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4 years ago

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The speed of sound through oxygen at 0°C is 316 meters per second. The speed of sound through solid copper is 5,010 meters per s

econd. Which of the following could be the speed of sound through liquid mercury? A. 52 meters per second B. 1,450 meters per second C. 6,130 meters per second D. 228 meters per second

2 answers:

vivado [14]4 years ago

5 0

Answer:

1,450 meters per second

Explanation:

Ray Of Light [21]4 years ago

3 0

Answer:1,450 meters per second

Explanation:

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If a bar of copper is brought near a magnet, the copper bar will be

Alecsey [184]

It will be unaffected by the magnet because it has no magnetic field. If you were to maybe have electricity going through it is the only way it would have anything to do with the magnet.
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3 years ago

Read 2 more answers

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 %.

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3 years ago

A person is sitting at the very back of a canoe of length L, when the front just bumps into the dock. show answer No Attempt 50%

Pavel [41]

The distance of the canoeist from the dock is equal to length of the canoe, L.

<h3>Conservation of linear momentum</h3>

The principle of conservation of linear momentum states that the total momentum of an isolated system is always conserved.

v(m₁ + m₂) = m₁v₁ + m₂v₂

where;

v is the velocity of the canoeist and the canoe when they are together

u₁ is the velocity of the canoe
u₂ velocity of the canoeist
m₁ mass of the canoe
m₂ mass of the canoeist

<h3>Distance traveled by the canoeist</h3>

The distance traveled by the canoeist from the back of the canoe to the front of the canoe is equal to the length of the canoe.

Thus, the distance of the canoeist from the dock is equal to length of the canoe, L.

Learn more about conservation of linear momentum here: brainly.com/question/7538238

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2 years ago

A plane electromagnetic wave travels northward. At one instant, its electric field has a magnitude of 9.6 V/m and points eastwar

lara31 [8.8K]

Answer:

The values is $B = 3.2 *10^{-8} \ T$

The direction is out of the plane

Explanation:

From the question we are told that

The magnitude of the electric field is $E = 9.6 \ V/m$

The magnitude of the magnetic field is mathematically represented as

$B = \frac{E}{c}$

where c is the speed of light with value

$B = \frac{ 9.6}{3.0 *10^{8}}$

$B = 3.2 *10^{-8} \ T$

Given that the direction off the electromagnetic wave( c ) is northward(y-plane ) and the electric field(E) is eastward(x-plane ) then the magnetic field will be acting in the out of the page (z-plane )

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3 years ago

The volume of an ideal gas changes from 0.40 to 0.55 m3 although its pressure remains constant at 50,000 Pa. What work is done o

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Answer:

w= p∆v 50000 ( 0.55-0.40) and calculate and you get it

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3 years ago