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Trava [24]
3 years ago
12

A 5.1-m-diameter merry-go-round is initially turning with a 4.1 s period. It slows down and stops in 25 s . Part A Before slowin

g, what is the speed of a child on the rim? Express your answer to two significant figures and include the appropriate units. vv = nothing nothing Request Answer Part B How many revolutions does the merry-go-round make as it stops? Express your answer using two significant figures. ΔθΔ θ = nothing rev
Physics
1 answer:
Elden [556K]3 years ago
6 0

Answer:

a)

1.53 rad s⁻¹

b)

3 rev

Explanation:

d = diameter of the merry-go-round = 5.1 m

r = radius of the merry-go-round = \frac{d}{2} =  \frac{5.1}{2} = 2.55 m

T = time period of the merry-go-round = 4.1 s

w_{i} = initial angular velocity of merry-go-round

Initial angular velocity is given as

w_{i} = \frac{2\pi }{T}

w_{i} =  \frac{2\pi }{4.1}

w_{i} = 1.53 rad s⁻¹

b)

w_{f} = final angular velocity of merry-go-round = 0 rad s⁻¹

t = time taken to stop = 25 s

\theta = angular displacement = ?

n = Number of revolutions made before stopping

Angular displacement is given as

\theta =\frac{(w_{f}+ w_{i})t}{2}

\theta =\frac{(0 + 1.53)(25)}{2}

\theta =19.125 rad

Number of revolutions are given as

n = \frac{\theta}{2\pi }

n = \frac{19.125}{2\pi }

n = 3 rev

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An aluminum calorimeter with a mass of 100 g contains 250 g of water. The calorimeter and water are in thermal equilibrium at 10
Nadya [2.5K]

Answer:

1822.14 J/kg C

Explanation:

Mass of aluminum  = 0.100 kg

Mass of Water = 0.250 kg

Mass of copper = 0.300 kg

Mass of unknown object = 0.70 kg

Initial temperature of Aluminum = 10 C

Initial  temperature  of water = 10 C

Initial temperature of copper = 30 C

Initial temperature of = 100 C

Final temperature of all substances = 20 C

Change in temperature of each  of them Aluminum, water , copper and unknown material  respectively = 10 C , 10 C , -10 C, -80 C respectively

Specific heat of  each  of them Aluminum, water , copper and unknown material  respectively = 900,4186,387 and c respectively.

Heat gained by aluminum and water = (0.1)(900)(10) + (0.250)(4186)(10) = 11365 j

heat lost by copper and unknown  material  =(0.3)(387)(-10) = -1161

heat lost by unknown  material  =  (0.070)(c)(-80) = -5.6 c

Now solve for c using the calorimetry principle.

Heat lost + Heat gained = 0

⇒ -5.6 c = -11365 - (-1161) = -10204 j

Specific heat of unknown material = - 10204 / -5.6

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4 0
3 years ago
Una manguera de agua de 1.3 cm de diametro es utilizada para llenar una cubeta de 24 Litros. Si la cubeta se llena en 48 s. A) ¿
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Answer:

We must translate this:

a 1.3 cm diameter water hose is used to fill a 24-liter bucket. If the bucket is filled in 48 s.  

A) What is the speed with which the water leaves the hose?

B) if the diameter of the hose is reduced to 0.63 cm and assuming the same flow, what will be the speed of the water leaving the hose?

A) If the velocity of the water is Xcm/s

and the radius of the hose is equal to half its diameter, so it is 1.3cm/2

Then in one second we can considerate that a cylinder of:

V = pi*(1.3cm/2)^2*X cm^3 of water.

So we have that quantity in one second of flow.

where pi = 3.14

then in 48 seconds, the amount of water in the bucket is:

V = 48*pi*(1.3/2)^2*X = 24 L = 24,000 cm^3

Now we need to solve this for X.

48*3.14*(1.3/2)^2*X = 24000

63.679*x = 24000

x = 24000/63.679 = 376.89

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B) if the diameter is 0.64cm, we have the equation:

48*3.14*(0.63/2)^2*x = 24000

14.955*X = 24000

X = 24000/14.955 = 1604.814 cm/s

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