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vichka [17]
3 years ago
7

A playground merry-go-round has radius 2.40 m and moment of inertia 2100 kg⋅m2 about a vertical axle through its center, and it

turns with negligible friction.
(a) A child applies an 18.0 N force tangentially to the edge of the merry-go-round for 15.0 s. If the merry-go-round is initially at rest, what is its angular speed after this 15.0 s interval?
(b) How much work did the child do on the merry-go-round?
(c) What is the average power supplied by the child?
Physics
1 answer:
daser333 [38]3 years ago
7 0

Answer:

a) 0.31 rad/s

b) 100 J

c) 6.67 W

Explanation:

(a) the force would generate a torque of:

T = FR = 18 * 2.4 = 43.2 Nm

According to Newton 2nd law, the angular acceleration would be

\alpha = \frac{T}{I} = \frac{43.2}{2100} = 0.021 rad/s^2

It starts from rest, then after 15s it would achieve a speed of

\omega = \alpha t = 0.021 * 15 = 0.31 rad/s

(b) The distance angle swept by it is:

\theta = \frac{\alpha t^2}{2} = \frac{0.021 * 15^2}{2} = 2.314 rad

Hence the work by the child

W = T\theta = 43.2 *2.314  \approx 100 J

c) Average power to work per time unit

P = \frac{W}{t} = \frac{100}{15} = 6.67 W

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Answer:

By calculation, it can be shown that;

K = \frac{F_{angular}}{2\times x\times c}

Whereby for constant K, as  {F_{angular}} increases,  x also increases.

Explanation:

The experiment set up consisted of the use of a spring force to maintain the object in  circular path.

The energy in the spring is given by

\frac{1}{2}\cdot k\cdot x^2.

Rotational kinetic energy = \frac{1}{2}·I·ω²

Inertia,  I =  \frac{1}{2}·m·r²

ω = \frac{v}{r}

Substituting gives

Rotational kinetic energy =  \frac{1}{2}·

=  \frac{1}{4}·m· v²

Equating both equations gives

K = \frac{2\times m\times v^2}{4\times x^{2} }  = \frac{1\times m\times v^2}{2\times x^{2} }  

Within the proportionality limit, x ∝ r

therefore we can write x = c·r which gives

\frac{ m\times v^2}{2\times x\times c\times r } = \frac{v^{2} }{r} \times\frac{m}{2\times x\times c}

Since \frac{v^{2} }{r} = angular acceleration, α, then

m× \frac{v^{2} }{r} = Angular force

Therefore K = \frac{F_{angular}}{2\times x\times c}

Therefore as Force, F increases, x also increases and the size of a spring force should be proportional to the amount of stretch in the spring.

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3 years ago
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Question: "<span>What is the half life of Strontium-90? Explain your answer"

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Block 1, of mass m₁ = 1.30 kg , moves along a frictionless air track with speed v₁ = 29.0 m/s. It collides with block 2, of mass
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Answer:

a. 37.7 kgm/s b. 0.94 m/s c. -528.85 J

Explanation:

a. The initial momentum of block 1 of m₁ = 1.30 kg with speed v₁ = 29.0 m/s is p₁ = m₁v₁ = 1.30 kg × 29.0 m/s = 37.7 kgm/s

The initial momentum of block 2 of m₁ = 39.0 kg with speed v₂ = 0 m/s since it is initially at rest is p₁ = m₁v₁ = 39.0 kg × 0 m/s = 0 kgm/s

So, the magnitude of the total initial momentum of the two-block system = (37.7 + 0) kgm/s = 37.7 kgm/s

b. Since the blocks stick together after the collision, their final momentum is p₂ = (m₁ + m₂)v where v is the final speed of the two-block system.

p₂ = (1.3 + 39.0)v = 40.3v

From the principle of conservation of momentum,

p₁ = p₂

37.7 kgm/s = 40.3v

v = 37.7/40.3 = 0.94 m/s

So the final velocity of the two-block system is 0.94 m/s

c. The change in kinetic energy of the two-block system is ΔK = K₂ - K₁ where K₂ = final kinetic energy of the two-block system = 1/2(m₁ + m₂)v² and K₁ = final kinetic energy of the two-block system = 1/2m₁v₁²

So, ΔK = K₂ - K₁ = 1/2(m₁ + m₂)v² - 1/2m₁v₁² = 1/2(1.3 + 39.0) × 0.94² - 1/2 × 1.3 × 29.0² = 17.805 J - 546.65 J = -528.845 J ≅ -528.85 J

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3 years ago
A 0.046 kg golf ball hit by a driver can accelerate from rest to 67 m/s in 1 ms while the driver is in contact with the ball. Ho
Flauer [41]

Answer:

Average force = 67 mn

Explanation:

Given:

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Time t = 1 ms = 0.001 sec.

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Using Momentum theory

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F × 0.001 = (67 - 0)/0.001

F= 67,000,000

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