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spayn [35]
4 years ago
13

What is a machine that continues to work without adding additional energy

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

Answer:

That would be a perpetual motion machine

Explanation:

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B is the right answer. Multiply numbers you get the answer
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A boy drops a coin down a well that is 225 m deep. How long does it take the coin to hit the bottom of the well? please help
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Which of the following is true of children with chronic illness? a.) They are all eligible to recievie special education service
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Answer:

c.) Their eligibility for social education services depends on whether their conditions adversely affect their educational functioning.

Explanation:

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The likeliest answer from the options given is option C because before social education services are given, it has to be decided if their health condition adversely affects their education.

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3 years ago
To determine which one of the products in a precipitation reaction is the precipitate, which of the following should you follow?
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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 = 312.5meter

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}

5 0
3 years ago
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