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qwelly [4]
3 years ago
9

The speed of sound in air is 345 m/s. A tuning fork vibrates above the open end of a sound resonance tube. If sound waves have w

avelengths 58-cm in the tube, what is the frequency (in Hertz) of the tuning fork
Physics
1 answer:
anzhelika [568]3 years ago
8 0

Answer:

594.8 Hz

Explanation:

Parameters given:

Speed of sound, v = 345 m/s

Wavelength = 58 cm = 0.58 m

Speed of a wave is given as:

Speed = wavelength * frequency

Therefore:

Frequency = Speed/Wavelength

Frequency = 345/0.58

Frequency = 594.8 Hz

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An 80-cm uniform 10-kg bar is resting on two scales, one at either end. A smaller 4-kg mass (m) is placed at a distance of d = 2
Varvara68 [4.7K]

Answer

given,

length of bar = 80 cm

mass of the bar = 10 kg

smaller mass = 4 kg

distance = 20 cm

s_1 + s_2 = 10 + 4

s_1 + s_2 = 14\ kg

taking moment about B

s_1 \times 0.8 - 10 \times 0.4 - 4 \times 0.6 = 0

s_1 \times 0.8 = 6.4

s_1 = 8\ N

s_2 = 14 - s_1

s_2 = 14 - 8

s_2 = 6 N

difference between two scale = 8 - 6

                                                  = 2 N

7 0
3 years ago
Which of the following is not a source of light? *<br> a) sun<br> b) star<br> c) mirror<br> d)cfl
Vladimir [108]

I think that would be c) mirror because mirrors reflects light and can't create it.

5 0
3 years ago
Red light of wavelength 633 nm from a helium-neon laser passes through a slit 0.400mm wide. The diffraction pattern is observed
kogti [31]

Answer:

a)y_{first}=5.3mm

b)y_{second}=10.6-5.3 =5.3 mm  

Explanation:

a)

The width of the central bright in this diffraction pattern is given by:

y=\frac{m\lambda D}{a} when m is a natural number.

here:

  • m is 1 (to find the central bright fringe)                
  • D is the distance from the slit to the screen
  • a is the slit wide
  • λ is the wavelength

So we have:

y_{first}=\frac{633*10^{9}*3.35}{0.0004}

y_{first}=5.3mm

b)

Now, if we do m=2 we can find the distance to the second minima.

y_{2}=\frac{2*633*10^{9}*3.35}{0.0004}

y_{2}=10.6 mm

Now we need to subtract these distance, to get the width of the first bright fringe :

y_{second}=10.6-5.3 =5.3 mm    

I hope it heps you!

     

4 0
3 years ago
An 805-kg race car can drive around an unbanked turn at a maximum speed of 54 m/s without slipping. the turn has a radius of cur
SIZIF [17.4K]
(a) Equating centripetal force to friction force, one finds the relation
  v² = kar
for car speed v, coefficient of friction k, radius of curvature r, and downward acceleration a.

There is already downward acceleration due to gravity. The additional accceleration due to the wing is
  a = F/m = 10600 N/(805 kg) ≈ 13.1677 m/s²
We presume this is added to the 9.80 m/s² gravity provides, so the coefficient of friction is
  k = v²/(ar) = (54 m/s)²/((13.1677 m/s² +9.80 m/s²)·(155 m))
  k ≈ 0.8191

(b) The maximum speed is proportional to the square root of the downward acceleration. Changing that by a factor of 9.80/(9.80+13.17) changes the maximum speed by the square root of this factor.
  max speed with no wing effect = (54 m/s)√(9.8/22.97) ≈ 35.27 m/s
8 0
3 years ago
When the material left behind has been carried by the glacier from its original location to a new location, this is called what
MAVERICK [17]

Answer:

C

Explanation:

Glaciers break rocks because of the thawing and icing of water in rock crevices. These rocks get carried in the glacier. The glaciers move because the base of the glaciers melts ever so slightly due to the weight of the glacier above (pressure reduces the melting point of ice). This smoothens the slipping of the glaciers . When the glaciers get to a warmer place, it melts and the rocks in it are deposited.

3 0
3 years ago
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