3 years ago

Getyour pointsssssssssssssssssssssss

Physics

2 answers:

vladimir2022 [97]3 years ago

8 0

Answer:

hi and thankss

Explanation:

Maurinko [17]3 years ago

3 0

Answer:

thankkkk youuuu soo muchhh

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How to understand that an inductor behaves like short circuit and capacitor like an open circuit in steady state in a network?

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Which of the following is a key assumption of the scientific method

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Explanation:

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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

IceJOKER [234]

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 -

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

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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3 years ago