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Wittaler [7]
2 years ago
12

A spring with spring constant 15 N/m hangs from the ceiling. A ball is attached to the spring and allowed to come to rest. It is

then pulled down 7.5 cm and released. The ball makes 31 oscillations in 15 s seconds.1) What is its the mass of the ball?Express your answer using two significant figures.m = _____ g2) What is its maximum speed?Express your answer using two significant figures.= _____ cm/s
Physics
1 answer:
olga nikolaevna [1]2 years ago
5 0

Explanation:

It is given that,

Spring constant of the spring, k = 15 N/m

Amplitude of the oscillation, A = 7.5 cm = 0.075 m

Number of oscillations, N = 31

Time, t = 15 s

(a) Let m is the mass of the ball. The frequency of oscillation of the spring is given by :

f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}

Total number of oscillation per unit time is called frequency of oscillation. Here, f=\dfrac{31}{15}=2.06\ Hz

m=\dfrac{k}{4\pi^2f^2}

m=\dfrac{15}{4\pi^2\times 2.06^2}

m = 0.0895 kg

or

m = 89 g

(b) The maximum speed of the ball that is given by :

v_{max}=A\times \omega

v_{max}=A\times 2\pi f

v_{max}=0.075\times 2\pi \times 2.06

v_{max}=0.970\ m/s

v_{max}=97\ cm/s

Hence, this is the required solution.

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Answer:

Mass of the wooden Block is 20g.

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Also we know from the question that

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Equating Eq-1 & Eq-2 we get

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6 0
2 years ago
You have a summer job at a company that developed systems to safely lower large loads down ramps. Your team is investigating a m
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Answer:

Note that the emf induced is

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---> v = emf / [B d cos (A)]

where

B = magnetic field

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A = angle of rails with respect to the horizontal

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Thus, the bar experiences a magnetic force of

F(B) = B I d = B^2 d^2 v cos (A) / R, horizontally, up the incline.

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As this is equal to the component of the weight parallel to the incline,

B^2 d^2 v cos^2 (A) / R = m g sin (A)

where m = the mass of the bar.

Solving for v,

v = [R m g sin (A) / B^2 d^2 cos^2 (A)]   [ANSWER, the constant speed, PART A]

******************************

v = [R m g sin (A) / B^2 d^2 cos^2 (A)]

Plugging in the units,

m/s = [ [ohm * kg * m/s^2] / [T^2 m^2] ]

Note that T = kg / (s * C), and ohm = J * s/C^2

Thus,

m/s = [ [J * s/C^2 * kg * m/s^2] / [(kg / (s * C))^2 m^2] ]

= [ [J * s/C^2 * kg * m/s^2] / [(kg^2 m^2) / (s^2 C^2)]

As J = kg*m^2/s^2, cancelling C^2,,

= [ [kg*m^2/s^2 * s * kg * m/s^2] / [(kg^2 m^2) / (s^2)]

Cancelling kg^2,

= [ [m^2/s^2 * s * m/s^2] / [(m^2) / (s^2)]

Cancelling m^2/s^2,

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Cancelling s,

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8 0
3 years ago
2.5 gram sample of a radioactive element was formed in a 1960 explosion of an atomic bomb at Johnson Island in the Pacific Test
kow [346]

Answer:

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Explanation:

Half life: It is related to the decay of radioactive material. The duration in which  half of the material will be degraded/decayed. That means after half life 50% of the radioactive material will be left. Here the half life is 28 years.

Initial quantity of the sample: 2.5 grams.

After 28 years, the leftover quantity = 1.25 grams

After 56 years, the leftover quantity = 0.625 grams

After 84 Years, the leftover quantity = 0.3125 grams

After 112 years, the leftover quantity = 0.15625 grams

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Answer:

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7 0
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