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andreyandreev [35.5K]
2 years ago
6

An object, experiencing no friction, keeps moving at a constant speed. What can we say about the net force on the object?

Physics
1 answer:
sleet_krkn [62]2 years ago
3 0

Answer:

Explanation:

Let's look at a mathematical representation of this. The equation for tis is just a souped up version of Newton's 2nd Law:

F - f = ma. It an object is moving at a constant speed, the acceleration of that object is 0. That changes this equation to

F = f which states that the applied Force equals the frictional force, choice a.

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An amusement park ride consists of a rotating circular platform 11.1 m in diameter from which 10 kg seats are suspended at the e
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To solve this problem we will use the relationship given between the centripetal Force and the Force caused by the weight, with respect to the horizontal and vertical components of the total tension given.

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v = Velocity

r = Radius

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\frac{\frac{mv^2}{r}}{sin\theta} =  \frac{mg}{cos\theta}

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r = \frac{\text{diameter of rotational circular platform}}{2} + \text{length of chains}

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6 0
3 years ago
Calculate the change in internal energy of the following system: A balloon is cooled by removing 0.659 kJ of heat. It shrinks on
bixtya [17]
<h2>Answer:</h2>

-310J

<h2>Explanation:</h2>

The change in internal energy (ΔE) of a system is the sum of the heat (Q) and work (W) done on or by the system. i.e

ΔE = Q + W       ----------------------(i)

If heat is released by the system, Q is negative. Else it is positive.

If work is done on the system, W is positive. Else it is negative.

<em>In this case, the system is the balloon and;</em>

Q = -0.659kJ = -695J    [Q is negative because heat is removed from the system(balloon)]

W = +385J  [W is positive because work is done on the system (balloon)]

<em>Substitute these values into equation (i) as follows;</em>

ΔE = -695 + 385

ΔE = -310J

Therefore, the change in internal energy is -310J

<em>PS: The negative value indicates that the system(balloon) has lost energy to its surrounding, thereby making the process exothermic.</em>

<em />

<em />

5 0
3 years ago
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