A student walks 1.0 mi west and then 1.0 mi north. afterward, how far is she from her starting point?

1 answer:

7 0

1.0 mi because as he has covered 1 the disrance of 1 m and then 1 m in north so he can go straight 1m north so the distance is actually 1m from the starting point

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A uniform narrow tube 1.90 m long is open at both ends. It resonates at two successive harmonics of frequencies 280 Hz and 294 H

Over [174]

Answer:

a)14Hz

b)26.6m/s

Explanation:

a)we were given

the first harmonics frequencies as 280 Hz

The second harmonic frequency as 294 Hz.

The fundamental frequency is equal to the gap which means the distance that exist between the harmonics, then

the fundamental frequency=(294 - 280 = 10 Hz)

= 14Hz

b) We know the frequency and the wavelength of the sound wave (

We were told that the wavelength must be twice the length of the tube then, velocity can be calculated as

And fundamental frequency= 14Hz, and distance of 1.90 m then

v = f*2L = (14Hz)*2*(1.90 m) = 26.6m/s

Therefore, the speed of sound in the gas in the tubes is 26.6m/s

7 0

3 years ago

8 POINTS AND MARK BRAINIEST:

lesya692 [45]

The answer is A i just took the test.

3 0

3 years ago

Read 2 more answers

a stone is thrown vertically upwards with a velocity of 20 m per second what will be its velocity when it reaches a height of 10

TEA [102]

Answer:

Explanation:

Here's the info we have:

initial velocity is 20 m/s;

final velocity is our unknown;

displacement is -10.2 m; and

acceleration due to gravity is -9.8 m/s/s. Using the one-dimensional equation

v² = v₀² + 2aΔx and filling in accordingly to solve for v:

$v=\sqrt{(20)^2+2(-9.8)(-10.2)}$ Rounding to the correct number of sig fig's to simplify:

$v=\sqrt{400+2.0*10^2}$ to get

v = $\sqrt{600}=20\frac{m}{s}$ If you don't round like that, the velocity could be 24, or it could also be 24.5 depending on how your class is paying attention to sig figs or if you are at all.