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2 years ago

The amount of diffraction that a sound wave undergoes depends on

1 answer:

5 0

The amount of diffraction of sound waves depends on the medium the sound wave travels to and the frequency. Diffraction happens as soon as it has been out of the source.

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2. a) If a bottle was filled with a liquid, tightly

kow [346]

Answer:

bottle might burst

Explanation:

because liquid inside will expand and exert pressure on walls of the bottle.

6 0

2 years ago

An airplane which intends to fly due south at 250 km/hr experiences a wind blowing westward at 40 km/hr. What is the actual spee

sleet_krkn [62]

Answer:

simple is rumple a daily ok I'll be

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2 years ago

Water flows straight down from an open faucet. The cross-sectional area of the faucet is 2.4 × 10-4m2 and the speed of the water

Ksenya-84 [330]

To solve this problem it is necessary to apply the continuity equations in the fluid and the kinematic equation for the description of the displacement, velocity and acceleration.

By definition the movement of the Fluid under the terms of Speed, acceleration and displacement is,

$v_2^2 = v_1^2 + 2gh$

Where,

$V_i =$ Velocity in each state

g= Gravity

h = Height

Our values are given as,

$A_1 = 2.4*10^{-4} m^2$

$v_1 = 0.8 m/s$

$h = 0.11m$

Replacing at the kinetic equation to find $V_2$ we have,

$v_2 = \sqrt{v_1^2 + 2gh}$

$v_2 = \sqrt{(0.8 m/s)^2 + 2(9.80 m/s2)(0.11 m)}$

$v_2= 1.67 m/s$

Applying the concepts of continuity,

$A_1v_1 = A_2v_2$

We need to find A_2 then,

$A_2= \frac{A_1v_1 }{v_2}$

So the cross sectional area of the water stream at a point 0.11 m below the faucet is

$A_2= \frac{A_1v_1 }{v_2}$

$A_2= \frac{(2.4*10^{-4})(0.8)}{(1.67)}$

$A_2= 1.14*10^{-4} m2$

Therefore the cross-sectional area of the water stream at a point 0.11 m below the faucet is $1.14*10^{-4} m2$

8 0

3 years ago

Convert the mass to kgs thanks

Natali [406]

Memorize this and you'll be able to do ALL of these: <em>1 kg = 1,000 g</em>

So if you have some grams, divide the number by 1,000 to get kilograms.

1,000 g = 1.000 kg

500 g = 0.500 kg

100 g = 0.100 kg

50 g = 0.050 kg

20 g = 0.020 kg

10 g = 0.010 kg

4 0

2 years ago

A particular planet has a moment of inertia of 9.74 × 1037 kg ⋅ m2 and a mass of 5.98 × 1024 kg. Based on these values, what is

malfutka [58]

Answer: A) $6.38(10)^{6} m$