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vaieri [72.5K]
2 years ago
9

Letícia leaves the grocery store and walks 150. 0 m to the parking lot. Then, she turns 90° to the right and walks an additional

70. 0 m to her car. What is the magnitude of the displacement of her car from the grocery store exit? round your answer to the nearest tenth.
Physics
2 answers:
slava [35]2 years ago
8 0

d = displacement

a triangle, displacement = hypotenuse = phytagoras

\tt d=\sqrt{70^2+150^2}=165.5~m

Eduardwww [97]2 years ago
6 0
165.5 m is the answer
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The height of a typical playground slide is about 1.8 m and it rises at an angle of 30 ∘ above the horizontal.
Salsk061 [2.6K]

Answer:

5.94\ \text{m/s}

1.7

0.577

Explanation:

g = Acceleration due to gravity = 9.81\ \text{m/s}^2

\theta = Angle of slope = 30^{\circ}

v = Velocity of child at the bottom of the slide

\mu_k = Coefficient of kinetic friction

\mu_s = Coefficient of static friction

h = Height of slope = 1.8 m

The energy balance of the system is given by

mgh=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 9.81\times 1.8}\\\Rightarrow v=5.94\ \text{m/s}

The speed of the child at the bottom of the slide is 5.94\ \text{m/s}

Length of the slide is given by

l=h\sin\theta\\\Rightarrow l=1.8\sin30^{\circ}\\\Rightarrow l=0.9\ \text{m}

v=\dfrac{1}{2}\times5.94\\\Rightarrow v=2.97\ \text{m/s}

The force energy balance of the system is given by

mgh=\dfrac{1}{2}mv^2+\mu_kmg\cos\theta l\\\Rightarrow \mu_k=\dfrac{gh-\dfrac{1}{2}v^2}{gl\cos\theta}\\\Rightarrow \mu_k=\dfrac{9.81\times 1.8-\dfrac{1}{2}\times 2.97^2}{9.81\times 0.9\cos30^{\circ}}\\\Rightarrow \mu_k=1.73

The coefficient of kinetic friction is 1.7.

For static friction

\mu_s\geq\tan30^{\circ}\\\Rightarrow \mu_s\geq0.577

So, the minimum possible value for the coefficient of static friction is 0.577.

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