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3 years ago

What wave have a frequency of less then 20hz

Physics

1 answer:

Andrei [34K]3 years ago

3 0

Answer: Infrasound

Explanation: Sounds with frequencies below 20 hertz are called infrasound. Infrasound is too low-pitched for humans to hear. Sounds with frequencies above 20,000 hertz are called ultrasound

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A loaded 375 kg toboggan is traveling on smooth horizontal snow at 4.50 m/s when it suddenly comes to a rough region. The region

zmey [24]

Answer:

a) The average friction force exerted on the toboggan is 653.125 newtons, b) The rough region reduced the kinetic energy of the toboggan in 92.889 %, c) The speed of the toboggan is reduced in 73.333 %.

Explanation:

a) Given the existence of non-conservative forces (friction between toboggan and ground), the motion must be modelled by means of the Principle of Energy Conservation and the Work-Energy Theorem, since toboggan decrease its speed (associated with due to the action of friction. Changes in gravitational potential energy can be neglected due to the inclination of the ground. Then:

$K_{1} = K_{2} + W_{f}$

Where:

$K_{1}$ , $K_{2}$ are the initial and final translational kinetic energies of the tobbogan, measured in joules.

$W_{f}$ - Dissipated work due to friction, measured in joules.

By applying definitions of translation kinetic energy and work, the expression described above is now expanded and simplified:

$f\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{1}^{2}-v_{2}^{2})$

Where:

$f$ - Friction force, measured in newtons.

$\Delta s$ - Distance travelled by the toboggan in the rough region, measured in meters.

$m$ - Mass of the toboggan, measured in kilograms.

$v_{1}$ , $v_{2}$ - Initial and final speed of the toboggan, measured in meters per second.

The friction force is cleared:

$f = \frac{m\cdot (v_{1}^{2}-v_{2}^{2})}{2\cdot \Delta s}$

If $m = 375\,kg$ , $v_{1} = 4.50\,\frac{m}{s}$ , $v_{2} = 1.20\,\frac{m}{s}$ and $\Delta s = 5.40 \,m$ , then:

$f = \frac{(375\,kg)\cdot \left[\left(4.50\,\frac{m}{s} \right)^{2}-\left(1.20\,\frac{m}{s}\right)^{2}\right]}{2\cdot (5.40\,m)}$

$f = 653.125\,N$

The average friction force exerted on the toboggan is 653.125 newtons.

b) The percentage lost by the kinetic energy of the tobbogan due to friction is given by the following expression, which is expanded and simplified afterwards:

$\% K_{loss} = \frac{K_{1}-K_{2}}{K_{1}}\times 100\,\%$

$\% K_{loss} = \left(1-\frac{K_{2}}{K_{1}} \right)\times 100\,\%$

$\% K_{loss} = \left(1-\frac{\frac{1}{2}\cdot m \cdot v_{2}^{2}}{\frac{1}{2}\cdot m \cdot v_{1}^{2}} \right)\times 100\,\%$

$\% K_{loss} = \left(1-\frac{v_{2}^{2}}{v_{1}^{2}} \right)\times 100\,\%$

$\%K_{loss} = \left[1-\left(\frac{v_{2}}{v_{1}}\right)^{2} \right]\times 100\,\%$

If $v_{1} = 4.50\,\frac{m}{s}$ and $v_{2} = 1.20\,\frac{m}{s}$ , then:

$\%K_{loss} = \left[1-\left(\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} }\right)^{2} \right]\times 100\,\%$

$\%K_{loss} = 92.889\,\%$

The rough region reduced the kinetic energy of the toboggan in 92.889 %.

c) The percentage lost by the speed of the tobbogan due to friction is given by the following expression:

$\% v_{loss} = \frac{v_{1}-v_{2}}{v_{1}}\times 100\,\%$

$\% v_{loss} = \left(1-\frac{v_{2}}{v_{1}} \right)\times 100\,\%$

If $v_{1} = 4.50\,\frac{m}{s}$ and $v_{2} = 1.20\,\frac{m}{s}$ , then:

$\% v_{loss} = \left(1-\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} } \right)\times 100\,\%$

$\%v_{loss} = 73.333\,\%$

The speed of the toboggan is reduced in 73.333 %.

5 0

3 years ago

If an object has a mass of 38 kg, what is its approximate weight on earth?

klasskru [66]

38*10=380 N
To be more exact, 38 should be multiplied by 9.8 instead of 10.

3 0

3 years ago

A graph titled Position versus time for with horizontal axis time (seconds) and vertical axis position (meters). The line runs i

Lelu [443]

Answer:

The first one is 3 m/s

The second one is 2 m/s

Explanation:

8 0

3 years ago

Read 2 more answers

How much heat would be needed to completely evaporate 31.5 g of boiling water at a temperature of 100 "C? Express your answer in

BaLLatris [955]

Answer:

Heat needed = 71.19 J

Explanation:

Here heat required can be calculated by the formula

H = mL

M is the mass of water and L is the latent heat of vaporization.

Mass of water, m = 31.5 g = 0.0315 kg

Latent heat of vaporization of water = 2260 kJ/kg

Substituting

H = mL = 0.0315 x 2260 = 71.19 kJ

Heat needed = 71.19 J

7 0

3 years ago

a 5.00 × 105 kg rocket is accelerating straight up. Its engines produce 1.50 × 107 of thrust, and air resistance is 4.50 × 106 N

Angelina_Jolie [31]

This one is simple :)

use the equation F = ma and then re-arrange for a.

a = F / m

But there is a trick, so be careful. the question gives you wind resistance. Simply subtract the wind resistance from the thrust of the rocket to get the net force upward.

$1.50 * 10^7N - 4.5 8 10^6N = 1.05 * 10^7N$

So,

$a = \frac{F}{m} = \frac{1.05*10^7N}{5.00*10^5kg}$

3 0

4 years ago