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Verdich [7]
3 years ago
9

A playground slide is in the form of an arc of a circle with a maximum height of 3.0 m, with a radius of 8.5 m, and with the gro

und tangent to the circle. Dr. Ritchey’s 12 kg daughter starts from rest at the top of the slide and has a speed of 4.5 m/s at the bottom. a. What is the length of the slide? b. What average frictional force acts on the child over this distance?
Physics
1 answer:
julia-pushkina [17]3 years ago
4 0

Answer:

a) s \approx 6.676\,m

Explanation:

a) Let assume that the ground is not inclined, since the bottom of the playground slide is tangent to ground. Then, the length of given by the definition of a circular arc:

s = \frac{\pi}{4}\cdot R

s=\frac{\pi}{4}\cdot (8.5\,m)

s \approx 6.676\,m

The bottom of the slide has a height of zero. The physical phenomenon around Dr. Ritchey's daughter is modelled after Principle of Energy Conservation. The child begins at rest:

U_{g,A} = K_{B} + W_{fr}

m\cdot g \cdot h_{A} = \frac{1}{2}\cdot m \cdot v_{B}^{2} + f\cdot s

The average frictional force is cleared within the expression:

f = \frac{m\cdot (g\cdot h_{A}-\frac{1}{2}\cdot v_{B}^{2})}{s}

f = \frac{(12\,kg)\cdot [(9.807\,\frac{m}{s^{2}} )\cdot (3\,m)-\frac{1}{2}\cdot (4.5\,\frac{m}{s} )^{2} ]}{6.676\,m}

f = 34.684\,N

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A bungee jumper with mass 65.0 kg jumps from a high bridge. After reaching his lowest point, he oscillates up and down, hitting
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Explanation:

It is given that,

Mass of a bungee jumper is 65 kg

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T=\dfrac{38}{8}\\\\T=4.75\ s

After many oscillations, he finally comes to rest 25.0 m below the level of the bridge.

For an oscillating object, the time period is given by :

T=2\pi \sqrt{\dfrac{m}{k}}

k = spring stiffness constant

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k=\dfrac{4\pi ^2m}{T^2}\\\\k=\dfrac{4\pi ^2\times 65}{(4.75)^2}\\\\k=113.43\ N/m

When the cord is in air,

mg=kx

x = the extension in the cord

x=\dfrac{mg}{k}\\\\x=\dfrac{65\times 9.8}{113.6}\\\\x=5.6\ m

So, the unstretched length of the bungee cord is equal to 25 m - 5.6 m = 19.4 m

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What was significant about the discovery of gallium
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It confirmed medeleeve's hypothesis (prediction) and showed the use of his table
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A car moves with constant acceieration of -0.s m/s? on a straight portion of the road. Att- 0s the car has a velocity of 69 mph,
Crazy boy [7]

Answer:

b) d = 0.71 Km

Explanation:

Car kinematics

Car 1 moves with uniformly accelerated movement

v_f^2=v_o^2+2a*d Formula (1)

d: displacement in meters (m)

v₀: initial speed in m/s

vf: final speed in m/s

a: acceleration in m/s²

Equivalences:

1mile = 1609.34 meters

1 hour = 3600s

1km = 1000m

Known data

v_o = 69\frac{mile}{hour} *1609.34\frac{meter}{mile} *\frac{1}{3600}\frac{hour}{s}=30.8 \frac{m}{s}

v_f= \frac{69}{2} \frac{mile}{hour} =34.5\frac{mile}{hour}=15.4 \frac{m}{s}

a = -0.5 m/s²

Distance calculation

We replace data in the Formula (1)

15.4^2 = 30.8^2+2(-0.5)*d

2(0.5)*d = 30.8^2 - 15.4^2

d =\frac{ 30.8^2 - 15.4^2}{2(0.5) }= 717.6m

d = 717.6 m\frac{1km}{1000m} = 0.7176Km

8 0
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Answer:

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r = 3.46 \times 10^8 m

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