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Illusion [34]
2 years ago
12

Sound waves with a constant frequency of 250 hertz are traveling through air at stp. what is the wavelength of the sound waves

Physics
1 answer:
Deffense [45]2 years ago
5 0

Answer:

wave length is 1.2m

Explanation:

since formula of wave length is v/f

v(speed of sound in air at stp is 300ms^-1)

f(frequency 250hertz)

then wave length is 300÷250 which give 1.2m

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The nucleus of most atoms is composed of which of the following sub-atomic particles?
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protons (+ charge) & neutrons (neutral charge)

however protons has a positive charge so it determined what atom it is.

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Who reported four “element” classifications, but included some substances that were combinations of elements rather than true el
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Chinook salmon are able to move upstream faster by jumping out of the water periodically; this behavior is called porpoising. Su
deff fn [24]

Answer:he formula for average speed is (total distance/total time)

the y-component does not matter in this problem. so do 6.26(cos45)=4.43m/s to find the x-component velocity which is constant throughout the duration of the flight. the total distance is 2L because he travels distance L twice.

the total time is ((time in water)+(time out of water)) since you dont have time you must eliminate it. to do this you need (distance)/(time)=velocity

solve for time and you get T=D/V

time in water is L/3.52 and time out of water is L/4.43

add them together and you get (4.43L+3.52L)/(15.59) = 7.95L/15.59

that value is your total time

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7 0
3 years ago
Two point charges, a +45nC charge X and a +12nC charge Y are separated by a distance of 0.5m.
Gnoma [55]

A) Calculate the resultant electric field strength at the midpoint between the charges.

Qx is the charge at X and Qy is the charge at Y.

E at midpoint = k×Qx/0.25² - k×Qy/0.25²

k = 9×10⁹Nm²C⁻², Qx = 45nC, Qy = 12nC

E = 4752N/C

Well done.

B) Calculate the distance from X at which the electric field strength is zero.

Let D be some point between X and Y for which the net E field is 0.

Let d be the distance from X to D.

Set up the following equation:

E at D = k×Qx/d² - k×Qy/(0.5-d)² = 0

Do some algebra to solve for d:

k×Qx/d² = k×Qy/(0.5-d)²

Qx/d² = Qy/(0.5-d)²

Qx(0.5-d)² = Qyd²

(0.5-d)√Qx = d√Qy

0.5√Qx-d√Qx = d√Qy

d(√Qx+√Qy) = 0.5√Qx

d = (0.5√Qx)/(√Qx+√Qy)

Plug in Qx = 45nC, Qy = 12nC

d ≈ 330mm

C) Calculate the magnitude of the electric field strength at the point P on the diagram below.

First determine the angles of the triangle. The sides of the triangle are 0.3m, 0.4m, and 0.5m, so this is a right triangle where the angle between the 0.3m and 0.4m sides is 90°

∠Y = tan⁻¹(0.4/0.3) = 53.13°

∠X = 90-∠Y = 36.87°

Determine the horizontal component of E at P:

Ex = E from Qx × cos(∠X) - E from Qy × cos(∠Y)

Ex = k×Qx/0.4²×cos(36.87°) - k×Qy/0.3²×cos(53.13°)

Ex = 1305N/C

Determine the vertical component of E at P:

Ey = E from Qx × sin(∠X) - E from Qy × sin(∠Y)

Ey = k×Qx/0.4²×sin(36.87°) - k×Qy/0.3²×sin(53.13°)

Ey = 2479N/C

Use the Pythagorean theorem to determine the magnitude of E at P:

E = √(Ex²+Ey²)

E ≈ 2802N/C

4 0
3 years ago
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