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3 years ago

1/012=1/0.05+1/d' hiiiiiiiiii

Physics

1 answer:

klasskru [66]3 years ago

8 0

Correct question is;

1/0.12 = (1/0.05) + (1/d')

Answer:

d' = -1/700

Explanation:

1/0.12 = (1/0.05) + (1/d')

Let's rearrange to get;

(1/d') = (1/0.12) - (1/0.05)

(1/d') = (1/(12/100)) - (1/(5/100))

(1/d') = 100/12 - 100/5

Let's multiply through by 60 to get rid of the denominators on the right side;

> (1/d') = 500 - 1200

> (1/d') = -700

> d' = -1/700

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2 years ago

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determine the greates possile acceleration of the 975 kg race car so that its front wheels do not leace the gorund

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<u>Answer</u>:

The greatest possible acceleration of the car is $a_G= 6.78 m/s^2$

<u>Explanation</u>:

$N_A+N_B-Mg = 0$

$-N_Aa +N_B(b-a)- \mu_s N_Bh - \mu_s N_Ah = 0$

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$-N_A(1.82) + N_B(2.20 -1.82) -0.9N_B(0.55)-0.8N_A(0.55)=0$

$-N_A(1.82) +0.38 N_B -0.44N_B -0.44N_A=0$

$-2.26N_A -0.06N_B= 0$ ----------------(2)

Solving the equation (1) and(2)

$N_A + N_B = 9564.75$

$-2.26N_A-0.06N_B=0$

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$N_B = 9825.60N$

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Next lets assume that the front wheels contact with the ground N_A = 0

$F_B = Ma_G$

$N_B = M_g$

$N_B - M_g = 0$

$N_B(b-a) –F_Bh = 0$

$F_B = 975a_G$

$N_B-975(9.8) = 0$

$N_B=9564.75N$

$9564.75(2.20 -1.82) -F_B(0.55)=0$

$\frac{3634.605}{0.55}=F_B$

$F_B = 6608.3$

$F_B = Ma_G$

$6608.3 = 975a_G$

$a_G = 6.7778 m/s^2$

$a_G_2 = 6.78m/s^2$

Choosing the critical case

$a_G = min(a_G_1 ,a_G_2)$

$a_G = min(7.848, 6.78)$

$a_G= 6.78 m/s^2$

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