All categories

3 years ago

Explain the difference between voluntary and involuntary muscles.

1 answer:

3 0

Muscles that are under your conscious control are called voluntary muscles, while muscles that are not under your conscious control are called involuntary muscles

You might be interested in

Where would you weigh the most?

uranmaximum [27]

Answer:

mars

Explanation:

4 0

3 years ago

Read 2 more answers

Kate must do 39 days of training each year as a naval reservist. How will she serve that time?

elixir [45]

Answer:

Explanation:

i did the test

3 0

3 years ago

During spring semester at MIT, residents of the parallel buildings of the East Campus dorms battle one another with large catapu

natali 33 [55]

Answer:

110.7 J

Explanation:

Hooke's law is represented by the formula:

F = ke where F is the force in Newton, K is force constant and e is extension in m

work done = 1/2ke² = 1/2 K ( e² - e₀²) and e₀ is the extension at relaxed length

e₀ =0

work done = 0.5 × 82N/m × (2.70 m)² = 110.7 J

4 0

3 years ago

Two 60.o-g arrows are fired in quick succession with an initial speed of 82.0 m/s. The first arrow makes an initial angle of 24.

olganol [36]

Answer:

a) The first arrow reaches a maximum height of 56.712 meters, whereas second arrow reaches a maximum height of 342.816 meters, b) Both arrows have a total mechanical energy at their maximum height of 201.720 joules.

Explanation:

a) The first arrow is launch in a parabolic way, that is, horizontal speed remains constant and vertical speed changes due to the effects of gravity. On the other hand, the second is launched vertically, which means that velocity is totally influenced by gravity. Let choose the ground as the reference height for each arrow. Each arrow can be modelled as particles and by means of the Principle of Energy Conservation:

First arrow

$U_{g,1} + K_{x,1} + K_{y,1} = U_{g,2} + K_{x,2} + K_{y,2}$

Where:

$U_{g,1}$ , $U_{g,2}$ - Initial and final gravitational potential energy, measured in joules.

$K_{x,1}$ , $K_{x,2}$ - Initial and final horizontal translational kinetic energy, measured in joules.

$K_{y,1}$ , $K_{y,2}$ - Initial and final vertical translational kinetic energy, measured in joules.

Now, the system is expanded and simplified:

$m \cdot g \cdot (y_{2} - y_{1}) + \frac{1}{2}\cdot m \cdot (v_{y, 2}^{2} -v_{y, 1}^{2}) = 0$

$g \cdot (y_{2}-y_{1}) = \frac{1}{2}\cdot (v_{y,1}^{2}-v_{y,2}^{2})$

$y_{2}-y_{1} = \frac{1}{2}\cdot \frac{v_{y,1}^{2}-v_{y,2}^{2}}{g}$

Where:

$y_{1}$ . $y_{2}$ - Initial and final height of the arrow, measured in meters.

$v_{y,1}$ , $v_{y,2}$ - Initial and final vertical speed of the arrow, measured in meters.

$g$ - Gravitational acceleration, measured in meters per square second.

The initial vertical speed of the arrow is:

$v_{y,1} = v_{1}\cdot \sin \theta$

Where:

$v_{1}$ - Magnitude of the initial velocity, measured in meters per second.

$\theta$ - Initial angle, measured in sexagesimal degrees.

If $v_{1} = 82\,\frac{m}{s}$ and $\theta = 24^{\circ}$ , the initial vertical speed is:

$v_{y,1} = \left(82\,\frac{m}{s} \right)\cdot \sin 24^{\circ}$

$v_{y,1} \approx 33.352\,\frac{m}{s}$

If $g = 9.807\,\frac{m}{s^{2}}$ , $v_{y,1} \approx 33.352\,\frac{m}{s}$ and $v_{y,2} = 0\,\frac{m}{s}$ , the maximum height of the first arrow is:

$y_{2} - y_{1} = \frac{1}{2}\cdot \frac{\left(33.352\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{9.807\,\frac{m}{s^{2}} }$

$y_{2} - y_{1} = 56.712\,m$

Second arrow

$U_{g,1} + K_{y,1} = U_{g,3} + K_{y,3}$

Where:

$U_{g,1}$ , $U_{g,3}$ - Initial and final gravitational potential energy, measured in joules.

