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2 years ago

13

The speed of sound in wood is ____ the speed of sound in water.

1 answer:

pishuonlain [190]2 years ago

8 0

Answer:

greater

Explanation:

the speed of sound in steel is greater than water. the speed of sound in wood is not, In water, the particles are much closer together, and they can quickly transmit vibration energy from one particle to the next. This means that the sound wave travels over four times faster than it would in air, but it takes a lot of energy to start the vibration.

Wood is less dense and force can make a sound.

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A branch falls from a tree how fast is the branch moving after drug .28 seconds

alexdok [17]

Acceleration=9.81m/s^2
initial velocity=0m/s
time=.28s

We have to find final velocity.

The equation we use is

Final velocity=initial velocity+acceleration x time

Vf=0m/s+(9.81m/s^2)(.28s)
Vf=2.7468m/s

We would round this to:

Vf (final velocity)=2.7m/s

6 0

2 years ago

As a woman walks, her entire weight is momentarily placed on one heel of her high-heeled shoes. Calculate the pressure exerted o

astraxan [27]

Answer:

3.6 x 10⁶ Pa

Explanation:

A = Area of the heel = 1.50 cm² = 1.50 x 10⁻⁴ m²

m = mass of the woman = 55.0 kg

g = acceleration due to gravity = 9.8 m/s²

Force of gravity on the heel is given as

F = mg

Inserting the values

F = (55) (9.8)

F = 539 N

Pressure exerted on the floor is given as

$P = \frac{F}{A}$

$P = \frac{539}{1.5\times 10^{-4}}$

P = 3.6 x 10⁶ Pa

6 0

2 years ago

Which of these energy transformations is not part of how a nuclear power plant

frutty [35]

I think the second one

7 0

2 years ago

A surgical microscope weighing 200 lb is hung from a ceiling by four springs with stiffness 25 lb/ft. The ceiling has a vibratio

Nikitich [7]

Answer:

If there is no damping, the amount of transmitted vibration that the microscope experienced is = $5.676*10^{-3} \ mm$

Explanation:

The motion of the ceiling is y = Y sinωt

y = 0.05 sin (2 π × 2) t

y = 0.05 sin 4 π t

K = 25 lb/ft × 4 sorings

K = 100 lb/ft

Amplitude of the microscope $\frac{X}{Y}= [\frac{1+2 \epsilon (\omega/ W_n)^2}{(1-(\frac{\omega}{W_n})^2)^2+(2 \epsilon \frac{\omega}{W_n})^2}]$

where;

$\epsilon = 0$

$W_n = \sqrt { \frac{k}{m}}$

= $\sqrt { \frac{100*32.2}{200}}$

= 4.0124

replacing them into the above equation and making X the subject of the formula:

$X =$ $Y * \frac{1}{\sqrt{(1-(\frac{\omega}{W_n})^2)^2})}}$

$X =$ $0.05 * \frac{1}{\sqrt{(1-(\frac{4 \pi}{4.0124})^2)^2})}}$

$X =$ $5.676*10^{-3} \ mm$

Therefore; If there is no damping, the amount of transmitted vibration that the microscope experienced is = $5.676*10^{-3} \ mm$

8 0

2 years ago

Please help on this one?

bezimeni [28]

Using the given equation you get:

E = 1.99x10^-25 / 9.0x10^-6

Divide 1.99 by 9.0: 1.99/9.0 = 0.22

For the scientific notation, when dividing subtract the two exponents:

25 -6 = 19

So you now have 0.22 x 10^-19

Now you need to change the 0.22 to be in scientific notation form:

2.2 x 10^-20

The answer is B.

3 0

2 years ago