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I am Lyosha [343]
2 years ago
12

You are given a copper bar of dimensions 3 cm × 5 cm × 8 cm and asked to attach leads to it in order to make a resistor. (a) If

you want to achieve the SMALLEST possible resistance, you should attach the leads to the opposite faces that measure A) 3 cm × 5 cm. B) 3 cm × 8 cm. C) 5 cm × 8 cm. D) Any pair of faces produces the same resistance. (b) If you want to achieve the LARGEST possible resistance, you should attach the leads to the opposite faces that measure A) 3 cm × 5 cm. B) 3 cm × 8 cm. C) 5 cm × 8 cm. D) Any pair of faces produces the same resistance.
Physics
1 answer:
dem82 [27]2 years ago
4 0

Answer:

A) If you want to achieve the SMALLEST possible resistance, you should attach the leads to the opposite faces that measure b) 5 cm by 8 cm.

B) If you want to achieve the LARGEST possible resistance, you should attach the leads to the opposite faces that measure a) 3 cm × 5 cm

Explanation:

Resistivity is directly proportional to lenght and inversely properly to cross sectional area.

For the first case, 5 cm by 8 cm gives the largest area and leave 3 cm as the lenght. The resistivity of the metal will be smallest in these dimensions.

For the second case, 3 cm by 5 cm gives the smallest area, leaving 8 cm as the lenght. This is the maximum arrangement that can give the largest resistance possible.

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Answer:

He traveled 9km

Explanation:

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2 years ago
"The International Space Station (ISS) orbits at a distance of 350 km above the surface of the Earth. (a) Determine the gravitat
vagabundo [1.1K]

Answer:

(a) g = 8.82158145m/s^2.

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Explanation:

(a) Strength of gravitational field 'g' by definition is

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r = 6721,000 meters, putting this value in above equation gives g = 8.82158145m/s^2.

(b) We have to essentially calculate centripetal acceleration that equals new 'g'.

a_{centripetal}=\frac{V^2}{r} =g here g is known, r is known and v is unknown.

plugging in r and g in above and solving for unknown gives V = 7699.990192m/s.

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3 years ago
Suppose a cup of coï¬ee is at 100 degrees Celsius at time t = 0, it is at 70 degrees at t = 10 minutes, and it is at 50 degrees
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Answer:

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Explanation:

Using Newton's Law of Cooling:

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We have two points to evaluate the above equation as follows:

T = 70 @ t = 10 using Eq 1  

70 = Tamb + (100 - Tamb)*e^(-10k)   ... Eq 2

T = 50 @ t = 20 using Eq 1

50 = Tamb + (100 - Tamb)*e^(-20k)   ... Eq 3

Solving the above Eq 2 and Eq 3 simultaneously:

Using Eq 2:

(70 - Tamb) / (100 - Tamb) = e^(-10k)  

Squaring both sides we get:

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Substitute Eq 4 into Eq 3

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After simplification:

50 = (Tamb (100-Tamb) + (70-Tamb)^2) / (100 - Tamb)

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