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garri49 [273]
3 years ago
6

A certain brand of hot-dog cooker works by applying a potential difference of 120 V across opposite ends of a hot dog and allowi

ng it to cook by means of the thermal energy produced. The current is 10.0 A, and the energy required to cook one hot dog is 60.0 kJ. If the rate at which energy is supplied is unchanged, how long will it take to cook three hot dogs simultaneously
Physics
1 answer:
ivanzaharov [21]3 years ago
5 0

Answer:

The time it will take to cook three hot dogs simultaneously is 2.5 minutes

Explanation:

Here we have, the Energy of electric heating given by Joule heating that is;

P = IV = 120×10 = 1200 J/s = 1.2 kJ/s

Since the energy required to cook one hotdog = 60.0 kJ we have

Energy required to cook three hot dogs = 3 × 60.0 kJ  = 180.0 kJ

Therefore, the time required to cook the three hot dogs is

(180.0 kJ)/(1.2 kJ/s) = 150 s

The time it takes to cook three hot dogs simultaneously is

150 seconds or 150/60 minutes which is 2 minutes 30 seconds or 2.5 minutes

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Answer:
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Explanation:
the mechanical advantage measures how much the system multiplies the input force to get the output.

In the given:
The input force (effort) is 20 Newton
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This means that the mechanical advantage is:
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Note that the mechanical advantage is unit-less (has no unit) since it is a ratio between two forces.

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A plastic rod of length d = 1.5 m lies along the x-axis with its midpoint at the origin. The rod carries a uniform linear charge
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Answer:

Explanation:

Let the plastic rod extends from - L to + L .

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If we consider the same strip along the x axis at the same position  on negative x axis , same result will be found . It is to be noted that the component of field in perpendicular to y axis will cancel out each other . Now for electric field due to whole rod at point p , we shall have to integrate the above expression from - L to + L

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A disk with radius R and uniform positive charge density s lies horizontally on a tabletop. A small plastic sphere with mass M a
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Answer:

a. F = Qs/2ε₀[1 - z/√(z² + R²)] b.  h =  (1 - 2mgε₀/Qs)R/√[1 - (1 - 2mgε₀/Qs)²]

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E = s/2ε₀[1 - z/√(z² + R²)]

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F = Qs/2ε₀[1 - z/√(z² + R²)]

b. At what height h does the sphere hover?

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So, F = mg

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expanding the bracket, we have

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collecting like terms, we have

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Factorizing, we have

[1 - (1 - 2mgε₀/Qs)²]h² = (1 - 2mgε₀/Qs)²R²

So, h² =  (1 - 2mgε₀/Qs)²R²/[1 - (1 - 2mgε₀/Qs)²]

taking square-root of both sides, we have

√h² =  √[(1 - 2mgε₀/Qs)²R²/[1 - (1 - 2mgε₀/Qs)²]]

h =  (1 - 2mgε₀/Qs)R/√[1 - (1 - 2mgε₀/Qs)²]

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