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garik1379 [7]
3 years ago
11

A brick of mass 4 kg hangs from the end of a spring. When the brick is at rest, the spring is stretched by 3 cm. The spring is t

hen stretched an additional 2 cm and released. Assume there is no air resistance. Note that the acceleration due to gravity, g, is g=980 cm/s^2.
Set up a differential equation with initial conditions describing the motion and solve it for the displacement s(t) of the mass from its equilibrium position (with the spring stretched 3 cm).
Physics
1 answer:
jeka57 [31]3 years ago
3 0

Answer:

Explanation:

Let s be displacement from equilibrium position . Restoring force

m d²s / dt² = - k s

d²s / dt² = - k /m  s

Put k /m  = ω

d²s / dt² + ω² s = 0

The solution of this differential equation

= s = A cosωt

Now when t = 0 ,  s = 2 cm

A =  2 cm

Putting the values we have

2 = A cos 0

A = 2 cm

s ( t) = 2 cos ωt

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Listed below are the measured radiation absorption rates​ (in W/kg) corresponding to 11 cell phones. Use the given data to const
Fantom [35]

Answer:

The 5-number summary is

1. Median = 0.93 W/kg

2. Lower quartile = 0.69 W/kg

3. Upper quartile = 1.16 W/kg

4. Minimum value = 0.54 W/kg

5. Maximum value = 1.42 W/kg

Explanation:

We are given the measured radiation absorption rates​ (in W/kg) corresponding to 11 cell phones.

1.16 0.85 0.69 0.75 0.95 0.93 1.18 1.17 1.42 0.54 0.57

What is 5-number summary?

A 5-number summary refers to a box plot that basically shows 5 statistical characteristics of a data set.

These statistical characteristics are:    

1. Median

2. Lower quartile

3. Upper quartile  

4. Minimum value  

5. Maximum value  

1. Median:

Arrange the data in ascending order

0.54 0.57 0.69 0.75 0.85 0.93 0.95 1.16 1.17 1.18 1.42

(n+1)/2 gives the median value of the data set.

(11 + 1)/2 = 6th position

Therefore, 0.93 W/kg is the median of the data set.

2. Lower quartile:

Divide the data set into two equal halfs (include median in both if n = odd)

Lower half = 0.54 0.57 0.69 0.75 0.85 0.93

Upper half = 0.93 0.95 1.16 1.17 1.18 1.42

The lower quartile is the median of the lower half of the data set.

Lower half = 0.54 0.57 0.69 0.75 0.85 0.93

The median is 6/2 = 3rd position

Therefore, the lower quartile of the data set is 0.69 W/kg

3. Upper quartile:

Divide the data set into two equal halfs (include median in both if n = odd)

Lower half = 0.54 0.57 0.69 0.75 0.85 0.93

Upper half = 0.93 0.95 1.16 1.17 1.18 1.42

The upper quartile is the median of the lower half of the data set.

Upper half = 0.93 0.95 1.16 1.17 1.18 1.42

The median is 6/2 = 3rd position

Therefore, the upper quartile of the data set is 1.16 W/kg

4. Minimum value:

The minimum value is the least value in the data set.

0.54 0.57 0.69 0.75 0.85 0.93 0.95 1.16 1.17 1.18 1.42

Therefore, the minimum value of the data set is 0.54 W/kg

5. Maximum value  

The maximum value is the least value in the data set.

0.54 0.57 0.69 0.75 0.85 0.93 0.95 1.16 1.17 1.18 1.42

Therefore, the maximum value of the data set is 1.42 W/kg

The box plot is illustrated in the attached diagram.

6 0
2 years ago
A man is standing on a weighing machine on a ship which is bobbing up and down with simple harmonic motion of period T=15.0s.Ass
STALIN [3.7K]

Well, first of all, one who is sufficiently educated to deal with solving
this exercise is also sufficiently well informed to know that a weighing
machine, or "scale", should not be calibrated in units of "kg" ... a unit
of mass, not force.  We know that the man's mass doesn't change,
and the spectre of a readout in kg that is oscillating is totally bogus.

If the mass of the man standing on the weighing machine is 60kg, then
on level, dry land on Earth, or on the deck of a ship in calm seas on Earth,
the weighing machine will display his weight as  588 newtons  or as 
132.3 pounds.  That's also the reading as the deck of the ship executes
simple harmonic motion, at the points where the vertical acceleration is zero.

If the deck of the ship is bobbing vertically in simple harmonic motion with
amplitude of M and period of 15 sec, then its vertical position is 

                                     y(t) = y₀ + M sin(2π t/15) .

