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Anastaziya [24]
3 years ago
6

Determine the automobile’s braking distance from 90 km/h when it is going up a 5° incline. (Round the final value to one decimal

place.) The automobile’s braking distance from 90 km/h when it is going up a 5° incline is:
Physics
1 answer:
NeX [460]3 years ago
8 0

Answer:

366 m

Explanation:

u = 90 km/h = 25 m/s,

theta = 5 degree

acceleration, a = g Sin theta = 9.8 x Sin 5 = 0.854 m/s^2

The final velocity os zero and let the braking distance be s.

Use third equation of motion

v^2 = u^2 - 2 a s

0 = 25 x 25 - 2 x 0.854 x s

s = 366 m

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Mrrafil [7]

Answer:

0.00903 rad

0.00926 rad

6.268\times 10^{-6}

Explanation:

s = Diameter of the object

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Angle subtended is given by

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For the Moon

\theta_m=\dfrac{3.48\times 10^6}{3.85\times 10^8}\\\Rightarrow \theta_m=0.00903\ rad

The angle subtended by the Moon is 0.00903 rad

For the Sun

\theta_s=\dfrac{1.39\times 10^9}{1.5\times 10^{11}}\\\Rightarrow \theta_s=0.00926\ rad

The angle subtended by the Sun is 0.00926 rad

Area ratio is given by

\frac{A_m}{A_s}=\dfrac{\pi r_m^2}{\pi r_s^2}\\\Rightarrow \frac{A_m}{A_s}=\dfrac{d_m^2}{d_s^2}\\\Rightarrow \frac{A_m}{A_s}=\dfrac{(3.48\times 10^{6})^2}{(1.39\times 10^9)^2}\\\Rightarrow \frac{A_m}{A_s}=6.268\times 10^{-6}

The area ratio is 6.268\times 10^{-6}

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Answer:

(b)

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Answer:

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(b) if the resultant force has a magnitude of 390 N and points west, it would be -390N is eastern reference, and the 1st force is 150N due East, then the additional force would also due east and has a magnitude of

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sineoko [7]
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