Find the current through a person and identify the likely effect on her if she touches a 120 V AC source in the following circumstances. (Note that currents above 10 mA lead to involuntarily muscle contraction.)
(a) if she is standing on a rubber mat and offers a total resistance of 300kΩ
(b) if she is standing barefoot on wet grass and has a resistance of only 4000kΩ

Answer:

A.

Explanation:

Amplitude measures how much a wave rises or falls. This is illustrated by A.

In the diagram, the amplitude of the wave is shown by A.

There are various definitions of amplitude, which are all functions of the magnitude of the differences between the variable's extreme values.

The amplitude of a variable is simply a measure of change relative to its central position. In contrast, magnitude is a measure of the distance or quantity of a variable irrespective of its direction.

Amplitude is a property that is unique to waves and oscillations.

Therefore, in the diagram, the amplitude of a wave is shown by A.

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Complete question is;

A circuit consists of four 100-W lamps connected in parallel across a 230-V supply. Inadvertently, a voltmeter has been connected in series with the lamps. The resistance of the voltmeter is 1500 Ω and that of the lamps under the conditions stated is six times their value then burning normally. What will be the reading of the voltmeter?

Answer:

150.42 V

Explanation:

We are told that the circuit consists of four 100W lamps.

We know that Power is given by the equation;

P = V²/R

Thus;

R = V²/P

Now, we are told that the four lamps are connected in parallel across a 230V supply.

Thus, V = 230 V

So resistance, R = 230²/100

R = 529 Ω

We are told that the resistance of the lamps under the conditions stated is six times their value when burning normally.

Thus, total resistance of each lamp under the conditions = 529 × 6 = 3174 Ω

So, since they are connected in parallel, equivalent resistance for each lamp = 3174/4 = 793.5 Ω

Now, since this resistance is connected in series with the voltmeter resistance of 1500 Ω

Therefore, total circuit resistance = 1500 + 793.5 = 2293.5 Ω

Thus;

circuit current = 230/2293.5 = 0.100283 A

Now, according to Ohm’s law, voltage drop across the voltmeter = 1500 × 0.100283 ≈ 150.42V

CHECK THE COMPLETE QUESTION BELOW

A circuit consists of four 100-W lamps connected in parallel across a 230-V supply. Inadvertently, a voltmeter has been connected in series with the lamps. The resistance of the voltmeter is 1500 Ω and that of the lamps under the conditions stated is six times their value then burning normally. What will be the reading of the voltmeter?

Answer:

the reading of the voltmeter=150.4V

Explanation:

We can determine the wattage of a lamp using below expression:

: W = I² R....................eqn(1)

But fro ohms law V=IR

then I= V/R

If we substitute I into equation (1)

We have W= V²/R

But W= 100W

V= 230V

Then

W=220²/R

100 = 2302/R

R = 529 Ω

We can as well calculate the Resistance of each lamp under given condition that they are sixtimes their value when burning normally.

R = 6 × 529 = 3174 Ω

We can also calculate quivalent resistance of the abovefour lamps connected in parallel then

R = 3174/4

= 793.5 Ω

total circuit resistance can be calculated since we know that resistance is connected to voltmeter of 1500 Ω resistance in series arrangement

Then

total circuit resistance = 1500 + 793.5

= 2293.5 Ω

Then from ohms law again

I= V/R

circuit current = 230/2293.5 A

The reading of the voltage drop across the voltmeter

= 1500 × 230/2293.5

= 150.4V

Answer:

4) oscillator

Explanation:

4) oscillator

Answer:

4.245s

Explanation:

Given that,

Hypothetical value of speed of light in a vacuum is 18 m/s

Speed of the car, 14 m/s

Time given is 6.76 s, and we're asked to find the observed time, T

The relationship between the two times can be given as

T = t / √[1 - (v²/c²)]

The missing variable were looking for is t, and we can find it if we rearrange the formula and make t the subject

t = T / √[1 - (v²/c²)]

And now, we substitute the values and insert into the equation

t = 6.76 * √[1 - (14²/18²)]

t = 6.76 * √[1 - (196/324)]

t = 6.76 * √(1 - 0.605)

t = 6.76 * √0.395

t = 6.76 * 0.628

t = 4.245 s

Therefore, the time the driver measures for the trip is 4.245s

Answer:

The rms current in the circuit is 3.513 A

Explanation:

