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NikAS [45]
3 years ago
6

What is the frequency of a wave having a period equal to 18 seconds? a. 6.6 × 10-2 hertz b. 5.5 × 10-2 hertz c. 3.3 × 10-2 hertz

d. 1.8 × 10-2 hertz e. 8.0 × 10-3 hertz
Physics
2 answers:
irakobra [83]3 years ago
6 0
Frequency = 1/period. ... 1 / 18 sec = (1/18) per sec. That's 0.056 per sec or 0.056 Hz. (rounded) (5.6 x 10^-2 Hz)
Karolina [17]3 years ago
6 0

Answer: Option (b) is the correct answer.

Explanation:

The time taken by a wave crest to travel a distance equal to the length of wave is known as wave period.

The relation between wave period and frequency is as follows.

                    T = \frac{1}{f}

where,        T = time period

                   f = frequency

It is given that wave period is 18 seconds. Therefore, calculate the wave period as follows.

                           T = \frac{1}{f}

 or,                      f = \frac{1}{T}

                                     = \frac{1}{18 sec}

                                     = 0.055 per second          (1 cycle per second = 1 Hertz)

or,                           f = 5.5 \times 10^{-2} hertz

Thus, we can conclude that the frequency of the wave is 5.5 \times 10^{-2} hertz.

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In which ancient civilization did the sport of cucj originate
Burka [1]
It originated in China.
3 0
3 years ago
A thin rod of length 0.75 m and mass 0.42 kg is suspended
MrRissso [65]

Answer:

a)  K = 0.63 J, b)  h = 0.153 m

Explanation:

a) In this exercise we have a physical pendulum since the rod is a material object, the angular velocity is

         w² = \frac{m g d}{I}

where d is the distance from the pivot point to the center of mass and I is the moment of inertia.

The rod is a homogeneous body so its center of mass is at the geometric center of the rod.

              d = L / 2

the moment of inertia of the rod is the moment of a rod supported at one end

              I = ⅓ m L²

we substitute

            w = \sqrt{\frac{mgL}{2}  \ \frac{1}{\frac{1}{3} mL^2} }

            w = \sqrt{\frac{3}{2}  \ \frac{g}{L} }

            w = \sqrt{ \frac{3}{2} \ \frac{9.8}{0.75}  }

            w = 4.427 rad / s

an oscillatory system is described by the expression

              θ = θ₀ cos (wt + Φ)

the angular velocity is

             w = dθ /dt

             w = - θ₀ w sin (wt + Ф)

In this exercise, the kinetic energy is requested in the lowest position, in this position the energy is maximum. For this expression to be maximum, the sine function must be equal to ±1

In the exercise it is indicated that at the lowest point the angular velocity is

           w = 4.0 rad / s

the kinetic energy is

           K = ½ I w²

           K = ½ (⅓ m L²) w²

           K = 1/6 m L² w²

           K = 1/6 0.42 0.75² 4.0²

           K = 0.63 J

b) for this part let's use conservation of energy

starting point. Lowest point

             Em₀ = K = ½ I w²

final point. Highest point

             Em_f = U = m g h

energy is conserved

             Em₀ = Em_f

             ½ I w² = m g h

             ½ (⅓ m L²) w² = m g h

             h = 1/6 L² w² / g

             h = 1/6 0.75² 4.0² / 9.8

             h = 0.153 m

5 0
3 years ago
Let’s say I am in a bumper car and have a velocity of 14 m/s, driving in the positive x-direction. I and my bumped car have a ma
AlekseyPX

Answer:

160 kg

12 m/s

Explanation:

m_1 = Mass of first car = 120 kg

m_2 = Mass of second car

u_1 = Initial Velocity of first car = 14 m/s

u_2 = Initial Velocity of second car = 0 m/s

v_1 = Final Velocity of first car = -2 m/s

v_2 = Final Velocity of second car

For perfectly elastic collision

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}\\\Rightarrow m_2v_2=m_{1}u_{1}+m_{2}u_{2}-m_{1}v_{1}\\\Rightarrow m_2v_2=120\times 14+m_2\times 0-(120\times -2)\\\Rightarrow m_2v_2=1920\\\Rightarrow m_2=\frac{1920}{v_2}

Applying in the next equation

v_2=\frac{2m_1}{m_1+m_2}u_{1}+\frac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\frac{2\times 120}{120+\frac{1920}{v_2}}\times 14+\frac{m_2-m_1}{m_1+m_2}\times 0\\\Rightarrow \left(120+\frac{1920}{v_2}\right)v_2=3360\\\Rightarrow 120v_2+1920=3360\\\Rightarrow v_2=\frac{3360-1920}{120}\\\Rightarrow v_2=12\ m/s

m_2=\frac{1920}{v_2}\\\Rightarrow m_2=\frac{1920}{12}\\\Rightarrow m_2=160\ kg

Mass of second car = 160 kg

Velocity of second car = 12 m/s

5 0
4 years ago
10.
myrzilka [38]

Answer:

<em>The new period of oscillation is D) 3.0 T</em>

Explanation:

<u>Simple Pendulum</u>

A simple pendulum is a mechanical arrangement that describes periodic motion. The simple pendulum is made of a small bob of mass 'm' suspended by a thin inextensible string.

The period of a simple pendulum is given by

T=2\pi \sqrt{\frac{L}{g}}

Where L is its length and g is the local acceleration of gravity.

If the length of the pendulum was increased to 9 times (L'=9L), the new period of oscillation will be:

T'=2\pi \sqrt{\frac{L'}{g}}

T'=2\pi \sqrt{\frac{9L}{g}}

Taking out the square root of 9 (3):

T'=3*2\pi \sqrt{\frac{L}{g}}

Substituting the original T:

T'=3*T

The new period of oscillation is D) 3.0 T

4 0
3 years ago
Can somebody help me on finding the constants if this experiment?
ki77a [65]
The constants could be the day because the experiment is all on the same day or the person throwing the pumpkin because it will always be the same perosn
8 0
3 years ago
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