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larisa86 [58]
4 years ago
6

A spring with spring constant 17.0 N/m hangs from the ceiling. A ball is suspended from the spring and allowed to come to rest.

It is then pulled down 4.00 cm and released. The ball makes 35.0 oscillations in 18.0 seconds.
1) What is its the mass of the ball?
Express your answer using two significant figures.
m = _____ g

2) What is its maximum speed?
Express your answer using two significant figures.

= _____ cm/s
Physics
1 answer:
Sonbull [250]4 years ago
6 0

Explanation:

Given that,

Spring constant of the spring, k = 17 N/m

It is then pulled down 4.00 cm and released, A = 4 cm = 0.04 m

The ball makes 35.0 oscillations in 18.0 seconds.

The time required to complete one oscillation is :

T=\dfrac{18}{35}=0.514\ s

Angular velocity,

\omega=\dfrac{2\pi }{T}\\\\\omega=\dfrac{2\pi }{0.514}\\\\\omega=12.22\ rad/s

(1) The relation between spring constant, angular velocity and mass is given by :

\omega^2=\dfrac{k}{m}\\\\m=\dfrac{k}{\omega^2}\\\\m=\dfrac{17}{(12.22)^2}\\\\m=0.11\ kg

(2) The maximum speed is given by :

v=A\omega\\\\v=0.04\times 12.22\\\\v=0.49\ m/s            

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Answer:

220,500 Joules

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An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by a distance of 1.70 mm. A 21
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Answer:

12353 V m⁻¹ = 12.4 kV m⁻¹

Explanation:  

Electric field between the plates of the parallel plate capacitor depends on the potential difference across the plates and their distance of separation.Potential difference across the plates V over the distance between the plates gives the electric field between the plates. Potential difference is the amount of work done per unit charge and is given here as 21 V. Electric field is the voltage over distance.

E = V ÷ d = 21 ÷ 0.0017 = 12353 V m⁻¹

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Two objects (42.0 and 21.0 kg) are connected by a massless string that passes over a massless, frictionless pulley. The pulley h
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Answer:

a = 3.27 m/s²

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Explanation:

Given that:

Mass m₁ = 42.p0 kg

Mass m₂ = 21.0 kg

Consider both masses to be in a whole system, then:

The acceleration can be determined as:

(m_1+m_2)a = g(m_1-m_2)

Making acceleration the subject in the above formula;

a =\dfrac{g(m_1-m_2)}{(m_1+m_2)}

a =\dfrac{9.8(42.0-21.0)}{(42.0+21.0)}

a =\dfrac{9.8(21.0)}{(63.0)}

a =\dfrac{205.8}{(63.0)}

a = 3.27 m/s²

in the string, the tension is calculated using the formula:

T = \dfrac{2m_1m_2g}{(m_1+m_2)}

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A 23 kg body is moving through space in the positive direction of an x axis with a speed of 130 m/s when, due to an internal exp
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Answer:

a) Vx = 1088m/s

b) Vy = -162.93m/s

c) 5246745J

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Mass of unbroken body = 23kg

Its velocity along +ve X-axis = 130m/s

Mass of first broken body, m1= 9.4kg

Its velocity along +ve X-axis = 130m/s

Nass of 2nd broken body, m2 = 6.1kg

Its velocity long-lived X - axis = -550m/s

Mass of 3rd broken body = ?

m3 = (23 - 9.4 - 6.1)kg

m3 = 7.5kg

Let velocity along the x-axis = Vx

Let the velocity along the x-axis = Vy

Applying law of conservation of momentum along x-axis

a) m1×0 + m2×(-550) + m3×(Vx) =M × 130

9.4 × 0 + 6.1× (-550) + 7.5(Vx) = 23 ×130

0 + (-5170) + 7.5Vx = 2990

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8160 = 7.5Vx

Vx = 8160/7.5

Vx = 1088m/s

b) Aplying conservation of momentum along the x-axis

(m1×130) + (m2 × 0) + (m3× Vy) = 0

(9.4 × 130) + (6.1 ×550) + 7.5Vy = 0

1222 + 0 + 7.5Vy = 0

1222 = -7.5Vy

Vy = 1222/(-7.5)

Vy = -262.93m/s

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Change in KE = 1/2[ 9.4× 130^2 + 6.1 × 550^2 + 7.5 × 1088^2 ] - 1/2(23 × 130^2)

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