:

\frac{dh}{dt}=0.25m/s

3 0
3 years ago
If youre reading this goodluck in school homie <3
bazaltina [42]
I needed this ! Thanks a lot
7 0
3 years ago
Read 2 more answers
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