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Taya2010 [7]
3 years ago
6

A neighbor's child wants to go to a neighborhood carnival to experience the wild rides. The neighbor is worried about safety as

one of the rides looks dangerous. and asks for your advice. the ride in question has a 10-lb chair which hangs freely from a 30-ft long chain attached to a pivot on the top of a tall tower. when a child enters the ride, the chain is hanging straight down. the child is then attached to the chair with a seat belt and shoulder harness.
When the ride starts up the chain rotates about the tower and reaches a maximum speed and remains at thta speed. It rotates the tower once every 3 seconds.

When you ask the operator he says that the ride is safe and sits in a stationary chair which creaks but holds the 200 pounds.

Has the operator shown that the ride is safe for a 50 lb child?

Physics
2 answers:
seraphim [82]3 years ago
6 0

Answer:

No, the ride is not safe

Explanation:

From the diagram attached, it is seen that

\sum f_{y} = 0

\tau cos \theta - W = 0................(1)

\sum f_{x} = 0

-ma + \tau sin \theta = 0.......(2)

a = w^{2} r

From the diagram, r = Lsin \theta

a = w^{2} L sin \theta

w = \frac{2\pi }{T}

a = \frac{4\pi ^{2}L sin \theta }{T^{2} }............(3)

Put (3) into (2)

\tau sin \theta = \frac{4\pi ^{2}m L sin \theta }{T^{2} }

\tau  = \frac{4\pi ^{2}m L  }{T^{2} }

w = weight of chair + weight of child = 50 + 10 = 60 lb

g = 32 ft/s²

m = w/g = 60/32 = 1.875

L = 30 ft

T = 3 secs

\tau  = \frac{4\pi ^{2}*1.875*30  }{0.3^{2} }

\tau = 246.74 lbs

since 246.74 lbs > 200 lbs, it is not safe because the stationary chair will creak

Black_prince [1.1K]3 years ago
6 0

Answer:

According to the result obtained, the trip is not safe for a 50 lb child.

Explanation:

The sum of forces in y must be equal to zero:

∑Fy = 0

0 = W + Tcosθ

The sum of forces in x:

∑Fx = m*a = -m*W²*r = -T1sinθ

r = Lsinθ

m*W²*Lsinθ = T1sinθ

m\frac{4\pi ^{2}L }{T^{2} } =T\\\frac{60}{32} \frac{4\pi ^{2} 30}{3^{2} } =T\\T=250lb

Like 250 lb > 200 lb, the answer is no

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If Scoobie could drive a Jetson's flying car at a constant speed of 450.0 km/hr across oceans and space, approximately how long
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Answer:

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Explanation:

From the question we are told that

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t =  \frac{38960.8}{450.0}

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2 years ago
A fireworks rocket is fired vertically upward. At its maximum height of 90.0 m , it explodes and breaks into two pieces, one wit
Alex73 [517]

Answer:

Ai. Speed of the fragment with mass mA= 1.35 kg is 34.64 m/s

Aii. Speed of the fragment with mass mB = 0.270 kg is 77.46 m/s

B. 475.3 m

Explanation:

A. Determination of the speed of each fragment.

I. Determination of the speed of the fragment with mass mA = 1.35 kg

Mass of fragment (m₁) = 1.35 kg

Kinetic energy (KE) = 810 J

Velocity of fragment (u₁) =?

KE = ½m₁u₁²

810 = ½ × 1.35 × u₁²

810 = 0.675 × u₁²

Divide both side by 0.675

u₁² = 810 / 0.675

u₁² = 1200

Take the square root of both side.

u₁ = √1200

u₁ = 34.64 m/s

Therefore, the speed of the fragment with mass mA = 1.35 kg is 34.64 m/s

II. I. Determination of the speed of the fragment with mass mB = 0.270 kg

Mass of fragment (m₂) = 0.270 kg

Kinetic energy (KE) = 810 J

Velocity of fragment (u₂) =?

KE = ½m₂u₂²

810 = ½ × 0.270 × u₂²

810 = 0.135 × u₂²

Divide both side by 0.135

u₂² = 810 / 0.135

u₂² = 6000

Take the square root of both side.

u₂ = √6000

u₂ = 77.46 m/s

Therefore, the speed of the fragment with mass mB = 0.270 kg is 77.46 m/s

B. Determination of the distance between the points on the ground where they land.

We'll begin by calculating the time taken for the fragments to get to the ground. This can be obtained as follow:

Maximum height (h) = 90.0 m

Acceleration due to gravity (g) = 10 m/s²

Time (t) =?

h = ½gt²

90 = ½ × 10 × t²

90 = 5 × t²

Divide both side by 5

t² = 90/5

t² = 18

Take the square root of both side

t = √18

t = 4.24 s

Thus, it will take 4.24 s for each fragments to get to the ground.

Next, we shall determine the horizontal distance travelled by the fragment with mass mA = 1.35 kg. This is illustrated below:

Velocity of fragment (u₁) = 34.64 m/s

Time (t) = 4.24 s

Horizontal distance travelled by the fragment (s₁) =?

s₁ = u₁t

s₁ = 34.64 × 4.24

s₁ = 146.87 m

Next, we shall determine the horizontal distance travelled by the fragment with mass mB = 0.270 kg. This is illustrated below:

Velocity of fragment (u₂) = 77.46 m/s

Time (t) = 4.24 s

Horizontal distance travelled by the fragment (s₂) =?

s₂ = u₂t

s₂ = 77.46 × 4.24

s₂ = 328.43 m

Finally, we shall determine the distance between the points on the ground where they land.

Horizontal distance travelled by the 1st fragment (s₁) = 146.87 m

Horizontal distance travelled by the 2nd fragment (s₂) = 328.43 m

Distance apart (S) =?

S = s₁ + s₂

S = 146.87 + 328.43

S = 475.3 m

Therefore, the distance between the points on the ground where they land is 475.3 m

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Answer:

The ocean is 6485.6m deep when measured from the vessel

Explanation:

v=1474m/s

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an echo travels a distance equivalent to 2d, that is to and fro after it reflects from the obstacle.

velocity=\frac{distance}{time}\\ v=\frac{2d}{t} \\vt=2d\\d=\frac{vt}{2}

d=\frac{1474*8.8}{2}

d= 6485.6m

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