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Ymorist [56]
2 years ago
14

If the period of oscillation of a simple pendulum is 4s, find its length. If the velocity of the bob

Physics
1 answer:
Natalija [7]2 years ago
4 0

Answer:

Explanation:

Because we assume the pendulum is a "mathematical pendulum" (neglecting the moment of inertia of the bob), we can find:

T=2\pi\sqrt{\frac{L}{g}} \rightarrow 4=2\pi\sqrt{\frac{L}{9.81}} \rightarrow \frac{4}{\pi^{2}}=\frac{L}{9.81} \rightarrow L \approx 3.97 m

By using the y=A\sin(\omega t)  \rightarrow v = \frac{dy}{dt}=\omega A \cos\omega t = \omega\sqrt{A^{2}-y^{2}}

The mean position is the position when <em>y</em> = 0, so:

\omega = \frac{2\pi}{T}=\frac{2\pi}{4}=0.5\pi rad/s

and v = \omega A \rightarrow A=\frac{40}{0.5\pi}=\frac{80}{\pi} in centimeters (cm).

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You use 8x binoculars were used on a warbler (14cm long) in a tree 18cm away. What angle (in degrees) does the image of the warb
mafiozo [28]

Answer:

The angle it subtend on the retina is  \theta_z = 0.44586^o    

Explanation:

From the question we are told that

     The length of the warbler is  L = 14cm = \frac{14}{100} = 0.14m

      The distance from the binoculars is    d = 18cm = \frac{18}{100} = 0.18m

        The magnification of the binoculars is  M =8

Without the 8 X binoculars the  angle made with the angular size of the object  is mathematically represented as

          \theta = \frac{L}{d}

        \theta  = \frac{0.14}{0.18}

           = 0.007778 rad

Now magnification can be represented mathematically as

         M = \frac{\theta _z}{\theta}

Where \theta_z is the angle the image of the warbler subtend on your retina when the   binoculars i.e the  binoculars zoom.

So

      \theta_z = M * \theta

=>    \theta_z =8 * 0.007778

            = 0.0622222224

Generally the conversion to degrees can be mathematically evaluated as

             \theta_z = 0.062222224 * (\frac{360 }{2 \pi rad} )

              \theta_z = 0.44586^o  

7 0
3 years ago
After the pendulum is dropped, determine the height at which the kinetic energy is equal to the potential energy.
Fofino [41]
0.5 m v² = m g h
⇅
h = 0.5 v²/g
4 0
3 years ago
The speed of the center of the earth as it orbits the sun is 107257 kmph and the absolute angular velocity of the earth about it
Sindrei [870]

Answer with Explanation:

We are given that

Speed ,v_0=107257 kmph

Angular velocity,\omega=7.292\times 10^{-5} rad/s

Radius of earth,r=6371 km=6371000 m

1 km=1000 m

Linear velocity,v=r\omega=6371000\times 7.292\times 10^{-5}=464.57 m/s

Linear velocity,v=464.57\times \frac{18}{5}=1672.46 km/h

Velocity at point A,v_A=-vi+v_0 j=-1672.46 i+107257j kmph

Velocity at point B,v_B=v_0j-vj=107257j-1672.46j=105584.54j kmph

Velocity at point C,v_C=v_0j+vi=1672.46 i+107257j kmph

Velocity at point  D,v_D=v_0j+vj=107257j+1672.46j=108929.46jkmph

3 0
3 years ago
A cross-country skier slides horizontally along the snow and comes to rest after sliding a distance of 11 m. If the coefficient
Basile [38]

Answer:

v_o = 4.54 m/s  

Explanation:

<u>Knowns  </u>

From equation, the work done on an object by a constant force F is given by:  

W = (F cos Ф)S                                   (1)  

Where S is the displacement and Ф is the angle between the force and the displacement.  

From equation, the kinetic energy of an object of mass m moving with velocity v is given by:  

K.E=1/2m*v^2                                       (2)

From The work- energy theorem , the net work done W on an object equals the difference between the initial and the find kinetic energy of that object:  

W = K.E_f-K.E_o                                 (3)

<u>Given </u>

The displacement that the sled undergoes before coming to rest is s = 11.0 m and the coefficient of the kinetic friction between the sled and the snow is μ_k = 0.020  

<u>Calculations</u>

We know that the kinetic friction force is given by:

f_k=μ_k*N

And we can get the normal force N by applying Newton's second law to the sled along the vertical direction, where there is no acceleration along this direction, so we get:  

∑F_y=N-mg

     N=mg

Thus, the kinetic friction force is:  

f_k = μ_k*N  

Since the friction force is always acting in the opposite direction to the motion, the angle between the force and the displacement is Ф = 180°.  

Now, we substitute f_k and Ф into equation (1), so we get the work done by the friction force:  

W_f=(f_k*cos(180) s

      =-μ_k*mg*s

Since the sled eventually comes to rest, K.E_f= 0 So, from equation (3), the net work done on the sled is:  

W= -K.E_o    

Since the kinetic friction force is the only force acting on the sled, so the net work on the sled is that of the kinetic friction force  

W_f= -K.E_o  

From equation (2), the work done by the friction force in terms of the initial speed is:  

W_f=-1/2m*v^2  

Now, we substitute for W_f= -μ_k*mg*s, and solving for v_o so we get:  

-μ_k*mg*s = -1/2m*v^2  

v_o = √ 2μ_kg*s

Finally, we plug our values for s and μ_k, so we get:  

v_o = √2 x (0.020) x (9.8 m/s^2) x (11.0 m) = 4.54 m/s  

v_o = 4.54 m/s  

6 0
3 years ago
Read 2 more answers
Which type of mixture could this illustration represent?
Oksanka [162]

Answer:

A homogeneous mixture

Explanation:

4 0
3 years ago
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