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Helen [10]
2 years ago
10

The ball will oscillate along the z axis between z=d and z=−d in simple harmonic motion. What will be the angular frequency ω of

these oscillations? Use the approximation d≪a to simplify your calculation; that is, assume that d2+a2≈a2.
Physics
1 answer:
Alinara [238K]2 years ago
5 0

Any kind of frequency, including the angular kind, is closely involved with
time. Still, for some unknown reason,you've given us no time information
whatsoever ... a peculiar decision on your part, since we can be sure that
it's right there, inexorably intertwined with the part of the question that you
DID copy and share with us. 

Furthermore and moreover, for one with no prior experience with simple
harmonic motion, the many symbols in this question such as ' d ', ' a ',
' << ', ' d₂ ', and ' a₂ ' would be of no help at all to guide him toward a
solution.  On the contrary, he would conclude that the question itself
had been posted by some alien life form.

To sum up:  Come back and post the drawing that goes along with the
question, make sure you have presented all of the information that the
question includes, and then we'll talk.


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You and your friend Peter are putting new shingles on a roof pitched at 20degrees . You're sitting on the very top of the roof w
Anit [1.1K]

Answer:

v₀ =3.8 m/s

Explanation:

Newton's second law of the box:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Known data

m=2.1 kg  mass of the box

d= 5.4m  length of the roof

θ = 20° angle θ of the roof with respect to the horizontal direction

μk= 0.51 : coefficient of kinetic friction between the box and the roof  

g = 9.8 m/s² : acceleration due to gravity

Forces acting on the box

We define the x-axis in the direction parallel to the movement of the box on the roof  and the y-axis in the direction perpendicular to it.

W: Weight of the box  : In vertical direction

N : Normal force : perpendicular to the direction the  roof

fk : Friction force: parallel to the direction to the roof

Calculated of the weight  of the box

W= m*g  =  (2.1 kg)*(9.8 m/s²)= 20.58 N

x-y weight components

Wx= Wsin θ= (20.58)*sin(20)° =7.039 N

Wy= Wcos θ =(20.58)*cos(20)°= 19.34 N

Calculated of the Normal force

∑Fy = m*ay    ay = 0

N-Wy= 0

N=Wy =19.34 N

Calculated of the Friction force:

fk=μk*N= 0.51* 19.34 N = 9.86 N

We apply the formula (1) to calculated acceleration of the block:

∑Fx = m*ax  ,  ax= a  : acceleration of the block

Wx-f = ( 2.1)*a

7.039 - 9.86  = ( 2.1)*a

-2.821 = ( 2.1)*a

a=(-2.821) /( 2.1)

a= -1.34  m/s²

Kinematics of the box

Because the box moves with uniformly accelerated movement we apply the following formula to calculate the final speed of the block :

vf²=v₀²+2*a*d Formula (2)

Where:  

d:displacement  = 5.4 m

v₀: initial speed  

vf: final speed  = 0

a : acceleration of the box = -1.34  m/s²

We replace data in the formula (2)

0²=v₀²+2*(-1.34)*(5.4)

2*(1.34)*(5.4)= v₀²

v_{o} =\sqrt{14.472}

v₀ = 3.8 m/s

7 0
2 years ago
An insurance company hired your group to help investigate an insurance claim following a car accident. In the accident, two cars
Romashka-Z-Leto [24]

Answer:

v=8m/s

Explanation:

To solve this problem we have to take into account, that the work done by the friction force, after the collision must equal the kinetic energy of both two cars just after the collision. Hence we have

W_{f}=E_{k}\\W_{f}=\mu N=\mu(m_1+m_1)g\\E_{k}=\frac{1}{2}[m_1+m_2]v^2

where

mu: coefficient of kinetic friction

g: gravitational acceleration

We can calculate the speed of the cars after the collision by using

W_f=(0.7)(1650kg+1900kg)(9.8\frac{m}{s^2})=24353J\\24353J=\frac{1}{2}(1650kg+1900kg)v^2\\v=\sqrt{\frac{24353J}{1775kg}}=3.70\frac{m}{s}

