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4 years ago

What sound frequency could a human detect?

Physics

2 answers:

katovenus [111]4 years ago

7 0

Answer: 50 Hertz

I took the quiz and got 100%.

I hope this helps! (:

Crazy boy [7]4 years ago

4 0

Good human ears can detect sound from about 50 Hz to about 20,000Hz (20 kHz); this can be called the audible range of frequencies.

so c.50 hertz

hope this helps:)sorry if it doesnt

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What is gravitational field intensity?

mezya [45]

<h3>Answers :</h3>

The gravitational field intensity (EG) or (g) is the force on a unit mass at a point in the field.

Since the force (F) on a body of mass m in a gravitational field of a body of mass M = GMm/r2 .

you can see that the force per unit mass is given by F/m. So:

Gravitational Field Intensity (EG) = (g) = F/m = GM/r2

The units for EG are N kg -1 or since F = ma, F/m = a and so EG may be expressed in ms -2

Hope this will help you :))

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7 0

3 years ago

Two loudspeakers emit sound waves along the x-axis. The sound has maximum intensity when the speakers are 19 cm apart. The sound

Vanyuwa [196]

Answer:

Explanation:

The former is the case of constructive interference and the later case relates to destructive interference.

For listener, path difference is the separation of loudspeaker as listener is not standing in between the speaker.

If λ be the wave length

For constructive interference

19 = n λ

For destructive interference

29 = (2n+1) λ / 2

= n λ + λ / 2

= 19 + λ / 2

10 = λ / 2

λ = 20 cm

= 0. 20 m

7 0

3 years ago

Floating piece of wood - buoyancy question.?

sashaice [31]

No. because the weight is the same. no m9re water displacement

7 0

3 years ago

Calculate the force it would take to accelerate a 650-kg vehicle at a rate of 3 m/s^2

aleksklad [387]

Explanation:

Newton's second law:

F = ma

F = (650 kg) (3 m/s²)

F = 1950 N

3 0

4 years ago

Find the height or length of these natural wonders in km, m, and cm.A. A cave system with a mapped length of 354 miles.B. A wate

cricket20 [7]

Explanation:

Some standard unit conversion are 1 mile = 1.609344 km, 1 ft.= 30.48 cm,

1 km= 1000 m or 1 m = 0.001 km and 1 m= 100 cm or 1 cm=0.01 m.

Now, use these values to convert the given lengths.

A. length of cave = 5 miles (given)

From standard value 1 mile = 1.609344 km

$\Rightarrow$ 5 miles = $\times$ 1.609344 km= 8.04672 km,

$\Rightarrow$ 5 miles =8.04672 $\times$ 1000 m= 8046.72 m [ as 1 km = 1000 m]

$\Rightarrow$ 5 miles = 8046.72 $\times$ 100 cm=804672 cm

B. Height of the waterfall = 1235 ft. (given)

1 ft.= 30.48 cm [ from standard value]

$\Rightarrow$ 1235.2 ft.= 1235.2 $\times$ 30.48 cm=37648.896 cm,

$\Rightarrow$ 1235.2 ft.=37648.896 $\times$ 0.01 m =376.48896 m [ as 1 cm = 0.01 m]

$\Rightarrow$ 1235.2 ft.=376.48896 $\times$ 0.001 km =0.37648896 km [ as 1 m = 0.001 km]

C. Height of the mountain= 21320 ft. (given)

From standard value: 1 ft.= 30.48 cm

$\Rightarrow$ 21320 ft.= 21320 $\times$ 30.48 cm=649833.6 cm,

$\Rightarrow$ 21320 ft.=649833.6 $\times$ 0.01 m =6498.336 m [ as 1 cm = 0.01 m]

$\Rightarrow$ 21320 ft.=6498.336 $\times$ 0.001 km =6.498336 km [ as 1 m = 0.001 km].

D. Depth of canyon =6630 ft.

From standard value: 1 ft.= 30.48 cm

$\Rightarrow$ 6630 ft.= 6630 $\times$ 30.48 cm=202082.4 cm,

$\Rightarrow$ 6630 ft.=202082.4 $\times$ 0.01 m =2020.824 m [ as 1 cm = 0.01 m]

$\Rightarrow$ 6630 ft.=2020.824 $\times$ 0.001 km =2.020824 km [ as 1 m = 0.001 km].

4 0

4 years ago