Given;

angular frequency of the inductor, ω = 363 rad/s

maximum voltage of the inductive AC, V₀ = 169 V

Inductance of the inductor, L = 0.0937 H

Inductive reactance is given by;

[tex]X_L = 2\pi f L= \omega L[/tex]

[tex]X_L = 363 *0.0937\\\\X_L = 34.0131 \ ohms[/tex]

The rms voltage is given by;

[tex]V_{rms} = \frac{V_o}{\sqrt{2} } \\\\V_{rms} =\frac{169}{\sqrt{2} } \\\\V_{rms} = 119.5 \ V[/tex]

The rms current in the circuit is given by;

[tex]I_{rms} = \frac{V_{rms}}{X_L} \\\\I_{rms} = \frac{119.5}{34.0131} \\\\I_{rms} = 3.513 \ A[/tex]

Therefore, the rms current in the circuit is 3.513 A

Answer:

b. Third Law entropy  

Explanation:

Third law entropy: In physics, the term "third law entropy" or "the third law of thermodynamics" states that the specific entropy of a particular system at "absolute zero" is considered as a "well-defined constant". It occurs because any system at "zero temperature"  tends to exists or persists in its "ground state" in order for the entropy to be determined or described only by the "degeneracy" of the given ground state.

In the question above, the correct answer is option b.

Answer:

4.5x 10^ -9m

Explanation:

See attached file

Answer:

The radius is  [tex]r_{min} = 0.00226 \ m[/tex]

Explanation:

   From the question we are told that

      The  elastic limit(stress) is [tex]\sigma = 5.0*10^{8} \ N /m^2[/tex]

      The length is  [tex]L = 4.0 \ m[/tex]

      The weight of the commercial sign is    [tex]F_s = 8000 \ N[/tex]

       The maximum extension of the wire is  [tex]\Delta L = 5.0 \ cm = 0.05 \ m[/tex]

Generally the elastic limit of an alloy (stress) is is mathematically represented as

            [tex]\sigma = \frac{ F_s }{ A }[/tex]

Where A is the cross-sectional area of the wire which is mathematically represented as

         [tex]A = \pi r^2[/tex]

here [tex]r = r_{min}[/tex] which is the minimum radius of the wire that support the commercial sign

So

          [tex]\sigma = \frac{ F_s }{ \pi r_{min}^2 }[/tex]

=>       [tex]r_{min} = \sqrt{\frac{F_s}{\sigma * \pi} }[/tex]

substituting values

             [tex]r_{min} = \sqrt{\frac{8000}{ 5.0* 10^8 * 3.142} }[/tex]

           [tex]r_{min} = 0.00226 \ m[/tex]

Answer:

Everything in the Universe is made up of matter and energy. Matter is anything that has mass and occupies space. ... Energy is the ability to cause change or do work. Some forms of energy include light, heat, chemical, nuclear, electrical energy and mechanical energy.

Explanation:

As No diagram attached I am taking all are connected in series

We know

[tex]\boxed{\sf C_{eq}=C_1+C_2\dots}[/tex]

[tex]\\ \sf \longmapsto C_{eq}=12+12+7+15+15+15[/tex]

[tex]\\ \sf \longmapsto C_{eq}=24+7+45[/tex]

[tex]\\ \sf \longmapsto C_{eq}=76\mu F[/tex]

Complete Question

At what rate must Uranium 235 undergo fission by neutron bombardment to generate energy at a rate of 100 W (1 W = 1 J/s)? Assume each fission reaction releases 200 MeV of energy.

Answer

a. Approximately [tex]5*10^{10}[/tex] fissions per second.

b. Approximately [tex]6*10^{12 }[/tex]fissions per second.

c. Approximately [tex]4*10^{11}[/tex] fissions per second.

d. Approximately [tex]3*10^{12}[/tex] fissions per second.

e. Approximately[tex]3*10^{14}[/tex] fissions per second.

Answer:

The correct option is  d

Explanation:

From the question we are told that

       The energy released by each fission reaction [tex]E = 200 \ MeV = 200 *10^{6} * 1.60 *10^{-19} =3.2*10^{-11} \ J /fission[/tex]

Thus to generated  [tex]100 \ J/s[/tex] i.e  (100 W  ) the rate of fission is  

              [tex]k = \frac{100}{3.2 *10^{-11} }[/tex]

              [tex]k =3*10^{12} fission\ per \ second[/tex]

Answer:

0.99Hz

Explanation:

Using F= -mx ( spring force)

At equilibrium the gravitational force will be balanced by the spring force so mg= kx

K= mg/ 0.25 N/m

But

Frequency f= 1/2pi √g/0.25

Frequency is 0.99Hz

The block is pulled down slightly and released so, Frequency of oscillation is 3.15 Hz

What information do we have?

