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3 years ago

An object has a. Kinetic energy of 17j and a mass of 27 kg, how fast is the object moving?

Physics

1 answer:

kolezko [41]3 years ago

4 0

Answer:

1.1 m/s

Explanation:

W = (mv^2)/2

W = 17J

m= 27kg

v = $\sqrt{\frac{2W}{m} }$ = $\sqrt{\frac{2*17}{27} }$ ≈ 1.1 m/s

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A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum

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Answer:

241.7 s

Explanation:

We are given that

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Final potential difference= $V_2=0.351 V$

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

$qV=K.E$

Using the formula

$2.74\times 10^{-6}V_1=6.65\times 10^{-10} J$

$V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V$

Initial voltage= $V_1=2.43\times 10^{-4} V$

$\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2$

Using the formula

$\frac{V_1}{V_2}=(\frac{6.36}{t})^2$

$\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}$

$t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}$

$t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}$

$t=241.7 s$

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

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3 years ago

A very good insulator has an R-value of 29. The material's heat transfer coefficient is

Fittoniya [83]

Answer:

b. 0.034

Explanation:

The heat transfer coefficient of a material (U-value) is equal to the reciprocal of its R-value, therefore:

$U = \frac{1}{R}$

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R is the R-value of the material

For the insulator in this problem,

R = 29

Substituting into the equation, we find the heat transfer coefficient:

$U=\frac{1}{29}=0.034$

4 0

2 years ago