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Dmitriy789 [7]
3 years ago
12

What is the speed of a 2000 Hz sound wave in vacuum?

Physics
1 answer:
Aleks [24]3 years ago
8 0
A sound wave can not value in a vacuum, so its speed is 0.
Sound waves are longitudinal and require a medium to pass through.
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Determine the position in the oscillation where an object in simple harmonic motion: (Be very specific, and give some reasoning
vladimir1956 [14]

Answer:

Explanation:

The greatest speed is attained at middle point or equilibrium point or where displacement from equilibrium point is zero .

When the object remains at one of the extreme point it experiences greatest acceleration but at that point velocity is zero . Due to acceleration , its velocity goes on increasing till it come to equilibrium point . At this point acceleration becomes zero . After that its velocity starts decreasing because of negative acceleration . Hence at middle point velocity is maximum .

The greatest acceleration is attained at maximum displacement or at one of the two extreme end .

Greatest restoring force too will be at position where acceleration is maximum because acceleration is produced by restoring force .

Restoring force is proportional to displacement or extension against restoring force . So it will be maximum when displacement is maximum .

Zero restoring force exists at equilibrium position or middle point or at point where displacement is zero . It is so because acceleration at that point is zero .

8 0
3 years ago
Conduct research to identify claims made about the effects that certain frequencies of EMR have when absorbed by matter. Find tw
svetoff [14.1K]

Answer:

The effects that certain frequencies of EMR have when absorbed by matter is explained below in complete detail.

Explanation:

Electromagnetic radiation of distinct frequencies associates with material adversely. ... Gamma rays, though commonly of somewhat greater frequency than X rays have the equivalent creation. When the power of gamma rays is consumed in material, its influence is practically indistinguishable from the outcome generated by X rays.

7 0
3 years ago
Read 2 more answers
Who thinks this a ufo? The following images are from a rare footage.
Galina-37 [17]

Answer:

That might be a sled ngl but it looks like a ufo but it's definitely not

5 0
3 years ago
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Steam at 0.6 MPa, 200 oC, enters an insulated nozzle with a velocity of 50 m/s. It leaves at a pressure of 0.15 MPa and a veloci
Rudiy27

Answer:

x2 = 0.99

Explanation:

from superheated water table

at pressure p1 = 0.6MPa and temperature 200 degree celcius

h1 = 2850.6 kJ/kg

From energy equation we have following relation

\dot m( h1+\frac{v1^2}{2}+ gz1 )+ Q = \dot m( h2+\frac{v2^2}{2}+ gz1) + W

\dot m( h1+\frac{v1^2}{2}) = \dot m( h2+\frac{v2^2}{2})

h1+\frac{v1^2}{2} = h2+\frac{v2^2}{2}

2850.6 + [\frac{50^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}] = h2 +[ \frac{600^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}]

h2 = 2671.85 kJ/kg

from superheated water table

at pressure p2 = 0.15MPa

specific enthalpy of fluid hf = 467.13 kJ/kg

enthalpy change hfg = 2226.0 kJ/kg

specific enthalpy of the saturated gas hg = 2693.1 kJ/kg

as it can be seen from above value hf>h2>hg, so phase 2 is two phase region. so we have

quality of steam x2

h2 = hf + x2(hfg)

2671.85 = 467.13 +x2*2226.0

x2 = 0.99

6 0
3 years ago
The figure shows the arrangement of master cylinder and slave cylinder of a part of braking system. The master cylinder piston,
Neko [114]

Answer:

a) 35 kPa

d) 140 N

c) 1) Increasing the brake fluid  pressure

2)  Increasing the slave piston surface area.

Explanation:

The parameters given are;

a) Force applied to the master cylinder piston = 28 N

Cross sectional area of the master cylinder piston = 8 cm² = 8 × 10⁻⁴ m²

The pressure P on the brake fluid is given by the formula for pressure as follows;

P= \dfrac{Applied \ force}{Area \over \ which \  force \ is \ applied} = \dfrac{28 \, N}{8 \times 10^{-4} \, m^2} = 35,000 \, N/m^2 = 35,000 \, Pa

The pressure on the brake fluid, P, produced by the master cylinder piston = 35,000 Pa = 35 kPa

b) Given that the area of the slave piston = 40 cm² = 0.004 m², we have from the formula for pressure, P;

P= \dfrac{Applied \ force}{Area \over \ which \  force \ is \ applied} = \dfrac{Applied \ force}{4 \times 10^{-3} \, m^2} = 35,000 \, N/m^2

Therefore;

Applied force on the slave piston = 4 × 10⁻³ m² × 35,000 N/m² = 140 N

c) The force, F produced by the slave cylinder piston is given by the relation;

F = Pressure × Area

Therefore, the two ways of increasing the force produced by the slave cylinder piston is as follows;

1) Increasing the pressure in the brake fluid by increasing the force exerted by the master cylinder piston

2) Increasing the surface area of the slave piston.

4 0
3 years ago
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