$K_{y,1}$ , $K_{y,3}$ - Initial and final vertical translational kinetic energy, measured in joules.

$m \cdot g \cdot (y_{3} - y_{1}) + \frac{1}{2}\cdot m \cdot (v_{y, 3}^{2} -v_{y, 1}^{2}) = 0$

$g \cdot (y_{3}-y_{1}) = \frac{1}{2}\cdot (v_{y,1}^{2}-v_{y,3}^{2})$

$y_{3}-y_{1} = \frac{1}{2}\cdot \frac{v_{y,1}^{2}-v_{y,3}^{2}}{g}$

If $g = 9.807\,\frac{m}{s^{2}}$ , $v_{y,1} = 82\,\frac{m}{s}$ and $v_{y,3} = 0\,\frac{m}{s}$ , the maximum height of the first arrow is:

$y_{3} - y_{1} = \frac{1}{2}\cdot \frac{\left(82\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{9.807\,\frac{m}{s^{2}} }$

$y_{3} - y_{1} = 342.816\,m$

The first arrow reaches a maximum height of 56.712 meters, whereas second arrow reaches a maximum height of 342.816 meters.

b) The total energy of each system is determined hereafter:

First arrow

The total mechanical energy at maximum height is equal to the sum of the potential gravitational energy and horizontal translational kinetic energy. That is to say:

$E = U + K_{x}$

The expression is now expanded:

$E = m\cdot g \cdot y_{max} + \frac{1}{2}\cdot m \cdot v_{x}^{2}$

Where $v_{x}$ is the horizontal speed of the arrow, measured in meters per second.

$v_{x} = v_{1}\cdot \cos \theta$

If $v_{1} = 82\,\frac{m}{s}$ and $\theta = 24^{\circ}$ , the horizontal speed is:

$v_{x} = \left(82\,\frac{m}{s} \right)\cdot \cos 24^{\circ}$

$v_{x} \approx 74.911\,\frac{m}{s}$

If $m = 0.06\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $y_{max} = 56.712\,m$ and $v_{x} \approx 74.911\,\frac{m}{s}$ , the total mechanical energy is:

$E = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (56.712\,m)+\frac{1}{2}\cdot (0.06\,kg)\cdot \left(74.911\,\frac{m}{s} \right)^{2}$

$E = 201.720\,J$

Second arrow:

The total mechanical energy is equal to the potential gravitational energy. That is:

$E = m\cdot g \cdot y_{max}$

$m = 0.06\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ and $y_{max} = 342.816\,m$

$E = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (342.816\,m)$

$E = 201.720\,J$

Both arrows have a total mechanical energy at their maximum height of 201.720 joules.

7 0

3 years ago

(a) Calculate the rate of heat conduction through house walls that are 13.0 cm thick and that have an average thermal conductivi

Mkey [24]

Answer:

a. 960 W b. One 1 kW room heater

Explanation:

a. The rate of heat conduction P = kA(T₂ - T₁)/d where k = 2 × 0.040 W/m-K = 0.080 W/m-K since the thermal conductivity of glass wool is 0.040 W/m-K and that of the material is twice the thermal conductivity of glass wool, A = area of walls = 120 m², T₁ = outside surface temperature = 5.0 °C, T₂ = inside surface temperature = 18.0 °C and d = thickness of wall = 13.0 cm = 0.13 m

P = kA(T₂ - T₁)/d

= 0.080 W/m-K × 120 m²(18.0 °C - 5.0 °C)/0.13 m

= 9.6 Wm/K × 13 K/0.13 m

= 124.8 Wm/0.13 m

= 960 W

b. The number of 1 kW room heater required will be

n = rate of heat conduction/power of one room heater = 960 W/ 1 kW = 960 W/1000 W = 0.96 ≅ 1

So we need only one 1 kW room heater.

6 0

3 years ago