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1 year ago

Pot holder should have high insulation and low _____.

Physics

1 answer:

kipiarov [429]1 year ago

3 0

Potholder should have high insulation and low conductivity, therefore the correct answer is the option B

<h3>What is insulation?</h3>

Insulation is a type of material used to create barriers to the transmission of the form of energy which either is in form of heat or electricity.

For outdoor trips in cold weather, several thin layers act as better insulating barriers for heat transfer.

The ability of an electric charge or heat to pass through a material is measured by its conductivity. A material is considered a conductor if it offers very little resistance to the flow of thermal or electric energy.

Thus, Potholders should be highly insulated and have low conductivity, therefore the correct answer is the option B

Learn more about insulation from here

brainly.com/question/14363642

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your question seems incomplete, the complete question is

To be effective, a pot holder should have low _____. viscosity conductivity malleability density

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An airplane dropped a flare from a height of 2860 feet above a lake. How many seconds did it take for the flare to reach the wat

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Answer: 13.2 seconds.

Explanation: using equation of motion; S= ut +1/2at² where u = initial velocity=0

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a = acceleration due gravity

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S= 2860 feet =872.3m

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t = 13.2 seconds to one decimal place.

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2 years ago

An elevator provides 21 000 w of power during a 12 s ride. how much work does the elevator do?

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2 years ago

A hanging weight, with a mass of m1 = 0.365 kg, is attached by a string to a block with mass m2 = 0.825 kg as shown in the figur

morpeh [17]

The speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.

<h3>Angular Speed of the pulley </h3>

The angular speed of the pulley after the block m1 fall through a distance, d, is obatined from conservation of energy and it is given as;

K.E = P.E

$\frac{1}{2} mv^2 + \frac{1}{2} I\omega^2 = mgh\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2(m_1R^2_2 + m_2R_2^2) + \frac{1}{2} \omega^2( \frac{1}{2} MR_1^2 + \frac{1}{2} MR_2^2) = m_1gd- \mu_km_2gd\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2[R_2^2(m_1 + m_2)+ \frac{1}{2} M(R_1^2 + R_2^2)] = gd(m_1 - \mu_k m_2)\\\\$

$\frac{1}{2} m_2v_0 + \frac{1}{4} \omega^2[2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = gd(m_1 - \mu_k m_2)\\\\2m_2v_0 + \omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = 4gd(m_1 - \mu_k m_2)\\\\\omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2\\\\\omega^2 = \frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)} \\\\\omega = \sqrt{\frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)}} \\\\$

Substitute the given parameters and solve for the angular speed;

$\omega = \sqrt{\frac{ 4(9.8)(0.7)(0.365 \ - \ 0.25\times 0.825) - 2(0.825)(0.82)^2}{2(0.03)^2(0.365 \ + \ 0.825)\ \ +\ \ 0.35(0.02^2\ + \ 0.03^2)}} \\\\\omega = \sqrt{\frac{3.25}{0.00214\ + \ 0.000455 } } \\\\\omega = 35.39 \ rad/s$

<h3>Linear speed of the block</h3>

The linear speed of the block after travelling 0.7 m;

v = ωR₂

v = 35.39 x 0.03

v = 1.1 m/s

Thus, the speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.

Learn more about conservation of energy here: brainly.com/question/24772394

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2 years ago