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Dimas [21]
3 years ago
7

Ursula has something that she thinks is a metal. If she places it above the flame from a candle, it should...? A. quickly get ve

ry cold. B. shock her because it carries a charge C. start to heat up D. become dull and put out the flame?
Physics
2 answers:
AlladinOne [14]3 years ago
5 0
The answer would be C since metals have electrons that help in heating up the metal
SVEN [57.7K]3 years ago
5 0

Answer:

start to heat up

Explanation:

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The acceleration of a particle as it moves along a straight line is given by a (2t 1) m/s2, where t is in seconds. Suppose that
Cerrena [4.2K]

Answer:

 v = 158 m / s

Explanation:

In this kinematic problem they give us the acceleration of the body, the position and speed of the initial moment, let's use the definition of acceleration

       a = dv / dt

        dv = a dt

We integrate

        ∫ d v = ∫ (2t + 1) dt

        v = 2 (2t²) + 1t

We evaluate between the initial time t = 0 that has speed (vo) 8 m/s and the final time t with speed v

        v -8 = 4t2 + t

We already have the velocity formula as a function of time, let's calculate for t = 6 s

         v = 8 + 4 6² +6

         

         v = 158 m / s

3 0
4 years ago
A cat swats a ball on the coffee table. The ball rolls horizontally off the edge of a table. Which statement is true?
Solnce55 [7]

Answer:

D

Explanation:

8 0
3 years ago
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How much must an object weigh to be able to float in water
dolphi86 [110]

Answer:

B

Explanation:

7 0
3 years ago
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15 POINTS PLZ HELP IM ON A TIMER! IM ON EDGE EXAM
sleet_krkn [62]

Answer:

B.) Pressure traps dissolved gases in magma

Explanation:

I hope this answer is correct

8 0
3 years ago
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Let k be the Boltzmann constant. If the configuration of the molecules in a gas changes so that the multiplicity is reduced to o
katen-ka-za [31]

Answer:

ΔS =  - k ln (3)

Explanation:

Using the Boltzmann's expression of entropy, we have;

S = k ln Ω

Where;

S = Entropy

Ω = Multiplicity

From the question, the configuration of the molecules in a gas changes so that the multiplicity is reduced to one-third its previous value. This also causes a change in the entropy of the gas as follows;

ΔS = k ln (ΔΩ)

ΔS = kln(Ω₂)  - kln(Ω₁)

ΔS = kln(Ω₂ / Ω₁)              -------------(i)

Where;

Ω₂ = Final/Current value of the multiplicity

Ω₁ = Initial/Previous value of the multiplicity

Ω₂ = \frac{1}{3} Ω₁         [since the multiplicity is reduced to one-third of the previous value]

Substitute these values into equation (i) as follows;

ΔS = k ln (\frac{1}{3} Ω₁ / Ω₁)

ΔS = k ln (\frac{1}{3})

ΔS = k ln (3⁻¹)

ΔS =  - k ln (3)

Therefore, the entropy changes by - k ln (3)

4 0
3 years ago
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