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padilas [110]
3 years ago
10

swimmers at a water park have a choice of two frictionless water slides as shown in the figure. although both slides drop over t

he same height, h, slide 1 is straight while slide 2 is curved, dropping quickly at first and then leveling out. how does the speed v1 of a swimmer reaching the end of slide 1 compares with v2, the speed of a swimmer reaching the end of slide 2?
Physics
1 answer:
LuckyWell [14K]3 years ago
4 0
If swimmers had a choice of the water slides shown in this figure,
they would all go home dry, since there is no figure.  I'll have to try to
answer this question based on only the words in the text, augmented
only by my training, education, life experience, and human logic.

-- Both slides are frictionless.  So no energy is lost as a swimsuit
scrapes along the track, and the swimmer's kinetic energy at the
bottom is equal to the potential energy he had at the top.

-- Both slides start from the same height.  So the same swimmer
has the same potential energy at the top of either one, and therefore
the same kinetic energy at the bottom of either one.

-- So the difference in the speeds of two different swimmers
on the slides depends only on the difference in the swimmers'
mass, and is not influenced by the shape or length of the slides
(as long as the slides remain frictionless).

If both swimmers have the same mass, then  v₁ = v₂ .
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<
goblinko [34]

Answer:

Explanation:

initial velocity u = 32.7 m /s

final velocity v = 50.3 m /s

displacement s = 44500 m

acceleration a = ?

v² = u² + 2 a s

50.3² = 32.7² + 2 x a x 44500

2530.09 = 1069.29 + 89000a

a .016 m /s²

time taken t = ?

v = u + at

50.3 = 32.7 + .016 t

t = 1100 s

6 0
2 years ago
A marble is dropped from rest and falls for 2.3 seconds. Find its final velocity.
juin [17]

Answer:

23 m/s downward

__________________________________________________________

<em>Taking the downward direction as positive</em>

<u>We are given:</u>

Initial velocity of the marble (u) = 0 m/s

Time interval (t) = 2.3 seconds

Final velocity (v) = x m/s

<u>Solving for the Final velocity:</u>

<u>Acceleration of the Marble:</u>

We know that gravity will make the marble accelerate at a constant acceleration of 10 m/s

<u>Final velocity:</u>

v = u + at                                              [First equation of motion]

x = 0 + (10)(2.3)                                    [replacing the given values]

x = 23 m/s

Hence, after 2.3 seconds, the marble will move at a velocity of 23 m/s in the downward direction

4 0
3 years ago
The weight of a block on the inclined plane is 500 N and the angle of incline is 30 degrees. What is the magnitude of the force
yanalaym [24]

Answer:

250 N

433 N

Explanation:

N = Normal force by the surface of the inclined plane

W = Weight of the block = 500 N

f = static frictional force acting on the block

Parallel to incline, force equation is given as

f = W Sin30

f = (500) Sin30

f = 250 N

Perpendicular to incline force equation is given  

N = W Cos30

N = (500) Cos30

N = 433 N

3 0
3 years ago
Bicyclists in the Tour de France do enormous amounts of work during a race. For example, the average power per kilogram generate
tigry1 [53]

Explanation:

It is given that,

Average power per unit mass generated by Lance, \dfrac{P}{m}=6.5\ W/kg

P=6.5\times 75=487.5\ W

(a) Distance to cover race, d = 160\ km =160\times 10^3\ m

Average speed of the person, v = 11 m/s

If t is the time taken to cover the race.

t=\dfrac{d}{v}

t=\dfrac{160\times 10^3\ m}{11\ m/s}

t = 14545.46 s

Let W is the work done. The relation between the work done and the power is given by :

P=\dfrac{W}{t}

W=P\times t

W=487.5\times 14545.46

W = 7090911.75 J

(b) Since, 1\ J=2.389\times 10^{-4}\ calories

So, in 7090911.75 J, W=7090911.75 \times 2.389\times 10^{-4}

W = 1694.01 J

Hence, this is the required solution.

6 0
3 years ago
A truck on a straight road starts from rest, accelerating at 2.00m/s^2 until it reaches a speed of 20.0m/s. Then the truck trave
Dennis_Churaev [7]

Answer:

Explanation:

a )

Time to reach the speed of 20 m/s with an acceleration of 2 m/s² can be calculated as follows .

v = u + a t

20 = 0 + 2 t

t = 20 /2 = 10 s .

Total time = 10 s + 20 s + 5 s = 35 s .

b) Average velocity = Total distance travelled / total time

Distance travelled in first 10 s

S₁ = ut + 1/2 a t²

= 0 + .5 x 2 x 10²

= 100 m

Distance travelled in next 20 s

S₂= 20s x 20 m/s  = 400 m

Distance travelled in last 5 s .

deceleration in last 5 s

v = u + at

0 = 20 m/s + a x 5

a = - 4 m/s²

v² = u² - 2 a s

0 = (20 m/s)² - 2 x 4 m/s² x s

s = 50 m

S₃ = 50 m

Total distance = S₁ + S₂ + S₃

= 100 m + 400 m + 50 m

= 550 m .

Average velocity = 550 m / 35 s

= 15.71 m /s .

3 0
3 years ago
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