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pychu [463]
2 years ago
13

Waves on a swimming pool propagate at 0.720 m/s. You splash the water at one end of the pool and observe the wave go to the oppo

site end, reflect, and return in 31.0 s. How far away is the other end of the pool?
Physics
1 answer:
SVEN [57.7K]2 years ago
5 0

Answer:

11.16 m

Explanation:

total distance traveled by the wave = speed * time

                                                            = 0.720 m/s * 31.0 s

                                                            = 22.32 m

Since this is an echo back to the starting point then the length of the swimming pool must be half of the distance traveled.

length of swimming pool  =  22.32 m÷2 = 11.16 m

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Pls help A car starts from rest and gains a velocity of 20m/s in 10 seconds calculate its acceleration and the distance covered
Soloha48 [4]

Answer:

\boxed{\sf Acceleration \ (a) = 2 \ m/s^{2}}

\boxed{\sf Distance \ covered \ (s) = 100 \ m}

Given:

Initial velocity (u) = 0 m/s

Final velocity (v) = 20 m/s

Time taken (t) = 10 sec

To Find:

(i) Acceleration (a)

(ii) Distance covered (s)

Explanation:

\sf (i) \ From \ 1^{st} \ equation \ of \ motion:

\sf \implies v = u + at

\sf \implies 20 = 0 + a(10)

\sf \implies 10a = 20

\sf \implies \frac{10a}{10}  =  \frac{20}{10}

\sf \implies a = 2 \: m/ {s}^{2}

\sf (ii) \ From \ 2^{nd} \ equation \ of \ motion:

\sf \implies s = ut +  \frac{1}{2} a {t}^{2}

\sf \implies s = (0)(10) +  \frac{1}{2}  \times 2 \times  {(10)}^{2}

\sf \implies s =  \frac{1}{ \cancel{2}}  \times  \cancel{2} \times  {(10)}^{2}

\sf \implies s =  {10}^{2}

\sf \implies s = 100 \: m

6 0
3 years ago
Alexis is studying how lenses work. She looks through a
Kryger [21]
It would be the 3rd one
6 0
3 years ago
How long would it take a machine to do 5.000
Gala2k [10]

Answer:

a

Explanation:

How long would it take a machine to do 5.000

joules of work if the power rating of the machine

is 100 watts?

6 0
3 years ago
Drag the correct labels to the images. Each label can be used more than once.
coldgirl [10]

Answer:

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Explanation:

4 0
3 years ago
Kevin is refinishing his rusty wheelbarrow. He moves his sandpaper back and forth 45 times over a rusty area, each time moving a
dmitriy555 [2]
W = _|....F*dx*cos(a)........With F=force, x=distance over which force acts on object,
.......0.............................and a=angle between force and direction of travel.

Since the force is constant in this case we don't need the equation to be an integral expression, and since the force in question - the force of friction - is always precisely opposite the direction of travel (which makes (a) equal to 180 deg, and cos(a) equal to -1) the equation can be rewritted like so:

W = F*x*(-1) ............ or ............. W = -F*x

The force of friction is given by the equation: Ffriction = Fnormal*(coeff of friction)

Also, note that the total work is the sum of all 45 passes by the sandpaper. So our final equation, when Ffriction is substituted, is:

W = (-45)(Fnormal)(coeff of friction)(distance)
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W = ................-11.178 Joules
5 0
3 years ago
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