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2 years ago

What happens when a sound wave passes from water to air?

2 answers:

6 0

Sound travells much faster in water than in air.

The wave is both reflected and tranmitted. Some goes into the water..and goes FASTER and some bounces back.

s344n2d4d5 [400]2 years ago

4 0

It slows down. The molecules/particles aren't as close together, resulting in slower movement.

You might be interested in

1pt Which is an observation? O A. A person hears the song of a bird. OB. A person suggests that bird calls are a means of commun

Doss [256]

Answer:

Explanation:

he describes as he writes them down

3 0

3 years ago

An Alaskan rescue plane traveling 41 m/s drops a package of emergency rations from a height of 192 m to a stranded party of expl

svet-max [94.6K]

Answer:

a)The package strikes 256.2 m in the ground relative to the point directly below where it was released

b) The horizontal component will not change it remains same as 41 m/s

c) Vertical component of velocity = 61.41 m/s

Explanation:

a) Consider the vertical motion of plane,

We have equation of motion, s = ut + 0.5 at²

Initial velocity, u = 0 m/s

Displacement, s = 192 m

Acceleration, a = 9.81 m/s²

Substituting

s = ut + 0.5 at²

192 = 0 x t + 0.5 x 9.81 x t²

t = 6.26 seconds

Now we need to find horizontal distance traveled in 6.26 seconds by the package.

We have equation of motion, s = ut + 0.5 at²

Initial velocity, u = 41 m/s

Time, t = 6.26 s

Acceleration, a = 0 m/s²

Substituting

s = ut + 0.5 at²

s = 41 x 6.26 + 0.5 x 0 x 6.26²

s = 256.52 m

The package strikes 256.2 m in the ground relative to the point directly below where it was released

b) The horizontal component will not change it remains same as 41 m/s

c) We have equation of motion, v = u+ at

Initial velocity, u = 0 m/s

Time, t = 6.26 s

Acceleration, a = 9.81 m/s²

Substituting

v = u+ at

v = 0 + 9.81 x 6.26 = 61.41 m/s

Vertical component of velocity = 61.41 m/s

4 0

3 years ago

A student hangs a weight on a newtonmeter. The energy currently stored in the spring in the newton meter is 0.045N. The student

castortr0y [4]

Answer:

5x10^-3

Explanation:

Hooke's Law states that the force needed to compress or extend a spring is directly proportional to the distance you stretch it.

Hooke's Law can be represented as

<em>where F is the force </em>

<em> k is the spring constant</em>

<em> x is the extension of the material </em>

<em />

Plug values in the equation

Step 1 find the original extension

0.045 = (400)x

x = 1.125x 10^-4 m d

Step 2 find the new extension

0.045+2 = 400(x)

2.045 = 400x

x = 5.1125x10^-3

Step 3 subtract the new extension with original

Total extension of the spring = 5.1125x10^-3 - 1.125x 10^-4 m = 5x10^-3

8 0

3 years ago

The tops of the towers of the Golden Gate Bridge in San franciscoo , are 227 m above the water. Suppose a worker drops a 655 g w

yaroslaw [1]

Answer:

Kinetic energy is 1425.11 J.

Explanation:

Given:

Mass of the wrench is, $m=655\ g=0.655\ kg$

Height of fall is, $h=227\ m$

Force of resistance is, $F=0.141\ N$

Now, the total energy at the top is equal to the potential energy of the wrench at the top since the kinetic energy at the top is 0.

Now, potential energy at the top is given as:

$U=mgh\\U=0.655\times 9.8\times 227\\U=1457.113\ J$

Now, the potential energy at the top is converted to kinetic energy at the bottom and some energy is wasted in overcoming the resistance force by air.

Potential Energy = Kinetic energy + Energy to overcome resistance.

⇒ Kinetic energy = Potential Energy - Energy to overcome resistance.

Energy to overcome resistance force is the work done by the wrench against the resistance force and is given as:

$W_{res}=Fh=0.141\times 227=32.007\ J$

Therefore, Kinetic energy at the bottom is given as:

$K=U-W_{res}\\\\K=1457.113\ J - 32.007\ J\\\\K=1425.106\approx1425.11\ J$

Hence, the kinetic energy of the wrench be when it hits the water is 1425.11 J.

7 0

3 years ago

A solid sphere of radius 40.0 cm has a total positive charge of 16.2 μC uniformly distributed throughout its volume. Calculate t

Jobisdone [24]

Answer:

(a) E=0 : 0 cm from the center of the sphere

(b) E= 227.8*10³ N/C : 10.0 cm from the center of the sphere

(c)E= 911.25*10³ N/C : 40.0 cm from the center of the sphere

(d)E= 411.84 * 10³ N/C : 59.5 cm from the center of the sphere

Explanation:

If we have a uniform charge sphere we can use the following formulas to calculate the Electric field due to the charge of the sphere

$E=\frac{K*Q}{r^{2} }$ : Formula (1) To calculate the electric field in the region outside the sphere r ≥ a

$E=k*\frac{Q}{a^{3} } *r$ :Formula (2) To calculate the electric field in the inner region of the sphere. r ≤ a

Where:

K: coulomb constant

a: sphere radius

Q: Total sphere charge

r : Distance from the center of the sphere to the region where the electric field is calculated

Equivalences

1μC=10⁻⁶C

1cm= 10⁻²m

Data

k= 9*10⁹ N*m²/C²

Q=16.2 μC=16.2 *10⁻⁶C

a= 40 cm = 40*10⁻²m = 0.4m

Problem development

(a)Magnitude of the electric field at 0 cm :

We replace r=0 in the formula (2) , then, E=0

(b) Magnitude of the electric field at 10.0 cm from the center of the sphere

r<a , We apply the Formula (2):

$E=9*10^{9} *\frac{16.2*10^{-6} }{0.4^{3} } *0.1$

E= 227.8*10³ N/C

r=a, We apply the Formula (1) :

$E=\frac{9*10^{9}*16.2*10^{-6} }{0.4^{2} }$

E= 911.25*10³ N/C

(d) Magnitude of the electric field at 59.5 cm from the center of the sphere

r>a , We apply the Formula (1) :

$E=\frac{9*10^{9}*16.2*10^{-6} }{0.595^{2} }$

E= 411.84 * 10³ N/C

4 0

3 years ago