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3 years ago

HELP!!!

2 answers:

5 0

We shall consider two properties:
1. Temperature difference
2. Thermal conductivity of the material

Use a cylindrical rod of a given material (say steel) which is insulated around its circumference.

One end of the rod is dipped in a large reservoir of water at 100 deg.C and the other end is dipped in water (with known volume) at 40 deg. C. The cold water if stored in a cylinder which is insulated on all sides. A thermometer reads the temperature of the cold water as a function of time.

This experiment will show that
(a) heat flows from a region of high temperature to a region of lower temperature.
(b) The thermal energy of a body increases when heat is added to it, and its temperature will rise.
(c) The thermal conductivity of water determines how quickly its temperature will rise. If mercury replaces water in the cold cylinder, its temperature will rise at a different rate because its thermal conductivity is different.

BabaBlast [244]3 years ago

5 0

How do mass and the type of material affect thermal energy transfer? This is the e2020 sample response.

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Polarizers #1 and #3 are "crossed" such that their transmission axes are perpendicular to each other. Polarizer #2 is placed bet

Tomtit [17]

Answer:

Explanation:

The angle between the polariser 1 and 2 is 60°

If I₁ be the intensity of light after polariser 1 , its intensity after polariser 2 that is I₂ can be expressed as follows

I₂ = I₁ cos² 60

= I₁ x 1/4

= I₁ x $\frac{1}{4}$

3 0

3 years ago

Also confused on this can someone please help??!

irga5000 [103]

The answer is 167 pounds.

7 0

2 years ago

A proton (mass m = 1.67 × 10-27 kg) is being accelerated along a straight line at 2.50 × 1012 m/s2 in a machine. If the proton h

Veronika [31]

Answer:

(A) Speed will be $44.18\times 10^4m/sec$

(b) Change in kinetic energy = $1560\times 10^{-19}$

Explanation:

We have given mass of proton $m=1.67\times 10^{-27}kg$

Acceleration of the proton $a=2.50\times 10^{12}m/sec^2$

Initial velocity u = $1.60\times 10^4$ m/sec

Distance traveled by proton s = 3.90 cm = 0.039 m

(a) From third equation of motion we know that

$v^2=u^2+2as$

$v^2=(1.60\times 10^4)^2+2\times 2.5\times 10^{12}\times 0.039$

$v=44.18\times 10^4m/sec$

(b) Initial kinetic energy $KE_I=\frac{1}{2}mv^2=\frac{1}{2}\times 1.67\times 10^{-27}\times (1.6\times 10^4)^2$

Final kinetic energy $KE_F=\frac{1}{2}mv^2=\frac{1}{2}\times 1.67\times 10^{-27}\times (44.18\times 10^4)^2$

So change in kinetic energy $\Delta KE=KE_F-KE_I=\frac{1}{2}\times 1.6\times 10^{-27}\times 10^8\times (44.18^2-1.6^2)=1560\times 10^{-19}J$

6 0

2 years ago

Given the nuclear equation:

Leokris [45]

This is a nuclear fission reaction, in which a larger nucleus is bombarded with a neutron to make it break down into two smaller nuclei and release energy.

6 0

2 years ago

PLEASE HELP!

lutik1710 [3]

Answer:

Thats her fault.........................b

Explanation:

6 0

2 years ago