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givi [52]
2 years ago
15

What factors affect the ability of a substance to transfer thermal energy to heat or be heated by it's surroundings?

Physics
1 answer:
xeze [42]2 years ago
4 0

The prime factors that affect the ability of substances to transfer the thermal energy to heat are the temperature difference between the two objects, area of cross-section, time, and distance travelled by the thermal energy.

<u>Explanation: </u>

The process of heat conduction takes place through contact between two or more objects. But this conduction depends on multiple factors that are responsible for thermal conduction. They are-

  • Temperature Difference(\Delta T) - The two objects must have a temperature difference else there will be no thermal conduction between them. The more the difference in their temperatures, the more thermal energy flows from one object to the other.
  • Area of Cross-section (A) - Larger areas of contact provide as better medium of thermal conduction.
  • Time (t) - The more time we give for the thermal conduction, the more energy is transmitted from one system to the other.
  • Distance Travelled (l) - The longer the distance, lesser the conduction. Means, the distance should be minimized in order to achieve the optimum thermal conduction between two objects.

Consider metal pot and its handle, it is being boiled for 15 m. The molecules present near the source of heat, showing fast vibration and bounce off. It actually indicates the heats of substance. That’s why, handle remains hot as heat conduction takes place.  It can be estimated by,

                                       Q=\frac{k A \Delta T t}{l}

k - Thermal conductivity of the material, measured in J/s.m.^{\circ} \mathrm{C}

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An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined
MrRa [10]

Incomplete question as the car's  speed is missing.I have assumed car's  speed as 6.0m/s.The complete question is here

An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined weight of the car and riders is 6.00 kN, and the radius of the circle is 15.0 m. At the top of the circle, (a) what is the force FB on the car from the boom (using the minus sign for downward direction) if the car's speed is v 6.0m/s

Answer:

F_{B}=-5755N

Explanation:

Set up force equation

∑F=ma

∑F=W+FB

\frac{mv^{2} }{R}=W+F_{B}\\  F_{B}=\frac{mv^{2} }{R}-W\\F_{B}=\frac{(W/g)v^{2} }{R}-W\\F_{B}=\frac{(6000N/9.8m/s^{2} )(6m/s)^{2} }{(15m)}-6000N\\F_{B}=-5755N

The minus sign for downward direction

6 0
2 years ago
A metal bar moves through a magnetic field. the induced charges on the bar are
Dmitry [639]
I would say its a positive cgarge
7 0
2 years ago
Starting at point 0, you travel 500 m on a straight road that slopes upward at a constant angle of 5 degrees. What is your heigh
ira [324]

Answer:

43.58 m

Explanation:

If you travel 500 m on a straight road that slopes upward at a constant angle of 5 degrees

Using trigonometry ratio

Sin 5 = opposite/hypothenus

Where the hypothenus = 500m

Opposite = height h

Sin 5 = h/500

Cross multiply

500 × sin 5 = h

h = 500 × 0.08715

h = 43.58m

Therefore, the height above the starting point is equal to 43.58m

5 0
2 years ago
8.4-1 Consider a magnetic field probe consisting of a flat circular loop of wire with radius 10 cm. The probe’s terminals corres
Vlad1618 [11]

Answer:

B_o = 1.013μT

Explanation:

To find B_o you take into account the formula for the emf:

\epsilon=-\frac{d\Phi_b}{dt}=-\frac{dBAcos\theta}{dt}=-Acos\theta\frac{dB}{dt}

where you used that A (area of the loop) is constant, an also the angle between the direction of B and the normal to A.

By applying the derivative you obtain:

\epsilon=-Acos\theta (2\pi f) B_ocos(2\pi f t+ \alpha)

when the emf is maximum the angle between B and the normal to A is zero, that is, cosθ = 1 or -1. Furthermore the cos function is 1 or -1. Hence:

\epsilon=2\pi fAB_o=2\pi (100*10^3Hz)(\pi (0.1m)^2)B_o=19739.20Hzm^2B_o\\\\B_o=\frac{20*10^{-3}V}{19739.20Hzm^2}=1.013*10^{-6}T=1.013\mu T

hence, B_o = 1.013μT

6 0
2 years ago
A student librarian lifts a 2.2 kg book from the floor to a height of 1.25 m. He carries the book 8.0 m to the stacks and places
Montano1993 [528]
The solution to your problem is as follows:


2.2Kg*9.8m/s = 21.56N 
<span>
21.56N*1.25m = 26.95J </span>


<span>We're only concerned with the work done against gravity, lifting the books to 1.25 meters. the distance walked has no effect on the problem, unless you take into account the wind resistance and the force needed to overcome it. Also, lowering the books onto the shelf doesnt count, because gravity does the work on the books.</span>
5 0
2 years ago
Read 2 more answers
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