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dem82 [27]
2 years ago
8

a sound has a frequency of 230 hz. what is the velocity of this sound if it has a wavelength of 46 cm?​

Physics
1 answer:
lapo4ka [179]2 years ago
6 0

Answer:

V=f×λ

Where,

V is the velocity of the wave measure using m/s.

f is the frequency of the wave measured using Hz.

λ is the wavelength of the wave measured using m.

transform 46 cm in m = 0.46m and there u have it

Explanation:

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Why are sunlight and gravity not considered matter?
Nady [450]

Sunlight and gravity are not considered matter because sunlight does not have  a rest mass that is characteristic of matter while gravity is just a vector component of matter.

<u>Explanation:</u>

Sunlight has energy and momentum that makes it similar to matter but it cannot be considered matter when you consider all the factors that are characteristic of matter. Light is an electromagnetic wave that is made up of photons and  the protons do not have rest mass.

Gravity is  a vector component of matter that has both magnitude as well as direction and the gravity increases with increase in mass of an object.

6 0
3 years ago
Read 2 more answers
Sam stands on a 20 m high cliff and throws a 45 g rock with an initial velocity of 5 m/s [forward] to the water below. Use the c
storchak [24]

Answer:

v = 12.52 [m/s]

Explanation:

To solve this problem we must use the energy conservation theorem. Which tells us that potential energy is transformed into kinetic energy or vice versa. This is more clearly as the potential energy decreases the kinetic energy increases.

Ep = Ek

where:

Ep = potential energy [J] (units of joules]

Ek = kinetic energy [J]

Ep = m*g*h

where:

m = mass of the rock = 45 [g] = 0.045 [kg]

g = gravity acceleration = 9.81 [m/s²]

h = elevation = (20 - 12) = 8 [m]

Ek = 0.5*m*v²

where:

v = velocity [m/s]

The reference level of potential energy is taken as the ground level, at this level the potential energy is zero, i.e. all potential energy has been transformed into kinetic energy. In such a way that when the Rock has fallen 12 [m] it is located 8 [m] from the ground level.

m*g*h = 0.5*m*v²

v² = (g*h)/0.5

v = √(9.81*8)/0.5

v = 12.52 [m/s]

5 0
2 years ago
If vector A =i+2j-k and vec A cross vec B =3i-j+5k. find vec B​​​
postnew [5]

Let <em>B</em> = <em>a</em> <em>i</em> + <em>b</em> <em>j</em> + <em>c</em> <em>k</em>. Then the cross product of <em>A</em> = <em>i</em> + 2<em>j</em> - <em>k</em> with <em>B</em> is

<em>A</em> × <em>B</em> = ( <em>i</em> + 2<em>j</em> - <em>k </em>) × ( <em>a</em> <em>i</em> + <em>b</em> <em>j</em> + <em>c</em> <em>k</em> )

<em>A</em> × <em>B</em>  = <em>a</em> ( <em>i</em> × <em>i</em> ) + 2<em>a</em> ( <em>j</em> × <em>i</em> ) - <em>a</em> ( <em>k</em> × <em>i </em>)

… … … + <em>b</em> ( <em>i</em> × <em>j</em> ) + 2<em>b</em> ( <em>j </em>× <em>j</em> ) - <em>b</em> ( <em>k</em> × <em>j</em> )

… … … + <em>c</em> ( <em>i</em> × <em>k</em> ) + 2<em>c</em> ( <em>j</em> × <em>k</em> ) - <em>c</em> ( <em>k</em> × <em>k</em> )

<em>A</em> × <em>B</em> = 0 - 2<em>a</em> <em>k </em>- <em>a</em> <em>j</em>

… … … + <em>b</em> <em>k</em> + 0 + <em>b</em> <em>i</em>

… … … - <em>c</em> <em>j</em> + 2<em>c</em> <em>i</em> - 0

<em>A</em> × <em>B</em> = (<em>b</em> + 2<em>c</em>) <em>i</em> + (-<em>a</em> - <em>c</em>) <em>j</em> + (<em>b</em> - 2<em>a</em>) <em>k</em>

So we have

3 <em>i</em> - <em>j</em> + 5 <em>k </em>= (<em>b</em> + 2<em>c</em>) <em>i</em> + (<em>c</em> - <em>a</em>) <em>j</em> + (<em>b</em> - 2<em>a</em>) <em>k</em>

which gives us the system of equations,

{ <em>b</em> + 2<em>c</em> = 3

{ -<em>a</em> - <em>c</em> = -1

{ -2<em>a</em> + <em>b</em> = 5

Solve for <em>a</em>, <em>b</em>, and <em>c</em>.

• Eliminate <em>c</em> from the first two equations:

(<em>b</em> + 2<em>c</em>) + 2 (-<em>a</em> - <em>c</em>) = 3 + 2 (-1)

-2<em>a</em> + <em>b</em> = 1

But -2<em>a</em> + <em>b</em> = 5, and 5 ≠ 1, so there is no such vector <em>B</em> that satisfies the cross product!

8 0
2 years ago
CHAPTER 6: KINETICS OF A PARTICLE
vekshin1

Explanation:

Work done by winch = kinetic energy of car

∫ T ds = ½ mv²

∫ 225s ds = ½ mv²

225/2 s² = ½ mv²

225 s² = mv²

v = 15s / √m

Given s = 10 m and m = 2500 kg:

v = 15 (10) / √2500

v = 3 m/s

8 0
3 years ago
A particle leaves the origin with a speed of 3.6 times 106 m/s at 34 degrees to the positive x axis. It moves in a uniform elect
gogolik [260]

Answer:

1.2X10^3NC

Explanation:

Pls see attached file

3 0
3 years ago
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