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galben [10]
3 years ago
15

HELP PLEASE 20 POINTS SHOW WORK, ALL EQUATIONS

Physics
1 answer:
nataly862011 [7]3 years ago
7 0

Answer:

s = 3 m

Explanation:

Let t be the time the accelerating car starts.

Let's assume the vehicles are point masses so that "passing" takes no time.

the position of the constant velocity and accelerating vehicles are

s = vt = 40(t + 2)  cm

s = ½at² = ½(20)(t)² cm

they pass when their distance is the same

½(20)(t)² = 40(t + 2)

10t² = 40t + 80

0 = 10t² - 40t - 80

0 = t² - 4t - 8

t = (4±√(4² - 4(1)(-8))) / 2(1)

t = (4± 6.928) / 2  ignore the negative time as it has not occurred yet.

t = 5.464 s

s = 40(5.464 + 2) = 298.564 cm

300 cm when rounded to the single significant digit of the question numerals.

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2 years ago
At time t=0 a positively charged particle of mass m=3.57 g and charge q=9.12 µC is injected into the region of the uniform magne
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Answer:

10.78 s

Explanation:

The force on the charge is computed by using the equation:

F^{\to}= qE^{\to} +q (v^{\to} + B^{\to}) \\ \\  F^{\to} = (9.12 \times 10^{-6}) *278 (-\hat k) +9.12 *10^{-6} *2.1 *0.18 (\hat i * \hat k) \\ \\  F^{\to} = -2.535 *10^{-3} \hat k -3.447*10^{-6} \hat j

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a ^{\to}= \dfrac{-1}{3.57\times 10^{-3}}(2.535*10^{-3}\hat k + 3.447*10^{-6} \hat j)

a ^{\to}=-0.710 \hat k -9.656*10^{-4} \hat j

At time t(sec; the partiCle velocity becomes v(t) = 3.78 v_o

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v^{\to}= v_{o \ \hat i} + a^{\to }t \\ \\  \implies (2.1)\hat i -0.710 t \hat k -9.656 \times 10^{-4} t \hat j = 3.78 v_o \\ \\ \implies (2.1)^2 +(0.710\ t)^2+ (9.656 *10^{-4}t )^2 = (3.78 *2.1^2 \\ \\ \implies 4.41 +0.5041 t^2 +9.324*10^{-7} t^2 = 63.012 \\ \\ \implies 4.41 +0.5041 t^2 = 63.012\\  \\  0.5041t^2 = 63.012-4.41 \\ \\ t^2 = \dfrac{58.602}{0.5041} \\ \\ t^2 = 116.25 \\ \\ t = \sqrt{116.25} \\ \\  \mathbf{t = 10.78 \ s}

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