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svetlana [45]
3 years ago
12

Calculate the vertical distance an object dropped from rest would cover in 12 seconds if it fell freely without air resistance

Physics
1 answer:
Nikolay [14]3 years ago
4 0

Answer:

705.6 m

Explanation:

time, t = 12 sends

initial velocity, u = 0

Acceleration due to gravity, g = - 9.8 m/s^2

As the object is falling freely so the initial velocity is zero.

Let the distance traveled by the body is h in time t  = 12 s.

Use second equation of motion

h = u t - 1/2  gt^{2}

Ad the distance is downwards so take it as negative

So, -h = 0 - 1/2\times 9.8 \times 12 \times 12

h = 705.6 m

Thus, the distance traveled in 12 second is 705.6 m

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3 years ago
If the car passes point A with a speed of 20 m/s and begins to increase its speed at a constant rate of at = 0.5 m/s2 , determin
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Answer:

1.68 \frac{m}{s^2}

Explanation:

Please find the image for the question as attached file.

Solution -

Given -

First of all we will calculate the velocity at point C,

As per newton's third law of motion-

V_C^2 = V_A^2 + 2 a_t (S_C - S_A)\\

Substituting the given values in above equation, we get -

V_C^2 = 20^2 + 2*0.5*(100-0)\\V_C = 22.361 \frac{m}{s}

Now we will determine the radius of curvature for the curve shown in the attached image

Y = 16 - \frac{1}{625} X^2\\

Differentiating on both the sides, we get -

\frac{dy}{dx} = -3.2 (10^-3) X\\\frac{d^2y}{d^2x} =  -3.2 (10^-3)\\Curve = \frac{[1+(\frac{dy}{dx})^2]^{\frac{3}{2}}  }{\frac{d^2y}{d^2x}} \\Curve = 312.5meter

Acceleration on curved path

a = \frac{V_C^2}{Curve} \\a = \frac{22.361^2}{312.5} \\a= 1.60 \frac{m}{s^2}

Final acceleration

a_f = \sqrt{0.5^2 + 1.6^2} \\a_f = 1.68\frac{m}{s^2}

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It is most likely A- an even number
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Read 2 more answers
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