Match each term to the best description. ::

What is the power of a parallel circuit with a resistance of 1,000 and a current of 0.03 A?

Kobotan [32]

i squared r = 0.03x0.03x1000=3x0.03x10=.9W

7 0

3 years ago

Will give correct answer brainliest

sineoko [7]

$200 \times 80\%$

<h2>calculate</h2>

$200 \times \frac{80}{100}$

<h2>reduce the numbers</h2>

$2 \times 80$

<h2>multiply</h2>

$= 160$

<h2>there for we have a solution to the equation</h2>

hope it helps

#carry on learning

mark me as brainlist plss

6 0

3 years ago

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A dog runs 30 feet to the north then 5 feet to the south what is the displacement of the dog

Nuetrik [128]

To look for displacement, just draw a vector from your beginning stage to your last position and settle for the length of this line. So we begin by drawing a line to the north which is 30 ft, since it is north, the line is going up, then it move 5 ft to the south, so put a line going down, so we are in 25 ft, North so that would be the answer.

5 0

3 years ago

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A projectile is launched at an angle above the

gtnhenbr [62]

The first rule of vectors is that the horizontal and vertical components are separate. Disregarding air resistance, the only thing we have to worry about is gravity.

The appropriate suvat to use for the vertical component is v = u +at
I will take a to be -9.81, you may have to change it to be 10 if your qualification likes g to be 10.

v = 30 + (-9.81x2)
v = 30 - 19.62
=10.38m/s

Therefore we know that after 2.0 s the vertical component will be 10.38ms^-1, ie 10m/s as the answers given are all to 2sf.

The horizontal component is completely separate to the vertical component and since there is no air resistance, it will remain constant throughout the projectiles trajectory. Therefore it will remain at 40ms^-1.

Combining this together we get:
(1) vx=40m/s and vy=10m/s

7 0

3 years ago

If the distance between two objects decreased, what would happen to the force of gravity between them?

evablogger [386]

Answer: A- It would increase

Explanation:

According to the law of universal gravitation:

$F=G\frac{m_{1}m_{2}}{r^2}$

Where:

$F$ is the module of the attraction force exerted between both objects

$G$ is the universal gravitation constant.

$m_{1}$ and $m_{2}$ are the masses of both objects

$r$ is the distance between both objects

As we can see, the gravity force is directly proportional to the mass of the bodies or objects and inversely proportional to the square of the distance that separates them.

In other words:

<h2>If we decrease the distance between both objects, the gravitational force between them will increase. </h2>

6 0

3 years ago

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