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Softa [21]
2 years ago
8

Why will the circuit stop working if the wites were made out of plastic instead of copper?

Physics
1 answer:
anastassius [24]2 years ago
4 0

Answer:

plastic isn't conductive

Explanation:

Correct me if I'm wrong but plastic isn't conductive copper is so if it's not conductive it will probably mess up the circuit or somthing

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Determine weather or not the equation below is balanced.
Liula [17]

Answer:

Yep.. It's balanced and its a combination reaction

Explanation:

Reactants : S₈ + 24F₂

Product 8SF₆

8 0
3 years ago
A bucket that has a mass of 20 kg when filled with sand needs to be lifted to the top of a 15 meter tall building. You have a ro
prohojiy [21]

Answer:

work done lifting the bucket (sand and rope) to the top of the building,

W=67.46 Nm

Explanation:

in this question we have given

mass of bucket=20kg

mass of rope=.2\frac{kg}{m}

height of building= 15 meter

We have to find the work done lifting the bucket (sand and rope) to the building =work done in lifting the rope + work done in lifting the sand

work done in lifting the rope is given as,W_{1}=Force \times displacement

=\int\limits^{15}_0 {.2x} \, dx ..............(1)

=.1\times 15^2

=22.5 Nm

work done in lifting the sand is given as,W_{2}=Force \times displacement

W_{2}=\int\limits^{15}_0 F \, dx.................(2)

Here,

F=mx+c

here,

c=20-18

c=2

m=\frac{20-18}{15-0}

m=.133

Therefore,

F=.133x+2

Put value of F in equation 2

W_{2}=\int\limits^{15}_0 (.133x+2) \, dx

W_{2}=.133 \times 112.5+2\times15\\W_{2}=14.96+30\\W_{2}=44.96 Nm

Therefore,

work done lifting the bucket (sand and rope) to the top of the building,W=W_{1}+W_{2}

W=22.5 Nm+44.96 Nm

W=67.46 Nm

4 0
3 years ago
A reversible Carnot engine has an efficiency of 30% and it operates between a heat source maintained at T 1 , and a refrigerator
bixtya [17]

To solve this problem we will apply the concepts related to the thermal efficiency given in an engine of the Carnot cycle. Here we know that efficiency is given under the equation

\eta = 1 - \frac{T_2}{T_1}

Where,

T_2 = Temperature of Cold Body

T_1 =Temperature of Hot Body

\eta= Efficiency

According to the statement our values are:

\eta = 0.3

T_2 = 210K

Replacing we have that

0.3 = 1- \frac{210}{T_1}

\frac{210}{T_1} = 0.7

T_1 = \frac{210}{0.7}

T_1 = 300K

Therefore the temperature of the heat source is 300K

6 0
3 years ago
Greg has decided to incorporate earth science into his teaching with snowflakes. As he walks around the room, he notices that ma
Nastasia [14]
A. By asking them to go outside and study real snowflakes.
4 0
3 years ago
Podrian ayudarme? Llevo dias intentando pero no se como 1. el profesor no explico
Paraphin [41]
You have to upload this in the area of mathematicians..!
8 0
3 years ago
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