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Sonja [21]
3 years ago
5

A mass weighing pounds is attached to a spring whose constant is lb/ft. The medium offers a damping force that is numerically eq

ual to the instantaneous velocity. The mass is initially released from a point foot above the equilibrium position with a downward velocity of ft/s. Determine the time at which the mass passes through the equilibrium position. Find the time at which the mass attains its extreme displacement from the equilibrium position. What is the position of the mass at this instant
Physics
1 answer:
andreyandreev [35.5K]3 years ago
6 0

Answer:

hello your question has some missing values attached below is the complete question with the missing values

answer :

a) 0.083 secs

b) 0.33 secs

c)  3e^-4/3

Explanation:

Given that

g = 32 ft/s^2 ,  spring constant ( k ) = 2 Ib/ft

initial displacement = 1 ft above equilibrium

mass = weight / g = 4/32 = 1/8

damping force = instanteous velocity  hence  β = 1

a<u>)Calculate the time at which the mass passes through the equilibrium position.</u>

time mass passes through equilibrium = 1/12 seconds = 0.083

<u>b) Calculate the time at which the mass attains its extreme displacement </u>

time when mass attains extreme displacement = 1/3 seconds = 0.33 secs

<u>c) What is the position of the mass at this instant</u>

position = 3e^-4/3

attached below is the detailed solution to the given problem

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Answer:

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Explanation:

A) Total distance covered by the dog = 20 + 15

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B) Since the other distance covered by the dog before chewing the stick, after the retrieval, was in an opposite direction to the initial direction, then;

total displacement of the dog = 20 - 15

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C) Net displacement = distance covered by the dog to retrieve the stick + distance covered before the dog starts chewing the stick

But, displacement involves a specified direction. The distance covered before the dog starts chewing the stick was in an opposite direction to the initial direction.

Thus,

Net displacement = distance covered by the dog to retrieve the stick - distance covered before the dog starts chewing the stick

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3 years ago
the base of a rectangular vessel measure 10m by 18cm. water is poured into a depth of 4cm. (a) what is the pressure on the base?
Alex787 [66]

Answer:

a) P =392.4[Pa]; b) F = 706.32[N]

Explanation:

With the input data of the problem we can calculate the area of the tank base

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W = width = 18[cm] = 0.18[m]

A = W * L = 0.18*10

A = 1.8[m^2]

a)

Pressure can be calculated by knowing the density of the water and the height of the water column within the tank which is equal to h:

P = density * g *h

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F = P*A

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Explanation:

Given:

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v = (3 m/s²) (4 s) + 0 m/s

v = 12 m/s

6 0
2 years ago
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