What is an experiment the variable that changes because of the experiment is called

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6 0

The independent (or manipulated) variable is something that the experimenter purposely changes or varies over the course of the investigational

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A 0.100-kilogram apple falls from a height of 1.50 meters to 1.00 meter. Ignoring frictional effects, the total mechanical energ

VashaNatasha [74]

The total mechanical energy is the sum of the kinetic energy and the potential energy. It is easier to solve for the potential energy in this case because we are given with the position of the apple.
PE = mgh
Substituting the mass, acceleration due to gravity and the height,
PE = (0.1 kg)(9.8 m/s²)(1 m) = 0.98 J
Then, we solve for the velocity at that point. The distance it had traveled is 0.5 meters.
d = v²/g ; 0.5 = v²/9.8 m/s²
v = 2.213 m/s
Then, solving for KE
KE = 0.5mv²
Substituting,
KE = 0.5(0.1 kg)(2.213 m/s)² = 0.245 J
Adding the two mechanical forces will give us an answer of 1.225 J. Thus, the answer is the first choice.

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4 years ago

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A leaky 10-kg bucket is lifted from the ground to a height of 16 m at a constant speed with a rope that weighs 0.7 kg/m. initial

IrinaVladis [17]

$\displaystyle W =\lim_{n\to\infty}{\sum_{i=0}^{n}{(678.16 - 36.26\;x_i)\cdot\dfrac{16}{n}}}$ .

<h3>Explanation</h3>

The mass comes in three parts:

the mass of the rope,
the mass of the water in the leaky bucket, and
the mass of the bucket.

Both the mass of the rope $m_\text{rope}$ and the mass of the water in the bucket $m_\text{water}$ varies with the height $x$ of the bucket. Express the two masses as a function of $x$ :

$m_\text{rope} = 0.7\;(16 - x)$ ,

The water in the bucket behaves like yet another rope of density $48 \;\text{kg}/ 16 \;\text{m}= 3.0\;\text{kg}\cdot\text{m}^{-1}$ . The mass of the water left in the bucket at height $x$ will be

$m_\text{water} = 3.0\;(16 - x)$ .

The mass of the bucket is $10\;\text{kg}$ . Combining the three:

$m(x) = m_\text{rope}(x) + m_\text{water}(x) + m_\text{bucket} \\\phantom{m(x)} = 0.7\;(16-x) + 3.0\;(16 - x) + 10\\\phantom{m(x)} = 69.2 - 3.7\;x$ .

Weight of the bucket at height $x$ :

$F_\text{weight}(x) = m(x)\cdot g = 9.8\;(69.2 - 3.7\;x) = 678.16 - 36.26\;x$ .

The bucket moves upward at a constant speed. As a result,

$F(x) = W(x) = 678.16 - 36.26\;x$ .

Express the work required as a definite integral:

$\displaystyle W = \int_{0}^{16}{F(x) \cdot dx} = \int_{0}^{16}{(678.16 - 36.26\;x)\cdot dx}$ .

Rewrite the definite integral as a Riemann Sum:

$\displaystyle W = \int_{0}^{16}{(678.16 - 36.26\;x)\cdot dx} \\\phantom{W}= \lim_{n\to\infty}{\sum_{i=0}^{n}{(678.16 - 36.26\;x_i)\cdot\frac{16-0}{n}}}\\\phantom{W} = \lim_{n\to\infty}{\sum_{i=0}^{n}{(678.16 - 36.26\;x_i)\cdot\frac{16}{n}}}$ .