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2 years ago

The speed of sound in aluminum is 5200 m/s. Can you hear a sound with a

Physics

1 answer:

romanna [79]2 years ago

5 0

Answer:

v = wavelength * frequency

frequency = 5200 m/s / .2 m = 26000 / sec

20,000 / sec is optimistic for the upper frequency of human hearing

So 26,000 is above the hearing range for human ears

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Two sources of light of wavelength 700 nm are 9 m away from a pinhole of diameter 1.2 mm. How far apart must the sources be for

Lelechka [254]

Answer:

The distance is $D = 0.000712 \ m$

Explanation:

From the question we are told that

The wavelength of the light source is $\lambda = 700 \ nm = 700 *10^{-9} \ m$

The distance from a pin hole is $x = 9\ m$

The diameter of the pin hole is $d = 1.2 \ mm = 0.0012 \ m$

Generally the distance which the light source need to be in order for their diffraction patterns to be resolved by Rayleigh's criterion is

mathematically represented as

$D = \frac{1.22 \lambda }{d }$

substituting values

$D = \frac{1.22 * 700 *10^{-9} }{ 0.0012 }$

$D = 0.000712 \ m$

5 0

2 years ago

What is the fastest motion that can be measured in any frame of reference?

Furkat [3]

Answer: The answer is D: 300,000km/s

Explanation:

7 0

2 years ago

I would love to stretch a wire from our house to the Shop so I can 'call' my husband in for meals. The wire could be tightened t

dezoksy [38]

Note: I'm not sure what do you mean by "weight 0.05 kg/L". I assume it means the mass per unit of length, so it should be "0.05 kg/m".

Solution:
The fundamental frequency in a standing wave is given by
$f= \frac{1}{2L} \sqrt{ \frac{T}{m/L} }$
where L is the length of the string, T the tension and m its mass. If we plug the data of the problem into the equation, we find
$f= \frac{1}{2 \cdot 24 m} \sqrt{ \frac{240 N}{0.05 kg/m} }=1.44 Hz$

The wavelength of the standing wave is instead twice the length of the string:
$\lambda=2 L= 2 \cdot 24 m=48 m$

So the speed of the wave is
$v=\lambda f = (48 m)(1.44 Hz)=69.1 m/s$

And the time the pulse takes to reach the shop is the distance covered divided by the speed:
$t= \frac{L}{v}= \frac{24 m}{69.1 m/s}=0.35 s$

7 0

3 years ago

In a game of egg-toss, you and a partner are throwing an egg back and forth trying not to break it. Given your knowledge of mome

lutik1710 [3]

F=dP/dt. So you want the momentum to change as slowly as possible in time to minimize the force. So as you catch the egg, let your hand move backward with it for awhile, slowly bringing it to a stop. If you hold your hand steady when you catch it the force due to the impact could break it.

3 0

2 years ago

Ryan is driving home from work and notices a deer leaping onto the road about 25 m in front of his car. He immediately applies t

Anvisha [2.4K]

Answer:

mu = 0.56

Explanation:

The friction force is calculated by taking into account the deceleration of the car in 25m. This can be calculated by using the following formula:

$v^2=v_0^2+2ax\\$

v: final speed = 0m/s (the car stops)

v_o: initial speed in the interval of interest = 60km/h

= 60(1000m)/(3600s) = 16.66m/s

x: distance = 25m

BY doing a the subject of the formula and replace the values of v, v_o and x you obtain:

$a=\frac{v^2-v_o^2}{2x}=\frac{0m^2/s^2-(16.66m/s)^2}{2(25m)}=-5.55\frac{m}{s^2}$

with this value of a you calculate the friction force that makes this deceleration over the car. By using the Newton second's Law you obtain:

$F_f=ma=(1490kg)(5.55m/s^2)=8271.15N$

Furthermore, you use the relation between the friction force and the friction coefficient:

$F_f= \mu N=\mu mg\\\\\mu=\frac{F_f}{mg}=\frac{8271.15N}{(1490kg)(9.8m/s^2)}=0.56$

hence, the friction coefficient is 0.56

6 0

2 years ago