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Mazyrski [523]
3 years ago
5

Compare and contrast light waves and sound waves

Physics
1 answer:
Naddika [18.5K]3 years ago
4 0
The difference between light waves and sound waves are that sound waves travel faster than light wave and have a frequency range while light waves have a to me range.
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Un coche de carreras de Fórmula 1 (véase Figura 2.30) acelera desde el reposo a razón de 18 m.s-2 . Suma que se mueve en línea r
Lynna [10]

Answer:

I will answer in English.

Ok, we know that the acceleration is a = 18m/s^2, and we have that the initial velocity and position are both zero. (because it starts at rest)

then we have:

a(t) = 18m/s^2

for the velocity, we integrate over time (because the initial velocity is equal to zero we do not have any integration constant)

v(t) = (18m/s^2)*t

for the position we integrate again over time, and again, we do not have any integration constant

p(t) = (1/2)(18m/s^2)*t^2 = (9m/s^2)*t^2

a) The speed at t= 3s can be found by replacing t = 3s in the velocity equation.

v(3s) =  (18m/s^2)*3s = 54m/s

b) the distance traveled by this time can be found by replacing t = 3s in the position equation.

p(3s) =(9m/s^2)*(3s)^2 = 81 m

c) first, we need to find what is the time when the position is equal to 200m.

p(t) = 200m =  (9m/s^2)*t^2

√(200/9) s = t = 4.7s

Now we replace that time in the velocity equation and we get:

v (4.7s) =   (18m/s^2)*4.7s = 84.6m/s

d) ok, to do this we know that.

1 hour has: 60*60 = 3600 seconds.

then we have the transformation  k = 1h/3600s

1 km has 1000 meters.

then we have the transformation c = 1km/1000m

so we have that:

84.6m/s = 84.6m/s*(c/k) = 84.6*(3600/1000)km/h = 304.56 km/h

6 0
3 years ago
When does a nebula become a star?
pashok25 [27]
When the contracting gas and dust from nebula become so dense and hot that nuclear fusion starts
7 0
4 years ago
An object has a mass of 50.0 g and a volume of 10.5 cm3. What is the object's density?
seraphim [82]
Volume=mass/density

volume=455.6/19.3

volume=23.6 mL

4 0
3 years ago
A cylindrical rod of copper (E = 110 GPa, 16 × 106 psi) having a yield strength of 240 MPa (35,000 psi) is to be subjected to a
Fynjy0 [20]

Answer:

d= 7.32 mm

Explanation:

Given that

E= 110 GPa

σ = 240 MPa

P= 6640 N

L= 370 mm

ΔL = 0.53

Area A= πr²

We know that  elongation due to load given as

\Delta L=\dfrac{PL}{AE}

A=\dfrac{PL}{\Delta LE}

A=\dfrac{6640\times 370}{0.53\times 110\times 10^3}

A= 42.14 mm²

πr² = 42.14 mm²

r=3.66 mm

diameter ,d= 2r

d= 7.32 mm

4 0
3 years ago
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What's the equation for voltage
Feliz [49]
Http://www.sengpielaudio.com/calculator-ohm.htm
8 0
3 years ago
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