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2 years ago

What must happen to the electric current in power lines before it can safely enter your home?

2 answers:

7 0

Answer: use the step down transformers to reduce these voltages in the power lines before they are supplied to the users.

Explanation:

lozanna [386]2 years ago

5 0

Answer:

It passes through a step-down transformer.

Explanation:

Electric current in power lines must be stepped down before they enter our homes because they carry a very high voltage.

The voltage can destroy appliances that have very low voltage.

In this regard, a step down transformer is used to achieve this effect.
The transformer brings the voltage to a level fit to consumers to use.
When voltage is stepped down, appliances can work perfectly.

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What sphere on earth includes mountains

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The mean diameters of Mars and Earth are 6.9 ✕ 103 km and 1.3 ✕ 104 km, respectively. The mass of Mars is 0.11 times Earth's mas

Roman55 [17]

Answer:

(a) Ratio of mean density is 0.735

(b) Value of g on mars 0.920 $m,/sec^2$

Explanation:

We have given radius of mars $R_{mars}=6.9\times 10^3km=6.9\times 10^6m$ and radius of earth $R_{E}=1.3\times 10^4km=1.3\times 10^7m$

Mass of earth $M_E=5.972\times 10^{24}kg$

So mass of mars $M_m=5.972\times\times 0.11 \times 10^{24}=0.657\times 10^{24}kg$

Volume of mars $V=\frac{4}{3}\pi R^3=\frac{4}{3}\times 3.14\times (6.9\times 10^6)^3=1375.357\times 10^{18}m^3$

So density of mars $d_{mars}=\frac{mass}{volume}=\frac{0.657\times 10^{24}}{1375.357\times 10^{18}}=477.69kg/m^3$

Volume of earth $V=\frac{4}{3}\pi R^3=\frac{4}{3}\times 3.14\times (1.3\times 10^7)^3=9.198\times 10^{21}m^3$

So density of earth $d_{E}=\frac{mass}{volume}=\frac{5.972\times 10^{24}}{9.198\times 10^{21}}=649.271kg/m^3$

(A) So the ratio of mean density $\frac{d_{mars}}{d_E}=\frac{477.69}{649.27}=0.735$

(B) Value of g on mars

g is given by $g=\frac{GM}{R^2}=\frac{6.67\times 10^{-11}\times0.657\times 10^{24}}{(6.9\times 10^6)^2}=0.920m/sec^2$

$v=\sqrt{\frac{2GM}{R}}=\sqrt{\frac{2\times 6.67\times 10^{-11}\times 0.657\times 10^{24}}{6.9\times 10^6}}=3.563\times 10^4m/sec$

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3 years ago

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What is the change in thermal energy E if the coefficent of kinetic friction between the box and floor is .4 , the distance the

Jet001 [13]

This question can be solved using the concept of friction energy.

The thermal energy change is b "258.4 J".

The change in thermal energy will be equal to the friction energy produced during the motion of the box.

$Change\ In\ Thermal\ Energy = E = Friction\ Energy\\\\E = \mu fd$

where,

μ = coefficient of kinetic friction = 0.4

f = force applied = 38 N

d = distance traveled by the box = 17 m

Therefore,

$E = (0.4)(38\ N)(17\ m)$

Learn more about friction energy here:

brainly.com/question/1343045?referrer=searchResults

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1 year ago