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3 years ago

How are surface waves similar to transverse waves?

1 answer:

7 0

In longitudinal and transverse waves<span>, all the particles in the entire bulk of the medium move in a parallel and a perpendicular direction (respectively) relative to the direction of energy transport. In a </span>surface wave, it is only the particles at the surface<span> of the medium that undergo the circular motion.</span>

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(SCIENCE)

mart [117]

Mean is the same thing as average, so to find this you would add all of the numbers together and divide them by the amount of numbers you added. For example, (6+9+11)/3

Median is the middle number of the list of numbers, but you have to put them in numerical order first. If there is not an exact middle, you would add the middle two together and divide by two.

Mode is the number that occurs most often within the list of numbers.

3 0

4 years ago

Which technique(s) are appropriate to completely factor the expression? A) Factor as a difference of squares. B) Factor out the

Ilya [14]

Answer:

The greatest common factor (GCF) is the greatest factor that is common to two or more numbers (they share it). The greatest common factor of two (or more) numbers is the product of all the prime factors the numbers have in common.

Explanation:

4 0

3 years ago

A student made a model of an onion skin cell she viewed using a microscope. The scale is 1:500.The students cell model has a len

r-ruslan [8.4K]

The original will be 15\500 to find the real length

6 0

3 years ago

Read 2 more answers

When separated by distance d, identically charged point-like objects A and B exert a force of magnitude F on each other. If you

Soloha48 [4]

Answer:

F = F₀ 0.2

Explanation:

For this exercise we apply Coulomb's law with the initial data

F₀ = k q_A q_B / d²

indicate several changes

q_A ’= ½ q_A

q_B ’= 1/10 q_B

d ’= ½ d

let's substitute these new values in the Coulomb equation

F = k q_A ’q_B’ / d’²

F = k ½ q_A 1/10 q_B / (1/2 d)²

F = (k q_A q_B / d2) ½ 1/10 2²

F = F₀ 0.2

5 0

4 years ago

The Problems: 1. Xavier starts at a position of 0 m and moves with an average speed of 0.50 m/s for 3.0 seconds. He normally mov

NemiM [27]

Answer:

(1). His final position is 1.5 m.

(2). The final position of the hedgehog is 3 m.

(3). The final position of the tortoise

(4). Her race time is 80 sec.

(5). It take to finish in 5 hr.

Explanation:

(1). Given that,

Initial position = 0 m

Average speed = 0.50 m/s

Time = 3.0 s

We need to calculate the final position

Using formula of average speed

$v_{av}=\dfrac{x_{f}+x_{i}}{t}$

Where, $x_{f}$ = final position

$x_{i}$ = Initial position

t = total time

Put the value into the formula

$0.50=\dfrac{x_{f}+0}{3.0}$

$x_{f}=0.50\times3.0$

$x_{f}=1.5\ m$

(2). Given that,

Initial position = 0 m

Average speed = 0.75 m/s

Time = 4.0 s

We need to calculate the final position

Using formula of average speed

$v_{av}=\dfrac{x_{f}+x_{i}}{t}$

Put the value into the formula

$0.75=\dfrac{x_{f}+0}{4.0}$

$x_{f}=0.75\times4.0$

$x_{f}=3\ m$

(3). Given that,

Average speed = 1.25 m/s

Time = 3.0 sec

Initial position = 1.0 m

We need to calculate the final position

Using formula of average speed

$v=\dfrac{x_{f}+x_{i}}{t}$

Put the value into the formula

$1.25=\dfrac{x_{f}+1.0}{3.0}$

$x_{f}=1.25\times3.0-1.0$

$x_{f}=2.75\ m$

(4). Given that,

Average speed = 1.25 m/s

Distance = 100 m

We need to calculate the time

Using formula of time

$t=\dfrac{d}{v}$

Put the value into the formula

$t=\dfrac{100}{1.25}$

$t=80 sec$

(5). Given that,

Average speed = 5 miles/hr

Suppose, distance = 25 miles

We need to calculate the time

Using formula of time

$t=\dfrac{d}{v}$

Put the value into the formula

$t=\dfrac{25}{5}$

$t=5\ hr$

Hence, (1). His final position is 1.5 m.

(2). The final position of the hedgehog is 3 m.

(3). The final position of the tortoise

(4). Her race time is 80 sec.

(5). It take to finish in 5 hr.

5 0

4 years ago