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3 years ago

Explain the 3 method of heat transfera. conductionb. convectionc. radiation

Physics

1 answer:

Aleks [24]3 years ago

8 0

Answer:

Heat can be transferred in 3 ways.

Explanation:

Conduction- The process of transmission of heat in solids,without the actual movement of particles.

Convection- A process of transmission of heat in fluids (liquids and gases)with the actual movement of particles.

Radiation- It is the process of transmission of heat without any medium.

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A roadrunner at rest suddenly spots a rattlesnake slithering directly away at a constant speed of 0.75 m/s. At the moment the sn

Shalnov [3]

Answer:

The roadrunner will take approximately 5.285 seconds to catch up to the rattlesnake.

Explanation:

From the statement we notice that:

1) Rattlesnake moves a constant speed ( $v_{S} = 0.75\,\frac{m}{s}$ ), whereas the roadrunner accelerates uniformly from rest. ( $v_{o, R} = 0\,\frac{m}{s}$ , $a = 1\,\frac{m}{s^{2}}$ )

2) Initial distance between the roadrunner and rattlesnake is 10 meters. ( $x_{o, R} = 0\,m$ , $x_{o,S} = 10\,m$ )

3) The roadrunner catches up to the snake at the end. ( $x_{S} = x_{R}$ )

Now we construct kinematic expression for each animal:

Rattlesnake

$x_{S} = x_{o,S}+v_{S}\cdot t$

Where:

$x_{o, S}$ - Initial position of the rattlesnake, measured in meters.

$x_{S}$ - Final position of the rattlesnake, measured in meters.

$v_{S}$ - Speed of the rattlesnake, measured in meters per second.

$t$ - Time, measured in seconds.

Roadrunner

$x_{R} = x_{o,R} +v_{o,R}\cdot t +\frac{1}{2}\cdot a\cdot t^{2}$

Where:

$x_{o, R}$ - Initial position of the roadrunner, measured in meters.

$x_{R}$ - Final position of the roadrunner, measured in meters.

$v_{o,R}$ - Initial speed of the roadrunner, measured in meters per second.

$a$ - Acceleration of the roadrunner, measured in meters per square second.

$t$ - Time, measured in seconds.

By eliminating the final positions of both creatures, we get the resulting quadratic function:

$x_{o,S}+v_{S}\cdot t = x_{o,R}+v_{o,R}\cdot t +\frac{1}{2}\cdot a \cdot t^{2}$

$\frac{1}{2}\cdot a \cdot t^{2} + (v_{o,R}-v_{S})\cdot t + (x_{o,R}-x_{o,S}) = 0$

If we know that $a = 1\,\frac{m}{s^{2}}$ , $v_{o, R} = 0\,\frac{m}{s}$ , $v_{S} = 0.75\,\frac{m}{s}$ , $x_{o, R} = 0\,m$ and $x_{o,S} = 10\,m$ , the resulting expression is:

$0.5\cdot t^{2}-0.75\cdot t -10=0$

We can find its root via Quadratic Formula:

$t_{1,2} = \frac{-(-0.75)\pm \sqrt{(-0.75)^{2}-4\cdot (0.5)\cdot (-10)}}{2\cdot (0.5)}$

$t_{1,2} = \frac{3}{4}\pm \frac{\sqrt{329}}{4}$

Roots are $t_{1} \approx 5.285\,s$ and $t_{2}\approx -3.785\,s$ , respectively. Both are valid mathematically, but only the first one is valid physically. Hence, the roadrunner will take approximately 5.285 seconds to catch up to the rattlesnake.

6 0

3 years ago

Two coins, made of different material, are placed next to each other so that they are touching. Both coins are 15mm in diameter.

nekit [7.7K]

The mass of the second coin for the given gravitation force is 20.9 g.

The given parameters;

distance between the two coins, r = 15 mm = 0.015 m
mass of one coin, m₁ = 6.5 g = 0.0065 kg
gravitational force between the coins = 4.03 × 10⁻¹¹ N

The mass of the second coin is determined by applying Newton's law of universal gravitation as shown below;

$F = \frac{Gm_1 m_2 }{R^2} \\\\m_2 = \frac{FR^2}{Gm_1} \\\\m_2 = \frac{(4.03 \times 10^{-11}) \times (0.015)^2}{(6.67 \times 10^{-11}) \times (0.0065)} \\\\m_2 = 0.0209 \ kg\\\\m_2 = 20.9 \ g$