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3 years ago

What is an non example of Cultural diffusion

Physics

1 answer:

belka [17]3 years ago

4 0

Beliefs is a non example

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When we use an analogy that represents the expanding universe with the surface of an expanding balloon, what does the inside of

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Answer: When we use an analogy that represents the expanding universe with the surface of an expanding balloon, what does the inside of the balloon represent? The inside of the balloon does not represent any part of our universe.

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3 years ago

If we want to see our full image then the minimum size of the plane mirror:

raketka [301]

It would be D I believe! Depending on the angle of the mirror and distance positioned!

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3 years ago

Calculate the work done on a Pressure / Volume (PV) isothermal expansion where 400 Pa and 0.08 volume in m3?

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Hope this helps

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3 years ago

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

Delvig [45]

To develop this problem it is necessary to apply the concept of Frequency based on speed and wavelength.

According to the definition the frequency can be expressed as

$f = \frac{v}{\lambda}$

Where,

v = Velocity

$\lambda =$ Wavelength

Our value are given by,

v = 345m/s

$\lambda = 63cm$

Replacing

$f = \frac{345}{0.63}$

$f = 547.61Hz$

Therefore the frequency of the tuning fork is 547.61Hz

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3 years ago

It takes a minimum distance of 57.46 m to stop a car moving at 13.0 m/s by applying the brakes (without locking the wheels). Ass

vivado [14]

Answer:

The minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

Explanation:

We know by equation of motion that,

$v^{2}=u^{2}+2\cdot a \cdot s$

Where, v= final velocity m/sec

u=initial velocity m/sec

a=Acceleration m/ $Sec^{2}$

s= Distance traveled before stop m

Case 1

u= 13 m/sec, v=0, s= 57.46 m, a=?

$0^{2} = 13^{2} + 2 \cdot a \cdot57.46$

a = -1.47 m/ $Sec^{2}$ (a is negative since final velocity is less then initial velocity)

Case 2

u=29 m/sec, v=0, s= ?, a=-1.47 m/ $Sec^{2}$ (since same friction force is applied)

$v^{2} = 29^{2} - 2 \cdot 1.47 \cdot S$

s = 285.94 m

Hence the minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

4 0

3 years ago