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3 years ago

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Thermograms are infrared photograms that show emission of infrared radiation emitted from objects. If you lived in a cold climat

e, how could a thermogram taken of your home be helpful to you as a homeowner?

1 answer:

labwork [276]3 years ago

5 0

You would be able to detect unwanted trespassers as they came within range of the thermogram.

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A 2120 kg car traveling at 13.4 m/s collides with a 2810 kg car that is initally at rest at a stoplight. The cars stick together

viktelen [127]

Answer:

The coefficient of friction between the cars and the road is 0.859.

Explanation:

The two cars collide each other inelastically, then we can determine the resulting velocity by the Principle of Momentum Conservation:

$m_{A}\cdot v_{A} + m_{B}\cdot v_{B} = (m_{A} + m_{B})\cdot v$ (1)

Where:

$m_{A}$ , $m_{B}$ - Masses of the cars, in kilograms.

$v_{A}$ , $v_{B}$ - Initial velocities of the cars, in meters per second.

$v$ - Velocity of the resulting system, in meters per second.

If we know that $m_{A} = 2120\,kg$ , $v_{A} = 13.4\,\frac{m}{s }$ , $m_{B} = 2810\,kg$ and $v_{B} = 0\,\frac{m}{s}$ , then the velocity of the resulting system:

$v = \frac{m_{A}\cdot v_{A}+m_{B}\cdot v_{B}}{m_{A}+m_{B}}$

$v = \frac{(2120\,kg)\cdot \left(13.4\,\frac{m}{s} \right)+(2810\,kg)\cdot \left(0\,\frac{m}{s} \right)}{2120\,kg + 2810\,kg}$

$v = 5.762\,\frac{m}{s}$

By Principle of Energy Conservation and Work-Energy Theorem, we understand that the initial translational kinetic energy ( $K$ ), in joules, is dissipated due to work done by friction ( $W_{f}$ ), in joules, that is to say:

$K = W_{f}$ (2)

$\frac{1}{2}\cdot (m_{A}+m_{B})\cdot v^{2} = \mu\cdot (m_{A}+m_{B})\cdot g \cdot s$

$\frac{1}{2}\cdot v^{2} = \mu \cdot g\cdot s$ (2b)

Where:

$\mu$ - Coefficient of friction, no unit.

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

$s$ - Travelled distance, in meters.

If we know that $v = 5.762\,\frac{m}{s}$ , $g = 9.807\,\frac{m}{s^{2}}$ and $s = 1.97\,m$ , then the coefficient of friction is:

$\mu = \frac{v^{2}}{2\cdot g\cdot s}$

$\mu = \frac{\left(5.762\,\frac{m}{s} \right)^{2}}{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (1.97\,m)}$

$\mu = 0.859$

The coefficient of friction between the cars and the road is 0.859.

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3 years ago

Please help :)<br><br><br> Microwaves are transmitted by ....... ?

Dima020 [189]

Microwaves are transmitted by Radio Waves.

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3 years ago

A penny is placed on a rotating turntable. where on the turntable does the penny require the largest centripetal force to remain

Mama L [17]

m = mass of the penny

r = distance of the penny from the center of the turntable or axis of rotation

w = angular speed of rotation of turntable

F = centripetal force experienced by the penny

centripetal force "F" experienced by the penny of "m" at distance "r" from axis of rotation is given as

F = m r w²

in the above equation , mass of penny "m" and angular speed "w" of the turntable is same at all places. hence the centripetal force directly depends on the radius .

hence greater the distance from center , greater will be the centripetal force to remain in place.

So at the edge of the turntable , the penny experiences largest centripetal force to remain in place.

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3 years ago

Which of the following is a strength training option?

Vilka [71]

Answer:

D. All of the above

PLZ MARK ME AS BRAINLEIST ;)

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3 years ago

Read 2 more answers

A 10kg object is 15 meters up a hill. Find its potential energy

Tema [17]

Answer:

Explanation:

Relative to an origin at the bottom of the hill,

PE = mgh = 10(9.8)(15) = 1470 J

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2 years ago