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UkoKoshka [18]
4 years ago
7

Answer this please. thanks in advance!! please tel me                                                                 a christma

s tree lights are set up in series
1. a set of lights has 50 bulbs  what will be the voltage across each bulb?
show your working out.
Physics
2 answers:
saw5 [17]4 years ago
6 0
If 50 identical light bulbs are connected in series across
a single power source, then the voltage across each bulb
is ( 1/50 ) of the voltage delivered by the power source.
4vir4ik [10]4 years ago
3 0
Presuming that this tree is connected into a mains power source in the UK. The total output of the UK mains source is 230V. So if we share this voltage equally between 50 bulbs:
230V \div 50 = 4.6V

I'm not 100 per cent sure this is correct though and different countries may have different mains power supplies. Hope it helps anyway.

EDIT: This also does not take into account resistance in the bulbs and wires but is all I can tell you from the information provided in the question.
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They use Scientific Law to support theories made by people who work in the Scientific field.

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Speed and velocity both measure an object’s rate of motion. However, speed is a scalar quantity, which means that it can be described with a numerical value. Velocity is a vector quantity, which depends on direction as well as magnitude. Read these definition for more information:

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Basically, an object’s speed tells you how fast it’s going. Its velocity tells you how fast it’s going in a certain direction. You use speed measurements in your daily life, but physicists depend on velocity measurements more frequently in their work.

Explanation:

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By what factor must we increase the amplitude of vibration of an object at the end of a spring in order to double its maximum sp
strojnjashka [21]

Answer:

A'=2A

Explanation:

According to the law of conservation of energy, the total energy of the system can be expresed as the sum of the potential energy and kinetic energy:

E=U+K=\frac{kA^2}{2}\\E=\frac{kx^2}{2}+\frac{mv^2}{2}=\frac{kA^2}{2}

When the spring is in its equilibrium position, that is x=0, the object speed its maximum. So, we have:

\frac{k(0)^2}{2}+\frac{mv_{max}^2}{2}=\frac{kA^2}{2}\\A^2=\frac{mv_{max}^2}{k}\\A=\sqrt{\frac{mv_{max}^2}{k}}

In order to double its maximum speed, that is v'{max}=2v_{max}. We have:

A'=\sqrt{\frac{m(v'_{max})^2}{k}}\\A'=\sqrt{\frac{m(2v_{max})^2}{k}}\\A'=\sqrt{\frac{4mv_{max}^2}{k}}\\A'=2\sqrt{\frac{mv_{max}^2}{k}}\\A'=2A

6 0
3 years ago
A projectile was launched horizontally with a velocity of 388 m/s, 2.89 m above the ground. How long did it take the projectile
tamaranim1 [39]

Answer:

Explanation:

Given

Velocity = 388m/s

Height S = 2.89m

Required

Time

Using the equation of motion

S =ut+1/2gt²

2.89 = 388t+1/2(9.8)t²

2.89 = 388t+4.9t²

Rearrange

4.9t²+388t-2.89 =0

Factorize

t = -388±√388²-4(4.9)(2.89)/2(4.9)

t= -388±√(388²-56.644)/9.8

t = -388±387.93/9.8

t =0.073/9.8

t = 0.00744 seconds

6 0
3 years ago
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