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2 years ago

17. Saan daw nakasakay ang mga Austronesyan nang dumating sa bansa? A. Bangka B. balangay C galyon D. barko

Physics

1 answer:

erastova [34]2 years ago

7 0

Answer:

D.barko po

Explanation:

I trying my best

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Trace fossils would include all of the following EXCEPT

Vadim26 [7]

Answer:

Im pretty sure its fossilized nests because nests arent tracings. hope this helps

Explanation:

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2 years ago

Three resistors connected in series have potential differences across them labeled /\V1 , /\V2 , and /\V3. What expresses the po

Brrunno [24]

Answer:

$\Delta V=\Delta V_1+\Delta V_2+\Delta V_3$

Explanation:

We are given that three resistors R1, R2 and R3 are connected in series.

Let

Potential difference across $R_1=\Delta V_1$

Potential difference across $R_2=\Delta V_2$

Potential difference across $R_3=\Delta V_3$

We know that in series combination

Potential difference , $V=V_1+V_2+V_3$

Using the formula

$\Delta V=\Delta V_1+\Delta V_2+\Delta V_3$

Hence, this is required expression for potential difference.

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3 years ago

Two pebbles are dropped into a pool of water as shown. If a new wave with a

Kazeer [188]

Answer:

Constructive interference

Explanation:

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2 years ago

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What is the primary function of a landfill?

RSB [31]

The most important function of a landfill is to store and get rid of solid waste which it provides an area where waste is disposed of and can be managed properly in order to reduce health and environmental risk<span />

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2 years ago

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A basketball player makes a jump shot. The 0.599 kg ball is released at a height of 2.18 m above the floor with a speed of 7.05

7nadin3 [17]

Answer:

$W_{drag} = 4.223\,J$

Explanation:

The situation can be described by the Principle of Energy Conservation and the Work-Energy Theorem:

$U_{g,A}+K_{A} = U_{g,B} + K_{B} + W_{drag}$

The work done on the ball due to drag is:

$W_{drag} = (U_{g,A}-U_{g,B})+(K_{A}-K_{B})$

$W_{drag} = m\cdot g\cdot (h_{A}-h_{B})+ \frac{1}{2}\cdot m \cdot (v_{A}^{2}-v_{B}^{2})$

$W_{drag} = (0.599\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (2.18\,m-3.10\,m)+\frac{1}{2}\cdot (0.599\,kg)\cdot [(7.05\,\frac{m}{s} )^{2}-(4.19\,\frac{m}{s} )^{2}]$

$W_{drag} = 4.223\,J$

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3 years ago

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