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3 years ago

Please help, What does a Transverse and longitude wave combine to form?

2 answers:

4 0

Electromagnetic waves transfer energy without going through a medium. ... Sometimes, a transverse wave and a longitudinal wave can combine to formanother kind of wave called a surface wave. Transverse Waves. Waves in which the particles vibrate in an up-and-down motion

yKpoI14uk [10]3 years ago

4 0

It is a surface wave i think, if that helps

You might be interested in

Properties that describe the appearance of matter are known as what properties

Marina86 [1]

Physical properties is the answer

3 0

3 years ago

A car traveling at 91.0 km/h approaches the turn off for a restaurant 30.0 m ahead. If the driver slams on the brakes with the a

eduard

Answer: 49.92 m

Explanation:

In this situation the following equation will be useful:

$V^{2}=V_{o}^{2} +2 a d$

Where:

$V=0 m/s$ is the final velocity of the car, when it finally stops

$V_{o}=91 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=25.27 m/s$ is the initial velocity of the car

$a=-6.4 m/s^{2}$ is the constant acceleration of the car after the driver slams on the brakes

$d$ is the stopping distance

Isolating $d$ :

$d=\frac{-V_{o}^{2}}{2a}$

$d=\frac{-(25.27 m/s)^{2}}{2(-6.4 m/s^{2})}$

$d=41.919 m \approx 41.92 m$

7 0

4 years ago

A book is sitting in a table. Jackie is pushing the book with a force of 9 N. If the force of friction is 4 N to the left, what

Bumek [7]

Answer:

The net force is 5N

Explanation:

A shown in the diagram,

we let P be the force Jackie applied and

$F_r$

be the frictional force.

To get the net force, we subtract the frictional force from the applied force.

This implies that,

$F_n= P-F_r$

$F_n= 9 - 4$

$F_n=5N$

4 0

3 years ago

A hot-air balloon is descending at a rate of 2.3 m/s when a pas- senger drops a camera. If the camera is 41 m above the ground w

monitta

Answer:

a) time taken = 2.66 s

b) v = 28.34

Explanation:

given,

rate of descending = 2.3 m/s

height of camera above ground = 41 m

using equation of motion

$h = u t + \dfrac{1}{2}gt^2$

$41 =2.3t + \dfrac{1}{2}\times 9.8\times t^2$

4.9 t² + 2.3 t - 41 =0

t = 2.66 ,-3.13

time taken = 2.66 s

b) v² = u² + 2 g h

v² = 0 + 2× 9.8 × 41

v = 28.34

3 0

4 years ago

a horizontal string is stretched between two points a Distance 0.80 metre apart the tension in the string is 90 Newtonand it is

notka56 [123]

Answer:

a) v = 126.5 m / s, b) λ₁ = 1.6 m, λ₂ = 0.8 m, λ₃ = 0.533 m, f1 = 79 Hz,

λ₂ = 0.8 m, f₃ = 237 Hz

Explanation:

This is an exercise we are going to solve in parts, let's start by looking for the speed of the wave in the string.

v = $\sqrt { \frac{T}{ \mu } }$

the rope tension is T = 90 N and the density can be calculated

μ = m / l

let's calculate

μ = 4.5 10⁻³ / 0.80

μ = 5.625 10⁻³ kg/m

let's calculate the speed

v = $\sqrt{ \frac{90 }{5.625 \ 10^{-3}} }$

v = 126.5 m / s

For the second part there is a process of resonance in the string. The points where it is attached are nodes so, if L is the length of the chord

L = ½ λ 1st harmonic

L = 2/2 λ 2nd harmonic

L = 3/2 λ 3rd harmonic

L = n/2 λ n harmonic

the wavelength of the first three harmonics is requested

let's calculate

λ₁ = 2L

λ₁ = 2 0.8

λ₁ = 1.6 m

λ₂ = L

λ₂ = 0.8 m

λ₃ = ⅔ L

λ₃ = ⅔ 0.8

λ₃ = 0.533 m

To calculate the frequency we use that the speed is related to the wavelength and the frequency

v = lam f

f = v / lam

we calculate

f1 = 126.5 / 1.6

f1 = 79 Hz

f2 = 126.5 / 0.8

f2 = 158 Hz

f3 = 126.5 / 0.533

f3 = 237 Hz

8 0

3 years ago