What are the two reactants in this equation?

A LOT OF POINTS! In your own words, compare and contrast transverse and longitudinal waves. Be sure to include at least one simi

tatiyna

Transverse Waves: Displacement of the medium is perpendicular to the direction of propagation of the wave.

To understand this it is good to think of a rope being held still by person B and being moved up and down by person A. The direction of propagation is from person A to B, so you will see the waves move along this way. But the displacement will be up and down.

Can travel in solids, but not in liquids and gas.

eg. Electromagnetic radiation

Longitudinal Waves: Displacement of the medium is parallel to the direction of propagation of the wave.

A good example for this is a slinky being pushed along the table, the propagation will be along the table and so will the displacement of all the 'rings'.

Can travel through all states of matter.

eg. Sound waves

5 0

3 years ago

When charged objects are brought close (not touching) to uncharged objects, what occurs?

seraphim [82]

They will become equally negative charges, I believe.

6 0

3 years ago

A 200kg bumper car moving 3 m/s collides head on with a 392kg bumper car moving 6m/s. After impact, the larger car continues tra

Maru [420]

Answer:

the final speed of the smaller car is 5.624 m/s

Explanation:

Given;

mass of the small car, m₁ = 200 kg

initial velocity of the small car, u₁ = 3 m/s

mass of the larger car, m₂ = 392 kg

initial velocity of the larger car, u₂ = 6 m/s

final velocity of the larger car, v₂ = 1.6 m/s

let the direction of the larger car be positive

let the direction of the smaller car be negative

Apply the principle of conservation of linear momentum to determine the final speed of the smaller car.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

200(-3) + 392(6) = 200v₁ + 392 x 1.6

-600 + 2352 = 200v₁ + 627.2

1752 = 200v₁ + 627.2

1752 - 627.2 = 200v₁

1124.8 = 200v₁

v₁ = 1124.8/200

v₁ = 5.624 m/s

Therefore, the final speed of the smaller car is 5.624 m/s

7 0

3 years ago

A ball is droped from a height of 16m how much time will pass before the ball hits the ground

sergey [27]

Answer:

The time is 1.8s

Explanation:

The ball droped, will freely fall under gravity.

Hence we use free fall formula to calculate the time by the ball to hit the ground

$h= \frac{1}{2}g{t}^{2}$

Where h is the height from which the ball is droped, g is the acceleration due to gravity that acted on the ball, and t is the time taken by the ball to hit the ground.

From the question,

h=16m

Also, let take

$g = 9.8m{s}^{-2}$

By substitution we obtain,

$16= \frac{1}{2}\times 9.8{t}^{2}$

$\implies32=9.8{t}^{2}$

Diving through by 9.8

$\frac{32}{9.8}= \frac{ 9.8{t}^{2} }{9.8}$

$\implies{t}^{2} =3.265$

square root both sides, we obtain

$\implies t= \sqrt{3.265}$

$t=1.8s$

4 0

4 years ago

Coherent light with wavelength = 600 nm falls on two very narrow closely spaced slits and the interference pattern is observed o

Mrac [35]

Answer:

The distance between the two slits is 1.2mm.

Explanation:

The physicist Thomas Young establishes, through its double slit experiment, a relationship between the interference (constructive or destructive) of a wave, the separation between the slits, the distance between the two slits to the screen and the wavelength.

$\Lambda x = L\frac{\lambda}{d}$ (1)

Where $\Lambda x$ is the distance between two adjacent maxima, L is the distance of the screen from the slits, $\lambda$ is the wavelength and d is the separation between the slits.

If light pass through two slits a diffraction pattern in a screen will be gotten, at which each bright region corresponds to a crest, a dark region to a trough, as consequence of constructive interference and destructive interference in different points of its propagation to the screen.

Therefore, d can be isolated from equation 1.

$d = L\frac{\lambda}{\Lambda x}$ (2)