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cricket20 [7]
3 years ago
11

What happens to the water particles in a wave?

Physics
2 answers:
lys-0071 [83]3 years ago
8 0

Answer: waves transport energy, not water. As a wave crest passes, the water particles move in circular paths. The movement of the floating inner tube is simulacra to the movement of the water particles. Water particles rise as a wave crest approaches.

Explanation:

bagirrra123 [75]3 years ago
7 0

Answer:

the answer is A.) The water particles move in a circular pattern.

Explanation:

You might be interested in
One particular descent goes from 2100m to 1600m. Assuming work done against friction is 90% of the potential energy change of th
alisha [4.7K]

Answer:

1/2 M V^2 = .1 M g H       where 10% of PE goes into KE

V^2 = .2 g H = .2 * 9.8 * (2100 - 1600) = 980 m^2 / s^2

V = 31.1 m/s       increase in speed during descent

1 km / hr = 1000 m / 3600 sec = .278 m/s

V = 31.1 m/s / (.278 m/s / km /hr)= 112 km/hr

7 0
2 years ago
Find the west component of 45 m 19º S of W
BaLLatris [955]

Answer:

The west component of the given vector is - 42.548 meters.

Explanation:

We need to translate the sentence into a vectoral expression in rectangular form, which is defined as:

(x, y) = (r_{x}, r_{y})

Where:

r_{x} - Horizontal component of vector distance, measured in meters.

r_{y} - Vertical component of vector distance, measured in meters.

Let suppose that east and north have positive signs, then we get the following expression:

(x, y) = (-45\cdot \cos 19^{\circ}, -45\cdot \sin 19^{\circ})\,[m]

(x, y) = (-42.548,-14.651)\,[m]

The west component corresponds to the first component of the ordered pair. That is to say:

x = -42.548\,m

The west component of the given vector is - 42.548 meters.

8 0
3 years ago
A cutting tool several forces acting on it. One force is F=-axy^2 j , a force in the negative y-direction whose magnitude depend
liq [111]
The force on the tool is entirely in the negative-y direction.
So no work is done during any moves in the x-direction.

The work will be completely defined by

                     (Force) x (distance in the y-direction),

and it won't matter what route the tool follows to get anywhere. 
Only the initial and final y-coordinates matter.

We know that    F = - 2.85 y².  (I have no idea what that ' j ' is doing there.)
Remember that 'F' is pointing down.

From  y=0  to  y=2.40  is a distance of  2.40  upward.

Sadly, since the force is not linear over the distance, I don't think
we can use the usual formula for  Work = (force) x (distance).
I think instead we'll need to integrate the force over the distance,
and I can't wait to see whether I still know how to do that.

        Work  =  integral of (F·dy) evaluated from  0  to  2.40

                  =  integral of (-2.85 y² dy) evaluated from  0  to  2.40

                 =  (-2.85) · integral of  (y² dy)  evaluated from  0  to  2.40 .


Now, integral of (y² dy)  =  1/3  y³ .

Evaluated from  0  to  2.40 , it's  (1/3 · 2.40³) - (1/3 · 0³)

                                            =  1/3 · 13.824  =  4.608 .

And the work  =  (-2.85) · the integral

                     =  (-2.85) · (4.608)

                     =      - 13.133  .

-- There are no units in the question (except for that mysterious ' j ' after the 'F',
which totally doesn't make any sense at all).
If the ' F ' is newtons and the 2.40 is meters, then the  -13.133 is joules.

-- The work done by the force is negative, because the force points
DOWN but we lifted the tool UP to 2.40.  Somebody had to provide
13.133 of positive work to lift the tool up against the force, and the force
itself did 13.133 of negative work to 'allow' the tool to move up. 

-- It doesn't matter whether the tool goes there along the line  x=y , or
by some other route.  WHATEVER the route is, the work done by ' F ' 
is going to total up to be  -13.133 joules at the end of the day.

As I hinted earlier, the last time I actually studied integration was in 1972,
and I haven't really used it too much since then.  But that's my answer
and I'm stickin to it.  If I'm wrong, then I'm wrong, and I hope somebody
will show me where I'm wrong.
3 0
3 years ago
The magnitude of the electric field between two parallel charged plates is 200. An electron moves to the negative plate 5. 0 cm
Mama L [17]

The potential difference between the two ends of the circuit is the electric potential difference. The electric potential difference and the work will be 10V and 1.6 x 10^-18 J respectively.

<h3>What is an electric field?</h3>

An electric field is an electric property that is connected with any location in space where a charge exists in any form. The electric force per unit charge is another term for an electric field.

The given data in the problem is given by;

E is the electric field = (200 N/C)

d is the distance = 5.0 cm.=0.05 m

Q is the charge of electrons= 1.602 x 10^-19 C

The formula for electric potential  is given by;

\rm V=Ed

\rm V=Ed \\\\ \rm V=200 \times 0.05 \\\\ \rm V=  10 \frac{Nm}{C} = 10 \frac{J}{C}  = 10 V.

The work is defined as the product of the potential difference and charge of an electron.

\rm W= 10 \times  1.602 x 10^{-19} \\\\\ \rm W=  1.6 x 10^{-18 }J

Hence the electric potential difference and the work will be 10V and 1.6 x 10^-18 J respectively.

To learn more about the electric field refer to the link;

brainly.com/question/15071884

8 0
2 years ago
Which are true about the proton?
Scrat [10]

Answer:

B

Explanation:

bIrish Kaleb picked Lucian usual

4 0
3 years ago
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