2 years ago

Pure water is a(n) _____.

Physics

1 answer:

Mashutka [201]2 years ago

6 0

The correct answer would be compound

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Why is electrical energy considered as potential energy as its the energy of moving electrons?

Veseljchak [2.6K]

Answer:

Explanation: When the electrons move in another direction, they convert this chemical potential energy to electricity in the circuit, thus discharging the battery. So, the battery is all potential energy.

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1 year ago

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A camper walks away from camp. He walks 23.72m north, then 39.25m south and

Arte-miy333 [17]

Answer:

89.53 North

Explanation:

Total N-S displacement

23.72 - 39.25 + 105.06 = + 89.53

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1 year ago

A force of 15 newtons is used to push a box along the floor a distance of 3 meters. How much work was done?

jeka57 [31]

Answer:

<h3>The answer is 45 J</h3>

Explanation:

The work done by an object can be found by using the formula

<h3>workdone = force × distance</h3>

From the question

distance = 3 meters

force = 15 newtons

We have

workdone = 15 × 3

We have the final answer as

Hope this helps you

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2 years ago

If a particle's position is given by x=4-12t+3t^2, where t is in seconds and x is in meters, what is its velocity at t=1 second?

andreyandreev [35.5K]

Answer:

v = -6m/s

Explanation:

$x=4-12t+3t^2$

$\frac{dx}{dt}=-12+6t$

For t = 1:

$\frac{dx}{dt}=-6$

3 0

3 years ago

A diverging lens (f1 = -12.0cm) is located 50.0 cm to the left of a converging lens (f2 = 34.0 cm). A 2.0 cm-tall object stands

pishuonlain [190]

Answer:

The final image relative to the converging lens is 34 cm.

Explanation:

Given that,

Focal length of diverging lens = -12.0 cm

Focal length of converging lens = 34.0 cm

Height of object = 2.0 cm

Distance of object = 12 cm

Because object at focal point

We need to calculate the image distance of diverging lens

Using formula of lens

$\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}$

$\dfrac{1}{v}=\dfrac{1}{-12}-\dfrac{1}{-12}$

$v=\infty$

The rays are parallel to the principle axis after passing from the diverging lens.

We need to calculate the image distance of converging lens

Now, object distance is ∞

Using formula of lens

$\dfrac{1}{v}=\dfrac{1}{34}-\dfrac{1}{\infty}$

$v=34$

The image distance is 34 cm right to the converging lens.

Hence, The final image relative to the converging lens is 34 cm.

5 0

3 years ago