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hjlf
3 years ago
6

Alice suffered a minor electrical shock as she switched on her bathroom light. She is concerned and wants to prevent it from hap

pening again.
What should Alice do to prevent getting an electrical shock again?
Physics
2 answers:
snow_tiger [21]3 years ago
7 0
Get a pull chord light switch installed in her bathroom by a qualified electrician asap. Meanwhile, keep hands as dry as possible, and try not to go near that switch until it's either been properly earthed, or whatever the problem actually is, and get qualified advice on what the problem is. Don't have wet feet either, and don't stand in puddles of water whilst operating - i she has to - the switch. 250V AC mains can be lethal, and at least painful.
SSSSS [86.1K]3 years ago
6 0

Explanation :

Prevention that must follow by Alice are as follows :

(1) The first way to solve any electrical problems is to turn off the main source of electricity in your home.

(2) The electric devices must be kept away from water and from any moisture.

(3) She should install GFCI i.e. Ground Fault Circuit Interrupters in her home. GFCI detects the leakage of current in the circuit.

(4)  When electric equipment fuses again and again. it is no longer only an accident - those are signs and symptoms that something is not going the right way.

These are some prevention that Alice must take in case of electric shock.

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Answer:

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5 0
3 years ago
Your GPS shows that your friend’s house is 10.0 km away. But there is a big hill between your houses and you don’t want to bike
kati45 [8]

Answer:

The value is c  = 8 \  km

Explanation:

From the question we are told that

The distance of friends house from your point is a =  10 \  km

The distance of your friends street from your street is b =  6 \  km \  in the \ direction \  towards \  the  \  north

The diagram illustrating this question is shown on the first uploaded image

From the diagram we can apply by Pythagoras theorem as follows

a^2 =  b^2 +  c^2

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6 0
3 years ago
Force F acts between a pair of charges, q1 and q2, separated by a distance d. For each of the statements, use the drop-down menu
lora16 [44]

The initial force between the two charges is given by:

F=k \frac{q_1 q_2}{d^2}

where k is the Coulomb's constant, q1 and q2 the two charges, d their separation. Let's analyze now the other situations:

1. F

In this case, q1 is halved, q2 is doubled, but the distance between the charges remains d.

So, we have:

q_1' = \frac{q_1}{2}\\q_2' = 2 q_2\\d' = d

So, the new force is:

F'=k \frac{q_1' q_2'}{d'^2}= k \frac{(\frac{q_1}{2})(2q_2)}{d^2}=k \frac{q_1 q_2}{d^2}=F

So the force has not changed.

2. F/4

In this case, q1 and q2 are unchanged. The distance between the charges is doubled to 2d.

So, we have:

q_1' = q_1\\q_2' = q_2\\d' = 2d

So, the new force is:

F'=k \frac{q_1' q_2'}{d'^2}= k \frac{q_1 q_2)}{(2d)^2}=\frac{1}{4} k \frac{q_1 q_2}{d^2}=\frac{F}{4}

So the force has decreased by a factor 4.

3. 6F

In this case, q1 is doubled and q2 is tripled. The distance between the charges remains d.

So, we have:

q_1' = 2 q_1\\q_2' = 3 q_2\\d' = d

So, the new force is:

F'=k \frac{q_1' q_2'}{d'^2}= k \frac{(2 q_1)(3 q_2)}{d^2}=6 k \frac{q_1 q_2}{d^2}=6F

So the force has increased by a factor 6.

8 0
3 years ago
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Answer:para enxergar

Explanation:

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