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2 years ago

How does a parallel circuit change when a branch is added?

Physics

1 answer:

Leto [7]2 years ago

3 0

Answer: The total resistance increases, so the current in the circuit decreases.

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A satellite in a nearly circular orbit is 2000 km above earth's surface. the radius of earth is approximately 6400 km. if the sa

Ann [662]

Velocity is computed using the formula:

$v= \frac{d}{t}$
Where:
V = speed
d = distance traveled
t = time/period

First you need to consider that the orbit is circular. To get the measurement or the distance going around Earth, you will need to get the circumference of the path.

$C = 2 \pi r$

Where:
C = circumference
π = 3.14
r = radius

The Earth has a radius of 6,400km, but you also need to consider that the satellite is orbiting above the surface of the Earth, so you add in the 2,000km to that radius.

r = 6,400Km + 2,000Km = 8,400Km

Next step is to insert that into our circumference formula:

$C = 2 \pi r$
$C = 2 \pi 8,400Km$
$C = 52,778.76 Km$

The distance traveled would then be 52,778.76Km

Now that we have the distance, we can then get the velocity:

$v= \frac{d}{t}$
$v= \frac{52,778.76Km}{12hrs}$
$v= 4,398.23km/hr$

The speed of the satellite is 4,398.23km/hr.

7 0

3 years ago

find the pressure exerted by a force of 8000 newtons on an area of 25m square..give your answer in Newtown/m2

baherus [9]

Answer:

Pressure = 320 N/m²

Explanation:

Given the following data;

Force = 8000 Newton

Area = 25 m²

To find the pressure, we would use the following formula;

Pressure = force/area

Substituting into the formula, we have;

Pressure = 8000/25

Pressure = 320 N/m²

5 0

3 years ago

Can someone check my answers please??

loris [4]

I checked them they are good.

3 0

4 years ago

Read 2 more answers

WILL GIVE BRAINLESS IF U ANSER ALL CORECTLY

erma4kov [3.2K]

Answer:

1-As winds rise up the windward side of a mountain range, the air cools and precipitation falls.

2-Mountains and mountain ranges can cast a rain shadow. As winds rise up the windward side of a mountain range, the air cools and precipitation falls.

3-Mountains and mountain ranges can cast a rain shadow. As winds rise up the windward side of a mountain range, the air cools and precipitation falls. On the other side of the range, the leeward side, the air is dry, and it sinks.

4-Rain shadow deserts are formed because tall mountain ranges prevent moisture-rich clouds from reaching areas on the lee, or protected side, of the range.

5-Mountains and mountain ranges can cast a rain shadow. As winds rise up the windward side of a mountain range, the air cools and precipitation falls. On the other side of the range, the leeward side, the air is dry, and it sinks. So there is very little precipitation on the leeward side of a mountain range.

6-Mountains and mountain ranges can cast a rain shadow. As winds rise up the windward side of a mountain range, the air cools and precipitation falls. On the other side of the range, the leeward side, the air is dry, and it sinks. So there is very little precipitation on the leeward side of a mountain range.

Explanation:

#6 and 5 are the same

5 0

3 years ago

Find time when boy catches the girl or when they are at their closest separation..

soldier1979 [14.2K]

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\implies{Time=10\:sec\:and\:30\:sec}}} \\\\$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}} \\ \\$

$\green{\underline{\bold{Given :}}} \\ \\ \:\:\:\: \bullet\:\:\tt\red{ Velocity \: of \: boy = 50 \: m/s} \\ \\ \:\:\:\: \bullet\:\: \tt\orange{Velocity \: of \: girl = 30 \: m/s }\\ \\ \:\:\:\: \bullet\:\:\tt\green{ Acceleration \: of \: boy = {1 \: m/s}^{2}} \\ \\ \:\:\:\:\bullet\:\: \tt\blue{Acceleration \: of \:girl= {2\: m/s}^{2} }\\ \\ \:\:\:\: \bullet \:\:\tt\purple{Sepration \: between \: them = 150 \: m }\\ \\ \\ \red{\underline{\bold{To \: Find :}}} \\ \\ \:\:\:\: \bullet\:\: \tt\blue{Time \: taken \: to \: catch \: the \: girl }\\$

<u>According to given question</u> :

$\\ \green{\star} \: \text{Using \: relative \: motion \: method} \\ \\ \green{ \circ} \: \tt Net \: velocity = 50 - 30 = 20 \: m/s \\ \\ \green{ \circ} \: \tt Net \: acceleration = 1 - 2 = - 1\: m /{s}^{2} \\\\ \\ \star\:\bold\red{\underline{\:As \: we \: know \: that\:}} \\\\ \tt\purple{: \implies s = ut + \frac{1}{2} {at}^{2}} \\ \\ \tt\green{: \implies 150 = 20 \times t + \frac{1}{2} \times -1 \times {t}^{2}} \\ \\ \tt\purple{: \implies 300 = 40t - {t}^{2}} \\ \\ \tt\green{: \implies {t}^{2} - 40t + 300 = 0} \\ \\ \tt\purple{: \implies t = \frac{ - ( - 40) \pm\sqrt{ { (- 40)}^{2} - 4 \times 1 \times 300} }{2 \times 1} }\\ \\ \tt\green{: \implies t = \frac{40 \pm \sqrt{1600 - 1200} }{2} }\\ \\ \tt\purple{: \implies t = \frac{40 \pm 20}{2} }\\ \\ \green{\tt: \implies t = 10 \: sec \: and \: 30 \: sec}$

7 0

3 years ago