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mezya [45]
3 years ago
11

Starting from rest at the top, a child slides down the water slide at a swimming pool and enters the water at a final speed of 4

.64 m/s. at what final speed would the child enter the water if the water slide were twice as high? ignore friction and resistance from the air and the water lubricating the slide.
Physics
1 answer:
yuradex [85]3 years ago
8 0
Energy conservation law states:
E_{total} = E_{potential} + E_{kinetic}
At the top potential energy is at maximum and kinetic energy is zero. At the bottom potential energy is zero and kinetic energ is at maximum.

We have:
E_{total-top} = E_{total-bottom} \\ E_{potential-top} + E_{kinetic-top} =E_{potential-bottom} + E_{kinetic-bottom} \\ E_{potential-top} =E_{kinetic-bottom} \\ m*g*h = \frac{1}{2} *m* v^{2} \\ g*h = \frac{1}{2} * v^{2}

We can use this formula to calculate height:
h= \frac{\frac{1}{2} * v^{2} }{g}  \\ h= \frac{\frac{1}{2} * 4.64^{2} }{9.81}  \\ h=1.09m

To calculate requested speed we just need to double the height and insert it into equation:
v= \sqrt{2*g*2*h}  \\ v=\sqrt{2*9.81*2*1.09} \\ v=6.54 m/s
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A car accelerates from rest at 1.0 m/s2 for 20.0 second along a straight road. It then moves at a constant speed for half an hou
Whitepunk [10]

Total distance = 36500 m

The average velocity = 19.73 m/s

<h3>Further explanation</h3>

Given

vo=initial velocity=0(from rest)

a=acceleration= 1 m/s²

t₁ = 20 s

t₂ = 0.5 hr = 1800 s

t₃= 30 s

Required

Total distance

Solution

State 1 : acceleration

\tt d=vo.t+\dfrac{1}{2}at^2\\\\d=\dfrac{1}{2}\times 1\times 20^2\rightarrow vo=0\\\\d=200~m

\tt vt=vo+at\\\\vt=at\rightarrow vo=0\\\\vt=1\times 20\\\\vt=20~m/s

State 2 : constant speed

\tt d=v\times t\\\\d=20\times 1800\\\\d=36000~m

State 3 : deceleration

\tt vt=vo+at\rightarrow vt=0(stop)\\\\vo=-at\\\\20=-a.30~s\\\\a=-\dfrac{2}{3}m/s^2(negative=deceleration)

\tt d=vot+\dfrac{1}{2}at^2\\\\d=20.30-\dfrac{1}{2}.\dfrac{2}{3}.30^2\\\\d=300~m

Total distance : state 1+ state 2+state 3

\tt 200 + 36000 + 300=36500~m

the average velocity = total distance : total time

\tt avg~velocity=\dfrac{36500}{20~s+1800~s+30~s}=19.73~m/s

4 0
2 years ago
PLEASE HELP ASAP
DaniilM [7]

Your answers:

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I'm sorry I cannot answer all your questions because there's a typing limit. :)

Thank you for your patience, and I hope I helped!

Would appreciate Brainliest ;)

4 0
3 years ago
NEED HELP FAST ILL MARK BRAINLIEST AND RATE 5 STARS AND SAY THANK YOU
daser333 [38]

Acceleration = (change in speed)/(time for the change)

Change in speed = (end speed) - (start speed)

Change in speed = (10 m/s) - (20 m/s) = -10 m/s

Time for the change = 5.00 seconds

Acceleration = (-10 m/s) / (5 sec)

<em>Acceleration = -2 m/s²</em>

That's choice-A .

8 0
3 years ago
Read 2 more answers
How long does it take a wheel that is rotating at 33.3 rpm to speed up to 78.0 rpm if it has an angular acceleration of 2.15 rad
Rzqust [24]

initial angular speed is given by 33.3 rpm

w_0 = 2\pi \frac{33.3}{60}

w_0 = 3.49 rad/s

final angular speed is given by 78 rpm

w_f = 2\pi \frac{78}{60}

w_f = 8.17 rad/s

now by using kinematics we will have

w_f = w_0 + \alpha * t

8.17 = 3.49 + 2.15 * t

t = 2.17 seconds

7 0
3 years ago
In a "Rotor-ride" at a carnival, people rotate in a vertical cylindrically walled "room." If the room radius is 5.5 m, and the r
Crank

Answer:

0.181

Explanation:

We can convert the 0.5 rps into standard angular velocity unit rad/s knowing that each revolution is 2π:

ω = 0.5 rps = 0.5*2π = 3.14 rad/s

From here we can calculate the centripetal acceleration

a_c = \omega^2r = 3.14^2*5.5 = 54.3 m/s^2

Using Newton 2nd law we can calculate the centripetal force that pressing on the rider, as well as the reactive normal force:

F = N = a_cm = 54.3 m

Also the friction force and friction acceleration

F_f = N\mu = 54.3 m \mu N

a_f = F_f / m = 54.3 \mu

For the rider to not slide down, friction acceleration must win over gravitational acceleration g = 9.81 m/s2:

g = a_f = 54.3 \mu

9.81 = 54.3 \mu

\mu = 9.81 / 54.3 = 0.181

6 0
3 years ago
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