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3 years ago

7

A student performs an experiment and must measure the lengths of four different objects: a textbook, a pencil, a cup, and a piec

e of bread. Which would be the most appropriate unit of measurement for her to use?

2 answers:

Strike441 [17]3 years ago

7 0

Most appropriate unit of measurement should be centimetres, cm.

ehidna [41]3 years ago

6 0

Answer:

Centimeter (cm)

Explanation:

A student performs an experiment and must measure the lengths of four different objects: a textbook, a pencil, a cup, and a piece of bread.

There are so many units of measuring length of an object like centimeter, millimeter, kilometer etc. Some of units are used to measure very large distances (i.e. the distance between sun and earth etc ) are light year, astronomical units etc.

In this case, a textbook, a pencil, a cup, and a piece of bread are small object. The most appropriate unit of measurement for her to use is centimeter or cm.

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g The potential energy of a pair of hydrogen atoms separated by a large distance x is given by U(x)=−C6/x6, where C6 is a positi

Arisa [49]

Answer:

$F_x = -\frac{6 C_6}{2^7}$

Attractive

Explanation:

Data provided in the question

The potential energy of a pair of hydrogen atoms given by $\frac{C_6}{X_6}$

Based on the given information, the force that one atom exerts on the other is

Potential energy μ = $\frac{C_6}{X_6}$

Force exerted by one atom upon another

$F_x = \frac{\partial U}{\partial X} = \frac{\partial}{\partial X} (-\frac{C_6}{X^6})$

or

$F_x = \frac{\partial}{\partial X} (\frac{C_6}{X^6})$

or

$F_x = -\frac{6 C_6}{2^7}$

As we can see that the $C_6$ comes in positive and constant which represents that the force is negative that means the force is attractive in nature

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3 years ago

The closer together a magnets field lines are,

elena55 [62]

The stronger they will be

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3 years ago

A 57.0 kg cheerleader uses an oil-filled hydraulic lift to hold four 120 kg football players at a height of 1.10 m. If her pisto

lapo4ka [179]

Answer:

The diameter of the piston of the players equals 55.136 cm.

Explanation:

from the principle of transmission of pressure in a hydraulic lift we have

$\frac{F_{1}}{A_{1}}=\frac{F_{2}}{A_{1}}$

Since the force in the question is the weight of the individuals thus upon putting the values in the above equation we get

$\frac{57.0\times 9.81}{\frac{\pi \times (19.0)^{2}}{4}}=\frac{4\times 120\times 9.81}{\frac{\pi \times D_{2}^{2}}{4}}$

Solving for $D_{2}$ we get

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3 years ago

(a) A load of coal is dropped (straight down) from a bunker into a railroad hopper car of inertia 3.0 × 104 kg coasting at 0.50

Firlakuza [10]

Answer:

a) m=20000Kg

b) v=0.214m/s

Explanation:

We will separate the problem in 3 parts, part A when there were no coals on the car, part B when there is 1 coal on the car and part C when there are 2 coals on the car. Inertia is the mass in this case.

For each part, and since the coals are thrown vertically, the horizontal linear momentum p=mv must be conserved, that is, $p=m_Av_A=m_Bv_B=m_Cv_C$ , were each velocity refers to the one of the car (with the eventual coals on it) for each part, and each mass the mass of the car (with the eventual coals on it) also for each part. We will write the mass of the hopper car as $m_h$ , and the mass of the first and second coals as $m_1$ and $m_2$ respectively

We start with the transition between parts A and B, so we have:

$m_Av_A=m_Bv_B$

Which means

$m_hv_A=(m_h+m_1)v_B$

And since we want the mass of the first coal thrown ( $m_1$ ) we do:

$m_hv_A=m_hv_B+m_1v_B$

$m_hv_A-m_hv_B=m_1v_B$

$m_1=\frac{m_hv_A-m_hv_B}{v_B}=\frac{m_h(v_A-v_B)}{v_B}$

Substituting values we obtain

$m_1=\frac{(3\times10^4Kg)(0.5m/s-0.3m/s)}{0.3m/s}=20000Kg=2\times10^4Kg$

For the transition between parts B and C, we can write:

$m_Bv_B=m_Cv_C$

Which means

$(m_h+m_1)v_B=(m_h+m_1+m_2)v_C$

Since we want the new final speed of the car ( $v_C$ ) we do:

$v_C=\frac{(m_h+m_1)v_B}{(m_h+m_1+m_2)}$

Substituting values we obtain

$v_C=\frac{(3\times10^4Kg+2\times10^4Kg)(0.3m/s)}{(3\times10^4Kg+2\times10^4Kg+2\times10^4Kg)}=0.214m/s$

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Lunar phase is the same wherever on Earth you observe
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3 years ago