Now , we can compute the speed of the second car by taking into account the conservation of the momentum

P_b=P_a\\m_1v_1+m_2v_2=(m_1+m_2)v\\\\v_2=\frac{(m_1+m_2)v-m_1v_1}{m_1}\\\\v_2=\frac{(1650kg+1900kg)(3.7\frac{m}{s})-(1650kg)(16\frac{m}{s})}{(1650kg)}\\\\v_2=8\frac{m}{s}

the car did not exceed the speed limit

Hope this helps!!

7 0
2 years ago
Read 2 more answers
A baseball thrown at an angle of 60.0° above the horizontal strikes a building 16.0 m away at a point 8.00 m above the point fro
yanalaym [24]

Answer:

a) v_{o} =16m/s

b) v=9.8m/s

c) \beta =-35.46º

Explanation:

From the exercise we know that the ball strikes the building 16m away and its final height is 8m more than the initial

Being said that, we can calculate the initial velocity of the ball

a) First we analyze its horizontal motion

x=v_{ox}t

x=v_{o}cos(60)t

v_{o}=\frac{x}{tcos(60)}=\frac{16m}{tcos(60)} (1)

That would be our first equation

Now, we need to analyze its vertical motion

y=y_{o}+v_{oy}t+\frac{1}{2}gt^2

y_{o}+8=y_{o}+v_{o}sin(60)t-\frac{1}{2}(9.8)t^2

Knowing v_{o} in our first equation (1)

8=\frac{16}{tcos(60)}sin(60)t-\frac{1}{2}(9.8)t^2

\frac{1}{2}(9.8)t^2=16tan(60)-8

Solving for t

t=\sqrt{\frac{2(16tan(60)-8)}{9.8} } =2s

So, the ball takes to seconds to get to the other building. Now we can calculate its <u>initial velocity</u>

v_{o}=\frac{16m}{(2s)cos(60)}=16m/s

b) To find the <u>magnitude of the ball just before it strikes the building</u> we need to calculate its x and y components

v_{x}=v_{ox}+at=16cos(60)=8m/s

v_{y}=v_{oy}+gt=16sin(60)-(9.8)(2)=-5.7m/s

So, the magnitude of the velocity is:

v=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{(8m/s)^2+(-5.7m/s)^2}=9.8m/s

c) The <u><em>direction of the ball</em></u> is:

\beta=tan^{-1}(\frac{v_{y} }{v_{x}})=tan^{-1}(\frac{-5.7}{8})=-35.46º

4 0
2 years ago
Suppose you sketch a model of an atom using the ones here as a guide. How would you build a model that is ionized? How would you
kap26 [50]

Answer:

Explanation:

An atom is constructed of three different particles known as electrons, protons and neutrons.

These particles have different mass and charges and are responsible for various characters than an atom posses.

An electron has a negative charge, a proton has positive charge and charge of neutron is neutral. Equal number of electrons and protons are present in an atom that make it electrically neutral but different conditions can occur if we remove these particles from an atom.

1 : Model of an ionized atom - an ionized atom is one which has some net charge on this. It can be either a positive charge or a negative charge.

If we need to sketch the model of an ionized atom then one should either keep the number of electrons less or proton.

2: Model of radioactive atom : A radioactive atom is one an unstable atom and has access of energy in its center. It can be caused by adding either neutrons or protons.

3 0
1 year ago
A hiker walks 9.4 miles at an angle of 60° south of west. Find the west and south components of the walk. Round your answers to
andrey2020 [161]
For this problem, we use our knowledge on trigonometric functions on a right triangle to be able to calculate for the x and y components of the walk. We do as follows:

sin 60 = ax / 9.4
ax = 8.14

cos 60 = ay / 9.4
ay = 4.7
3 0
2 years ago
Read 2 more answers
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