The vertical speed of the deck is     y'(t) = M (2π/15) cos(2π t/15)

and its vertical acceleration is          y''(t) = - (2πM/15) (2π/15) sin(2π t/15)

                                                                = - (4 π² M / 15²)  sin(2π t/15)

                                                                = - 0.1755 M sin(2π t/15) .

There's the important number ... the  0.1755 M.
That's the peak acceleration.
From here, the problem is a piece-o-cake.

The net vertical force on the intrepid sailor ... the guy standing on the
bathroom scale out on the deck of the ship that's "bobbing" on the
high seas ... is (the force of gravity) + (the force causing him to 'bob'
harmonically with peak acceleration of  0.1755 x amplitude).

At the instant of peak acceleration, the weighing machine thinks that
the load upon it is a mass of  65kg, when in reality it's only  60kg.
The weight of 60kg = 588 newtons.
The weight of 65kg = 637 newtons.
The scale has to push on him with an extra (637 - 588) = 49 newtons
in order to accelerate him faster than gravity.

Now I'm going to wave my hands in the air a bit:

Apparent weight = (apparent mass) x (real acceleration of gravity)

(Apparent mass) = (65/60) = 1.08333 x real mass.

Apparent 'gravity' = 1.08333 x real acceleration of gravity.

The increase ... the 0.08333 ... is the 'extra' acceleration that's due to
the bobbing of the deck.

                        0.08333 G  =  0.1755 M

The 'M' is what we need to find.

Divide each side by  0.1755 :          M = (0.08333 / 0.1755) G

'G' = 9.0 m/s²
                                       M = (0.08333 / 0.1755) (9.8) =  4.65 meters .

That result fills me with an overwhelming sense of no-confidence.
But I'm in my office, supposedly working, so I must leave it to others
to analyze my work and point out its many flaws.
In any case, my conscience is clear ... I do feel that I've put in a good
5-points-worth of work on this problem, even if the answer is wrong .

8 0
3 years ago
A force of 10 n acts on a mass of 5 kg producing an acceleration of 1.5 m/s2 what is the magnitude of the force of friction acti
prisoha [69]
The friction is 2.5N. The Net force is 10 N - 2.5 N .= 7.5 N.
acceleration = 7.5 / 5 = 1.5 m/s^2
8 0
3 years ago
Imagine you are mpho:how can you help your friend (phenyo) not to be influenced by media? you really feel she needs some help be
LUCKY_DIMON [66]

One of the ways to help someone not to be influenced by media is:

  • Encourage independent thinking.
  • Let her know that she should make sure the information she is consuming is free from bias and is factual.

<h3>What is Media Influence?</h3>

This refers to the effects and reactions which the mass media has on the people who listen to them by shaping their views and by controlling the narrative.

With this in mind, we can note that if a person is overtly influenced by the media, then such a person lacks independent thinking and becomes emotional by what he watches on the television or reads from an online source.

This can be changed through the encouragement of independent thinking, and the need to check for factual information from the media news and also to focus more on studies.

Read more about media here:
brainly.com/question/26152499

6 0
2 years ago
A roller coaster is traveling at 13 m/s when it approaches a hill that is 400 m long. Heading down the hill, it accelerates at 4
SIZIF [17.4K]

Answer:

The  value is    v = 47 \  m/s

Explanation:

From the question we are told that

   The initial  speed of the roller coaster is u =  13 \  m/s

    The  length of the hill is  l   = 400 \  m

    The  acceleration of the  roller coaster is a=4.0 \ m/s^2

Generally the acceleration is mathematically represented as

      a =  \frac{ v - u}{ t_f -  t_i }

Here  t_i is the initial time which is equal to zero

         v_f is the final velocity which is mathematically represented as

          v_f  =  \frac{d}{ t_f}

So  

     a =  \frac{ \frac{d}{d_f}  - u }{ t_f - t_i}

     4 = \frac{\frac{400}{ t_f}  - 13}{t_f - 0}

      4 =  \frac{400 - 13t_f}{ t_f} *  \frac{1}{t_f}

     4t_f ^2  +13f  + 400 =

Solving this using quadratic formula we obtain

    t_f =  8.5 \ s

     t_f =  -11.8 \ s

Generally  time cannot be negative so

       t_f =  8.5 \ s

Generally the  final velocity is mathematically represented as

         v = \frac{400}{8.5}

         v = 47 \  m/s

       

5 0
3 years ago
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