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2 years ago

The distance and displacement of a object in motion can be the same (true or false)

Physics

1 answer:

Rasek [7]2 years ago

4 0

Aswer:

False, the values of the distance traveled and the displacement only coincide when the trayectorie is a straight line. Otherwise, the distance will always be greater than the offset.

Although these terms are used synonymously in other cases, they are totally different. Since the distance that a mobile travels is the equivalent of the length of its trajectory. Whereas, the displacement will be a vector magnitude.

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An office building has a 24-volt branch circuit installed for landscape lighting around the front of the building. The circuit w

Arturiano [62]

The circuit was installed in UF cable which requires a minimum burial depth of 6 inches for this circuit.

<h3>UF cable</h3>

UF cable is used as an underground feeder cable to distribute power from an existing building to outdoor equipment. UF cable can also be used as direct burial cable.

For the 24-volt branch circuit installed, the minimum burial depth will be 6 inches.

Thus, the circuit was installed in UF cable which requires a minimum burial depth of 6 inches for this circuit.

Learn more about UF cable here: brainly.com/question/8591560

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2 years ago

What is true about solar energy hitting earth?

polet [3.4K]

B. It’s the same roughly at all latitudes

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2 years ago

Which two chemists organized elements based on properties such as how the elements react or whether they are solid or liquid

Lelechka [254]

Doberiner and lavoiser

5 0

3 years ago

Read 2 more answers

explain why cups of soup at a take away kiosk are often sold in white polystrene cups with a lid to stop spillage

Talja [164]

Answer:

polystyrene is a good insulater so less heat will escape from the cup and it will keep it warm.

the cup helps it become more insulated

5 0

3 years ago

Xelga [282]

Answer:

About 7.67 m/s.

Explanation:

Mechanical energy is always conserved. Hence:

$\displaystyle \begin{aligned} E_i & = E_f \\ \\ U_i + K_i &= U_f + K_f\end{aligned}$

Where <em>U</em> is potential energy and <em>K</em> is kinetic energy.

Let the bottom of the slide be where potential energy equals zero. As a result, the final potential energy is zero. Additionally, because the child starts from rest, the initial kinetic energy is zero. Thus:

$\displaystyle U_i = K_f$

Substitute and solve for final velocity:
$\displaystyle \begin{aligned} mgh_i &= \frac{1}{2}mv_f^2 \\ \\ 2gh_i &= v^2_f \\ \\ v_f &= \sqrt{2gh_i} \\ \\ & =\sqrt{2(9.8\text{ m/s$^2$})(3.00\text{ m})} \\ \\ & \approx 7.67\text{ m/s} \end{aligned}$

In conclusion, the child's speed at the bottom of the slide is about 7.67 m/s.

8 0

2 years ago