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siniylev [52]
3 years ago
9

A sinusoidal transverse wave of amplitude ym = 8.4 cm and wavelength = 5.3 cm travels on a stretched cord. Find the ratio of the

maximum particle speed (the speed with which a single particle in the cord moves transverse to the wave) to the wave speed.
Physics
1 answer:
Scilla [17]3 years ago
8 0

Answer:

The ratio is 9.95

Solution:

As per the question:

Amplitude, y_{m} = 8.4\ cm

Wavelength, \lambda = 5.3\ cm

Now,

To calculate the ratio of the maximum particle speed to the speed of the wave:

For the maximum speed of the particle:

v_{m} = y_{m}\times \omega

where

\omega = 2\pi f = angular speed of the particle

Thus

v_{m} = 2\pi fy_{m}

Now,

The wave speed is given by:

v = f\lambda

Now,

The ratio is given by:

\frac{v_{m}}{v} = \frac{2\pi fy_{m}}{f\lambda}

\frac{v_{m}}{v} = \frac{2\pi \times 8.4}{5.3} = 9.95

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Flapping flight is very energy intensive. A wind tunnel test
steposvetlana [31]

Answer:

The metabolic power for starting flight=134.8W/kg

Explanation:

We are given that

Mass of starling, m=89 g=89/1000=0.089 kg

1 kg=1000 g

Power, P=12 W

Speed, v=11 m/s

We have to find the metabolic power for starting flight.

We know that

Metabolic power for starting flight=\frac{P}{m}

Using the formula

Metabolic power for starting flight=\frac{12}{0.089}

Metabolic power for starting flight=134.8W/kg

Hence, the metabolic power for starting flight=134.8W/kg

4 0
3 years ago
A torque of 36.5 N · m is applied to an initially motionless wheel which rotates around a fixed axis. This torque is the result
vivado [14]

Answer:

21.6\ \text{kg m}^2

3.672\ \text{Nm}

54.66\ \text{revolutions}

Explanation:

\tau = Torque = 36.5 Nm

\omega_i = Initial angular velocity = 0

\omega_f = Final angular velocity = 10.3 rad/s

t = Time = 6.1 s

I = Moment of inertia

From the kinematic equations of linear motion we have

\omega_f=\omega_i+\alpha_1 t\\\Rightarrow \alpha_1=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha_1=\dfrac{10.3-0}{6.1}\\\Rightarrow \alpha_1=1.69\ \text{rad/s}^2

Torque is given by

\tau=I\alpha_1\\\Rightarrow I=\dfrac{\tau}{\alpha_1}\\\Rightarrow I=\dfrac{36.5}{1.69}\\\Rightarrow I=21.6\ \text{kg m}^2

The wheel's moment of inertia is 21.6\ \text{kg m}^2

t = 60.6 s

\omega_i = 10.3 rad/s

\omega_f = 0

\alpha_2=\dfrac{0-10.3}{60.6}\\\Rightarrow \alpha_1=-0.17\ \text{rad/s}^2

Frictional torque is given by

\tau_f=I\alpha_2\\\Rightarrow \tau_f=21.6\times -0.17\\\Rightarrow \tau=-3.672\ \text{Nm}

The magnitude of the torque caused by friction is 3.672\ \text{Nm}

Speeding up

\theta_1=0\times t+\dfrac{1}{2}\times 1.69\times 6.1^2\\\Rightarrow \theta_1=31.44\ \text{rad}

Slowing down

\theta_2=10.3\times 60.6+\dfrac{1}{2}\times (-0.17)\times 60.6^2\\\Rightarrow \theta_2=312.03\ \text{rad}

Total number of revolutions

\theta=\theta_1+\theta_2\\\Rightarrow \theta=31.44+312.03=343.47\ \text{rad}

\dfrac{343.47}{2\pi}=54.66\ \text{revolutions}

The total number of revolutions the wheel goes through is 54.66\ \text{revolutions}.

3 0
3 years ago
A roller coaster car is elevated to a height of 30 m and released from rest to roll along a track. At a certain time T it is at
Naddika [18.5K]

Explanation:

Initial energy = final energy + work done by friction

PE = PE + KE + W

mgH = mgh + 1/2 mv² + W

(800)(9.8)(30) = (800)(9.8)(2) + 1/2 (800) v² + 25000

v = 22.1 m/s

Without friction:

PE = PE + KE

mgH = mgh + 1/2 mv²

(800)(9.8)(30) = (800)(9.8)(2) + 1/2 (800) v²

v = 23.4 m/s

4 0
2 years ago
½H2(g) + ½I2(g) → HI(g), ΔH = + 26 kJ/mole 117 kJ/mole + C(s) + 2S(s) → CS2(g) The temperature of the surroundings will:
Alona [7]
From the above reaction the temperature of the surroundings will increase.
8 0
3 years ago
Read 2 more answers
The activity of a radioisotope is found to decrease 40% of its original value in 2.59 x 10 s.
Rainbow [258]

Answer: 0.0353\ s^{-1}

Explanation:

Given

Radioactive material is found to decrease 40% of its original value in 2.59\times 10\ s

Sample at any time is given by

N=N_oe^{-\lambda t}

where, \lambda=\text{decay constant}

Put values

\Rightarrow 0.4N_o=N_oe^{-\lambda\cdot 2.59\times 10}\\\Rightarrow 0.4=e^{-\lambda\cdot 2.59\times 10

Taking natural logarithm both side

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8 0
3 years ago
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