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3 years ago

11

A mechanic test driving a car that she has just given a tune-up accelerates from rest to 50.0 m/s in 9.8 s. How far (in meters)

does she travel in that time?

1 answer:

Stels [109]3 years ago

6 0

Answer:

245 m

Explanation:

v = at + v₀

50.0 m/s = a (9.8 s) + 0 m/s

a = 5.10 m/s²

x = x₀ + v₀ t + ½ at²

x = 0 m + (0 m/s) (9.8 s) + ½ (5.10 m/s²) (9.8 s)²

x = 245 m

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1. A pair of oppositely charged parallel plates is separated by 5.51 mm. A potential difference of 614 V exists between the plat

Fittoniya [83]

Answer:

Part a)

$E = 1.11 \times 10^5 N/C$

Part 2)

$F = 1.78 \times 10^{-14} N$

Part 3)

$W = 5 \times 10^{-17} J$

Explanation:

Part 1)

As we know that electric field and potential difference related to each other as

$E = \frac{\Delta V}{x}$

so we will have

$\Delta V = 614 V$

$x = 5.51 mm$

so we have

$E = \frac{614}{5.51 \times 10^{-3}}$

$E = 1.11 \times 10^5 N/C$

Part 2)

Charge of an electron

$e = 1.6 \times 10^{-19} C$

now force is given as

$F = qE$

$F = (1.6 \times 10^{-19})(1.11 \times 10^5)$

$F = 1.78 \times 10^{-14} N$

Part 3)

Work done to move the electron

$W = F.d$

$W = (1.78 \times 10^{-14})(5.51 - 2.7) \times 10^{-3}$

$W = 5 \times 10^{-17} J$

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3 years ago