Answer:
4009.21658986 Hz
Explanation:
v = Speed of sound in air = 348 m/s
L = Length of tube = 2.17 cm
Fundamental frequency for a tube where one end is closed and the other end open is given by
The fundamental frequency of the canal is 4009.21658986 Hz
1. a=(v-v0)/t
a=14 m/s / 2s
a=7 m/s^2
2. <span>a=(v-v0)/t
</span>a=(30 m/s - 0 m/s) / 12 s
a=2.5 m/s^2
3. <span>a=(v-v0)/t
a=(37 m/s - 22 m/s) / 2 s
a= 7.5 m/s^2
4. </span><span>a=(v-v0)/t
a=(12 km/s - 0 km/s)/8 s
a=1.5 km/s^2
etc</span>
Answer:
x = 0.327 m
Explanation:
Block 2 of mass 1.00 kg is at rest
spring constant = 200 N/m
Block 1 of mass 2 kg moving at 4 m/s
m₁ v₁ = (m₁ + m₂)V
2 x 4 = (2 + 1) V
V = 2.67 m/s
loss of kinetic energy = gain elastic potential energy
x = 0.327 m
hence, the spring compressed distance is equal to x = 0.327 m
<h3>Answer: 22.48°</h3><h3 /><h3>Explanation:</h3>
refractive index = sin i / sin e
where i is the angle of incidence
e is the angle of refraction
1.5 = sin 35 / sin e
1.5 = 0.5736/sin e
sin e = 0.5736/ 1.5
sin e = 0.3824
e = 22.48°
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