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16 days ago

Each of the following situations involves the flow of heat energy. For each scenario, specify the primary mode (conduction, conv

ection, or radiation) by which the energy is transferred. a)What is the primary mode of energy transfer from hot coffee inside a Thermos bottle to the environment? b)What is the primary mode of energy transfer from your bare feet to a cold ceramic bathroom floor in the middle of winter? c)What is the primary mode of energy transfer from your body to your hands when you blow on them to warm them? d)What is the primary mode of energy transfer into your sealed car parked in the sun on a hot summer day? e)What is the primary mode of the flow of energy from your tongue to a metal railing if you happen to lick it on a cold winter day? f)what is the primary mode of energy transfer from a hair dryer to your wet hair? g)What is the primary mode of energy transfer from your body on a cold day with no wind? h)What is the primary mode of energy transfer away from your body on a very windy day?

Physics

Answer;

Heat is generated
Light is emitted
Bubbles form

Explanation;

Chemical reactions are characterized by the interaction of two or more substances, leading to the formation of a new substance, and are typically irreversible.
Indicators of chemical reactions include gas production, release of energy in the form of light or heat, formation of precipitates, and color changes.

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3 months ago

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The parasailing system shown uses a winch to pull the rider in towards the boat, which is traveling with a constant velocity. Du

ValentinkaMS [3465]

Answer:

The force magnitude is $F_{net}= 1.837 *10^4N$

and it is directed at 57.98° from the horizontal in a counterclockwise manner.

Explanation:

The problem states that

At t = 0, $\theta = 20^o$

The angular rate of increase is $w = 2 \ ^o/s$

Converting to revolutions per second gives us $\theta ' = 2 \ ^o/s * \frac{\pi}{180} =0.0349\ rps$

The rope length is defined by

$r = 125- \frac{1}{3}t^{\frac{3}{2} }$

At $\theta =30^o$ , Tension T of the rope is 18 kN.

The weight of the para-sailor is $M_p = 75kg$

In analyzing the question, we observe that the equation for length can be represented as a linear displacement equation.

The derivative of displacement results in velocity.

Hence,

$r' = -\frac{1}{3} [\frac{3}{2} ] t^{\frac{1}{2} }$

signifies the velocity, and further differentiation yields acceleration.

Therefore,

$r'' = -\frac{1}{4} t^{-\frac{1}{2} }$

Now considering the moment when the rope forms a 30° angle with the water,

typically angular velocity is expressed as

$w = \frac{\Delta \theta}{\Delta t}$

where $\theta$ represents the angular displacement.

Next, evaluating the interval from $20^o \ to \ 30^o$ gives us

$2 = \frac{30 -20 }{t -0}$

making t the focal point.

$t = \frac{10}{2}$

$= 5s$

At this time, the displacement measures

$r = 125- \frac{1}{3}(5)^{\frac{3}{2} }$

$= 121.273 m$

The linear velocity computes to

$r' = -\frac{1}{3} [\frac{3}{2} ] (5)^{\frac{1}{2} }$

$= -1.118 m/s$

Whereas linear acceleration calculates as

$r'' = -\frac{1}{4} (5)^{-\frac{1}{2} }$

$= -0.112m/s^2$

Generally, radial acceleration is given by

$\alpha _R = r'' -r \theta'^2$

$= -0.112 - (121.273)[0.0349]^2$

$= 0.271 m/s^2$

Simultaneously, angular acceleration can be represented as

$\alpha_t = r \theta'' + 2 r' \theta '$

Then $\theta '' = \frac{d (0.0349)}{dt} = 0$

Thus,

$\alpha _t = 121.273 * 0 + 2 * (-1.118)(0.0349)$

$= -0.07805 m/s^2$

The resultant acceleration is mathematically denoted as

$a = \sqrt{\alpha_R^2 + \alpha_t^2 }$

$= \sqrt{(-0.07805)^2 +(-0.027)^2}$

$= 0.272 m/s^2$

Now the acceleration's direction is mathematically expressed as

$tan \theta_a = \frac{\alpha_R }{\alpha_t }$

$\theta_a = tan^{-1} \frac{-0.271}{-0.07805}$

$= 73.26^o$

The y-axis force acting on the para-sailor is mathematically shown as

$F_y = mg + Tsin 30^o + ma sin(90- \theta )$

$= (75 * 9.8) + (18 *10^3) sin 30 + (75 * 0.272)sin(90-73.26)$

$= 9.74*10^3 N$

The x-axis force acting on the para-sailor is represented as

$F_x = mg + Tcos 30^o + ma cos(90- \theta )$

$= (75 * 9.8) + (18 *10^3) cos 30 + (75 * 0.272)cos(90-73.26)$

$= 1.557 *10^4 N$

The overall force is calculated as

$F_{net} = \sqrt{F_x^2 + F_y^2}$

$=\sqrt{(1.557 *10^4)^2 + (9.74*10^3)^2}$

$F_{net}= 1.837 *10^4N$

The directional force is evaluated as

$tan \theta_f = \frac{F_y}{F_x}$

$\theta_f = tan^{-1} [\frac{1.557*10^4}{9.74*10^3} ]$

$= tan^{-1} (1.599)$

$= 57.98^o$

7 0

3 months ago