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3 years ago

What does AS>0 mean?| I

Physics

2 answers:

bulgar [2K]3 years ago

8 0

Significant other hope it helped

tigry1 [53]3 years ago

5 0

Answer:

Significant Other

Explanation:

that comes to mind when I see that

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If the wind or current is pushing your boat away from the dock as you prepare to dock, which line should you secure first?

Akimi4 [234]

Answer:

Bow Line

Explanation:

If the wind or current is pushing your boat away from the dock, bow line should be secured first.

1- We should cast off the bow and stern lines.

2-With the help of an oar or boat hook, keep the boat clear of the dock.

3-Leave the boat on its own for sometime and let the wind or current carry the boat away from the dock.

4 - As you see there is sufficient clearance, shift into forward gear and slowly leave the area.

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3 years ago

What is an ecosystem? Need answer asap

Evgesh-ka [11]

Answer:

it’s the second one

Explanation:

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2 years ago

How much heat is needed to bring 12.0 g of water from 28.3 °C to 43.87 °C, if the specific heat capacity of water is 4.184 J/(g•

Vika [28.1K]

Answer:

Heat capacity, Q = 781.74 Joules

Explanation:

Given the following data;

Mass = 12g

Initial temperature = 28.3°C

Final temperature = 43.87°C

Specific heat capacity of water = 4.184J/g°C

To find the quantity of heat needed?

Heat capacity is given by the formula;

$Q = mcdt$

Where;

Q represents the heat capacity or quantity of heat.

m represents the mass of an object.

c represents the specific heat capacity of water.

dt represents the change in temperature.

dt = T2 - T1

dt = 43.87 - 28.3

dt = 15.57°C

Substituting into the equation, we have;

$Q = 12*4.184*15.57$

Q = 781.74 Joules

7 0

3 years ago

A plastic rod has been bent into a circle of radius R = 8.00 cm. It has a charge Q1 = +2.70 pC uniformly distributed along one-q

vovikov84 [41]

Answer:

Part a)

$V = -1.52 V$

Part b)

$V = -1.16 V$

Explanation:

Part a)

Electric potential is a scalar quantity

so here we can say that total potential due to a ring on its center is given as

$V = \frac{kQ}{R}$

here we know that

$Q = Q_1 - 6Q_1$

$Q = - 5 Q_1$

$Q_1 = 2.70 pC$

now we have

$V = \frac{(9\times 10^9)(-5\times 2.70 \times 10^{-12})}{0.08}$

$V = -1.52 V$

Part b)

Potential on the axis of the ring is given as

$V = \frac{kQ}{\sqrt{r^2 + R^2}}$

$V = \frac{(9\times 10^9)(-5\times 2.70\times 10^{-12})}{\sqrt{0.08^2 + 0.0671^2}}$

$V = -1.16 V$

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3 years ago

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Kazeer [188]

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3 years ago