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Zina [86]
3 years ago
8

A strand of 10 lights is plugged into an outlet. How can you determine if the lights are connected in series or parallel? Select

all that apply.
Question 8 options:

1) Unscrew one light. If the other lights turn off, it's a series circuit.


2) Unscrew one light. If the other lights stay on, it's a series circuit.


3) Unscrew one light. If the other lights stay on, it's a parallel circuit.


4) Unscrew one light. If the other lights turn off, it's a parallel circuit.

Physics
1 answer:
ExtremeBDS [4]3 years ago
7 0

Answer:

1) True; 2) False; 3) True; 4) False

Explanation:

To better understand and solve this problem in an easier way, the attached diagram shows two circuits A & B. A is connected in series and B in parallel. In each of the circuits, we can see that different currents are generated, in the serial circuit, you have a single current, while the parallel circuit you have the main current plus two currents.

1)

This is true (for a series circuit) , when one of the bulbs is removed the only current that exists in the circuit is interrupted, so the other bulbs will not work.

2)

This is false, removing a light bulb involves interrupting the current, therefore no light bulb in this type of series circuits will continue its work.

3)

This is true, because when removing one of the bulbs, the current is not interrupted for the other bulbs, each has its own current.

4)

This is false, a circuit connected in parallel, allows the division of the current, in each of its loads, so each load will have its respective current.

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A catamaran with a mass of 5.44×10^3 kg is moving at 12 knots. How much work is required to increase the speed to 16 knots? (One
Andre45 [30]

The work that is required to increase the speed to 16 knots is 14,176.47 Joules

If a catamaran with a mass of 5.44×10^3 kg is moving at 12 knots, hence;

5.44×10^3 kg = 12 knots

For an increased speed to 16knots, we will have:

x = 16knots

Divide both expressions

\frac{5.44 \times 10^3}{x} = \frac{12}{16}\\12x = 16 \times 5.44 \times 10^3\\x = 7.23\times 10^3kg\\

To get the required work done, we will divide the mass by the speed of one knot to have:

w=\frac{7230}{0.51}\\w= 14,176.47Joules

Hence the work that is required to increase the speed to 16 knots is 14,176.47 Joules

Learn more here: brainly.com/question/25573786

8 0
2 years ago
Displacement vectors of 4km north, 2km south, 5km north, 5km south combine to a total displacement of
goldfiish [28.3K]

<u>Answer</u>

The combined displacement is 2km north


<u>Explanation</u>


Since displacement is a vector quantity, we take into account the direction.


Good for us all the displacement vectors are in the same dimension, so we can make north positive and south negative or vice-versa.


We now add to obtain,

4+-2+5+-5

This will simplify to

=4-2+5-5=2

Therefore the combined displacement is 2km north

5 0
3 years ago
A cyclical heat engine, operating between temperatures of 450º C and 150º C produces 4.00 MJ of work on a heat transfer of 5.00
gogolik [260]

Answer:

(a) Heat transfer to the environment is: 1 MJ and (b) The efficiency of the engine is: 41.5%

Explanation:

Using the formula that relate heat and work from the thermodynamic theory as:W=Q=Q_{in}-Q_{out} solving to Q_out we get:Q_{out}=Q_{in}-W=5(MJ)-4(MJ)=1(MJ) this is the heat out of the cycle or engine, so it will be heat transfer to the environment. The thermal efficiency of a Carnot cycle gives us: n=1-\frac{T_{Low} }{T_{High}} where T_Low is the lowest cycle temperature and T_High the highest, we need to remember that a Carnot cycle depends only on the absolute temperatures, if you remember the convertion of K=°C+273.15 so T_Low=150+273.15=423.15 K and T_High=450+273.15=723.15K and replacing the values in the equation we get:n=1-\frac{423.15}{723.15} =0.415=41.5%

5 0
3 years ago
A mass hanging from a spring oscillates with a period of 0.35 s. Suppose the mass and spring are swung in a horizontal circle, w
Annette [7]

Answer:

66 rpm

Explanation:

The period of oscillation is given by

T=2\pi \sqrt{\frac {m}{k}}

\frac {k}{m}=\frac {4\pi^{2}}{T^{2}} where  T is time period of oscillation which is given as 0.35 s, k s spring constant and m is the mass of the object attached to the spring.

Also, net force is given by

Net force=m\omega^{2} L

\omega=\sqrt{\frac {k\triangle L}{mL}} where \triangle L is the elongation, L is original length, \omega is the angular velocity

Substituting the equation of \frac {k}{m} into the above we obtain

\omega=\sqrt {\frac {4\pi^{2}\triangle L}{T^{2} L}}

\omega=\sqrt {4\pi^{2}\times 0.15L}{0.35^{2}\times L}}=6.952763\approx 6.95 rad/s

6.95\times\frac {60 s}{2\pi rad}\approx 66 rpm

6 0
3 years ago
Anyone know what the answer is?..
Licemer1 [7]

Answer:

sometimes harmful and sometimes beneficial

8 0
3 years ago
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