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3 years ago

Identifying Advantages of Parallel Circuits

Physics

2 answers:

I am Lyosha [343]3 years ago

8 0

Answer: The correct answers are B: If one bulb goes out, the other bulbs stay lit, and

C: If there is a break in one branch of the circuit, current can still flow through the other branches.

hope this helps you :3

melisa1 [442]3 years ago

3 0

Answer: If one bulb goes out the other bulbs stay lit.

If there is a break in one branch of the circuit, current can still flow through the other branches.

Explanation:

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Where do you think the government should put the greatest support: solar energy, wind energy, clean coal, oil exploration, gas e

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The government should put it support in a combination of sources, as no source in the present can fully provide all energy requirements.

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3 years ago

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Which best describes the current atomic theory?

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At the intersection of Texas Avenue and University Drive,

Zielflug [23.3K]

Answer:

The initial speed of the truck is 21.93 m/s, and the initial speed of the car is 19.524 m/s

Explanation:

We can use conservation of momentum to find the initial velocities.

Taking the unit vector $\hat{i}$ pointing north and $\hat{j}$ pointing east, the final velocity will be

$\vec{V}_f = 16.0 \frac{m}{s} \ ( \ cos(24.0 \°) \ , \ sin (24.0 \°) \ )$

$\vec{V}_f = ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )$

The final linear momentum will be:

$\vec{P}_f = (m_{car}+ m_{truck}) * V_f$

$\vec{P}_f = (950 \ kg \ + 1900 \ kg \ ) * ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )$

$\vec{P}_f = (2.850 \ kg \ ) * ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )$

$\vec{P}_f = ( \ 41,658.45 \frac{ kg \ m}{s} \ , \ 18,547.8 \frac{kg \ m}{s} \ )$

As there are not external forces, the total linear momentum must be constant.

So:

$\vec{P}_0= \vec{P}_f$

As initially the car is travelling east, and the truck is travelling north, the initial linear momentum must be

$\vec{P}_0= ( m_{truck} * v_{truck}, m_{car}* v_{car} )$

so:

$\vec{P}_0= \vec{P}_f$

$( m_{truck} * v_{truck}, m_{car}* v_{car} ) = ( \ 41,658.45 \frac{ kg \ m}{s} \ , \ 18,547.8 \frac{kg \ m}{s} \ )$

$\left \{ {{m_{truck} \ v_{truck} = 41,658.45 \frac{ kg \ m}{s} } \atop {m_{car} \ v_{car}=18,547.8 \frac{kg \ m}{s} }} \right.$

So, for the truck

$m_{truck} \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}$

$1900 \ kg \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}$

$v_{truck} = \frac{41,658.45 \frac{ kg \ m}{s}}{1900 \ kg}$

$v_{truck} = 21.93 \frac{m}{s}$

And, for the car

$950 \ kg \ v_{car}=18,547.8 \frac{kg \ m}{s}$

$v_{car}=\frac{18,547.8 \frac{kg \ m}{s}}{950 \ kg}$

$v_{car}=19.524 \frac{m}{s}$

5 0

3 years ago

Help me with this please

creativ13 [48]

Answer:

i dont understand, can you put it in word form please.

Explanation:

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3 years ago

In the Niagara Falls hydroelectric generating plant, the energy of falling water is converted into electricity. The height of th

Tamiku [17]

Answer:

2.124 kg of water

Explanation:

height of the falls is about 48 meters.

Mass of water needed is 1kg = 1000g

Power needed is 106 watts.

The amount of energy in 106 watts in one sec is 106 joules.

To calculate the energy of the 1kg falling water = Mgh

Energy = 1000*9.81*48

Energy = 470880 joules.

1 megawatt is = 1000000watts

The kilogram of water needed is 1000000/470880 = 2.124 kg of water

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3 years ago