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3 years ago

vibrations cause sounds yet if you move your hand back and forth through the air, you don't hear a sound. Why?

2 answers:

8 0

The movement of your hand is not considered vibration I think. It's just movement. Music is vibtration, your voice is a vibtration Becasue it makes sound. If I answered this question incorrectly let me know

Nesterboy [21]3 years ago

6 0

Answer:

One cannot hear sound when moving hands in air because moving hands in air is not considered as vibration. This is because the movement is very slow.

The slowest vibration that human ear an detect is 20 times a second. Then it can be considered as vibration as the movement of hand in air is less than this speed so it is detected by ear.

The fastest vibration that can be heard by ear is 20,000 time per second. So, the answer is moving hands in air is not vibration so it is not heard by ear.

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(a) What is the minimum width of a single slit (in multiples of λ ) that will produce a first minimum for a wavelength λ ? (b) W

Kitty [74]

Answer:

The minimum value of width for first minima is λ

The minimum value of width for 50 minima is 50λ

The minimum value of width for 1000 minima is 1000λ

Explanation:

Given that,

Wavelength = λ

For D to be small,

We need to calculate the minimum width

Using formula of minimum width

$D\sin\theta=n\lambda$

$D=\dfrac{n\lambda}{\sin\theta}$

Where, D = width of slit

$\lambda$ = wavelength

Put the value into the formula

$D=\dfrac{n\lambda}{\sin\theta}$

Here, $\sin\theta$ should be maximum.

So. maximum value of $\sin\theta$ is 1

Put the value into the formula

$D=\dfrac{1\times\lambda}{1}$

$D=\lambda$

(b). If the minimum number is 50

Then, the width is

$D=\dfrac{50\times\lambda}{1}$

$D=50\lambda$

(c). If the minimum number is 1000

Then, the width is

$D=\dfrac{1000\times\lambda}{1}$

$D=1000\lambda$

Hence, The minimum value of width for first minima is λ

The minimum value of width for 50 minima is 50λ

The minimum value of width for 1000 minima is 1000λ

4 0

4 years ago

A small box is held in place against a rough vertical wall by someone pushing on it with a force directed upward at 27 ∘ above t

ycow [4]

Answer:

Explanation:

Applied force, F = 18 N

Coefficient of static friction, μs = 0.4

Coefficient of kinetic friction, μs = 0.3

θ = 27°

Let N be the normal reaction of the wall acting on the block and m be the mass of block.

Resolve the components of force F.

As the block is in the horizontal equilibrium, so

F Cos 27° = N

N = 18 Cos 27° = 16.04 N

As the block does not slide so it means that the syatic friction force acting on the block balances the downwards forces acting on the block .

The force of static friction is μs x N = 0.4 x 16.04 = 6.42 N .... (1)

The vertically downward force acting on the block is mg - F Sin 27°

= mg - 18 Sin 27° = mg - 8.172 ... (2)

Now by equating the forces from equation (1) and (2), we get

mg - 8.172 = 6.42

mg = 14.592

m x 9.8 = 14.592

m = 1.49 kg

Thus, the mass of block is 1.5 kg.

6 0

4 years ago

When the medium is uniform, how do light waves pass through it?

LekaFEV [45]

The correct answer is, A) Straight line motion

I took the quiz

8 0

4 years ago

if a spring has a spring constant of 2 N/m and it is stretched 5 cm, what is the force of the spring?

djyliett [7]

Answer:

0.1 N

Explanation:

Considering the relationship between force,

spring constant and extension as defined by Hook's law

The force F=xk as from Hooke's law where F is the force of the spring, k is spring constant and x is extension or compression. Substituting 2 N/m for k and 5cm which is equivalent to 0.05 m for extention x then the force will be

F=2*0.05=0.1 N

4 0

3 years ago

the angular speed of an automobile engine is increased at a constant rate from 1300rev/min to 2000rev/min in 3s (a) what is its

GalinKa [24]

Answer:

please find attached pdf

Explanation:

Download pdf

8 0

3 years ago