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3 years ago

7

Please help me answer this question! Will appreciate help very much. a) Draw a circuit to show 3 lights in series b) Draw a circ

uit to show 2 lights in parallel c) Draw a circuit that shows 2 lights in parallel which are both in series with a third light. d) What would happen to the 2 other lights if the third light in question c) fails

1 answer:

Scorpion4ik [409]3 years ago

3 0

Answer:

C kind sir

Explanation:

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Kirchhoff's loop rule for circuit analysis is an expression of which of the following? Conservation of charge Conservation of en

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Kirchhoff's circuit laws are two equalities that deal with the current and potential difference (commonly known as voltage) in the lumped element model of electrical circuits. They were first described in 1845 by German physicist Gustav Kirchhoff. This generalized the work of Georg Ohm and preceded the work of Maxwell.

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3 years ago

What is the formula for lithium and chlorine?

siniylev [52]

LiCl, because lithium (Li) has one positively charged ion (1+), and
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3 years ago

Read 2 more answers

A 20 cm-radius ball is uniformly charged to 71 nC.

artcher [175]

Answer:

Part a)

$\rho = 2.12\mu C/m^3$

Part b)

$q_1 = 1.11 nC$

$q_2 = 8.88 nC$

$q_3 = 71 nC$

Part c)

$E_1 = 3996 N/C$

$E_2 = 7992 N/C$

$E_3 = 15975 N/C$

Explanation:

Part a)

As we know that charge density is the ratio of total charge and total volume

So here the volume of the charge ball is given as

$V = \frac{4}{3}\pi R^3$

$V = \frac{4}{3}\pi(0.20)^3$

$V = 0.0335 m^3$

now the charge density of the ball is given as

$\rho = \frac{71 nC}{0.0335} = 2.12\mu C/m^3$

Part b)

Now the charge enclosed by the surface is given as

$q = \rho V$

at radius of 5 cm

$q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.05)^3$

$q = 1.11 nC$

at radius of 10 cm

$q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.10)^3$

$q = 8.88 nC$

at radius of 20 cm

$q = 71 nC$

Part c)

As we know that electric field is given as

$E = \frac{kq}{r^2}$

so we have electric field at r = 5 cm

$E_1 = \frac{(9\times 10^9)(1.11 nC)}{0.05^2}$

$E_1 = 3996 N/C$

electric field at r = 10 cm

$E_2 = \frac{(9\times 10^9)(8.88 nC)}{0.10^2}$

$E_2 = 7992 N/C$

electric field at r = 20 cm

$E_3 = \frac{(9\times 10^9)(71 nC)}{0.20^2}$

$E_3 = 15975 N/C$

3 0

3 years ago

4 Suggest four ways in which participating in physical activities can build healthy relationship? 5.Show in two ways the influeb

valina [46]

Answer:

Increases levels of well-being neurotransmitters
Improves fitness
Improves mental health
Reduces the risk of developing diseases

Explanation:

Performing physical activities helps to build a healthy relationship with yourself, in the sense that physical activity promotes health at all levels. Health is a set of physical, mental and social well-being, so physical activity reaches all these levels and increases the quality of life for a person as a whole.

There are several types of physical activities that can be performed and you can choose the one that best fits your preferences and routine. A walk for example can be done in any safe place and at any time and it already helps in improving physical conditioning, in receiving serotonin and endorphins, in reducing blood pressure and protecting the heart and in improving mental health.

So the ideal is for each person to set their own goals in relation to their health and seek to achieve them, either to have a better quality of life, to socialize more with people, to prevent diseases, etc.

8 0

3 years ago

A 2.7-kg ball is thrown upward with an initial speed of 20.0 m/s from the edge of a 45.0 m high cliff. At the instant the ball i

scoray [572]

Answer:

The distance traveled by the woman is 34.1m

Explanation:

Given

The initial height of the cliff

yo = 45m final, positition y = 0m bottom of the cliff

y = yo + ut -1/2gt²

u = 20.0m/s initial speed

g = 9.80m/s²

0 = 45.0 + 20×t –1/2×9.8×t²

0 = 45 +20t –4.9t²

Solving quadratically or by using a calculator,

t = 5.69s and –1.61s byt time cannot be negative so t = 5.69s

So this is the total time it takes for the ball to reach the ground from the height it was thrown.

The distance traveled by the woman is

s = vt

Given the speed of the woman v = 6.00m/s

Therefore

s = 6.00×5.69 = 34.14m

Approximately 34.1m to 3 significant figures.

3 0

3 years ago