Length starched = 2.5 cm

F = Kx

We know that

F = mg

So,

mg = Kx

K/m = g/x

[tex]f=\frac{1}{2\pi}\sqrt{\frac{g}{x} }\\f=\frac{1}{2\pi}\sqrt{\frac{9.8}{0.025} }[/tex]

Frequency of oscillation = 3.15 Hz

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Answer:

Raw materials are most times gotten from the earth through various forms of extraction procedures.

A) Stainless steel utensils is made up of mainly Iron and other elements such as chromium , carbon etc.

B) Cat litter comprises of ceramic products which is made up of clay.

C) Tums brand antacid tablets comprises of calcium carbonate, magnesium hydroxide, aluminum hydroxide and sodium bicarbonate which could be extracted from the earth.

D)Lithium batteries are made up of elements in the earth such as lithium and carbon.

E)Aluminum beverage cans are made up of aluminum extracted from the ground.

Answer:

i) 545.2 N  upwards

ii) 828.2 N  downwards

Explanation:

mass of the roller = 70 kg

force exerted = 200 N

angle the force makes with the ground ∅ = 45°

weight of the roller W = mg

where

m is the mass of the roller

g is the acceleration due to gravity = 9.81 m/s^2

weight of the roller = 70 x 9.81 = 686.7 N

The effective vertical force exerted by the man = F sin ∅ = 200 x sin 45°

==> F = 200 x 0.707 = 141.5 N

i) if the man pulls, then the exerted force will be in opposite direction to the weight of the roller vertically upwards

Resultant vertical force = 686.7 N - 141.5 N = 545.2 N  upwards

ii) if he pushes, then the exerted force will be in the direction of the weight vertically downwards

Resultant vertical force = 686.7 N + 141.5 N = 828.2 N  downwards

Answer:

Scientific models are representations of objects, systems or events and are used as tools for understanding the natural world. Models use familiar objects to represent unfamiliar things. Models can help scientists communicate their ideas, understand processes, and make predictions.

Answer:

hhhbbbbbbbbbbbbbbnnnnnbbhb

Answer:

Therefore the characteristics to be found are:

* the focal length must be large and the focal length of the eyepiece must be small

* The diameter of the objective lens should be as large as possible, to be able to collect small without need from light

* The system must be configured to the far sight tip,

Explanation:

The length of the telescope is

         L = f_ocular + f_objetive

the magnification of the telescope is

         m = - f_objective / f_ocular

These are the two equations that describe the behavior of the telescope. Therefore the characteristics to be found are:

* the focal length must be large and the focal length of the eyepiece must be small

* The diameter of the objective lens should be as large as possible, to be able to collect small without need from light

* The system must be configured to the far sight tip,

Answer:

2)The Cell with OER 7

Explanation:

OER is the acronym for Oxygen Enhancement Ratio. It is the measure of the  enhancement of the effect of ionizing radiation due to the presence of oxygen. The ionization effect can be detrimental or therapeutic (use in cancer treatment). OER is the ratio of radiation dose during the lack of oxygen (hypoxia), to the dosage in the presence of oxygen (air is used as a reference). From the definition, one can see that the higher the OER the higher the sensitivity of the cell.

Answer:

200cm

Explanation:

Answer:

100cm

Explanation:

Using

F= ( N/2L)(√T/u)

F1 will now be (0.5*2)( √600/0.015)

=> L( wavelength)= 200/2cm = 100cm

Answer:

e. It is neither attracted nor repelled.

Explanation:

Electrostatic attraction or repulsion occurs between two or more charged particles or conductors. In this case, if the negatively charged ball is brought close to the neutral end A of the rod, there would be no attraction or repulsion between the rod end A and the negatively charged ball. This is because a charged particle or conductor has no attraction or repulsion to a neutral particle or conductor.

Answer:

96 m

Explanation:

Given:

v₀ = 0 m/s

a = 3 m/s²

t = 8 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (0 m/s) (8 s) + ½ (3 m/s²) (8 s)²

Δx = 96 m

Answer:

Period is 86811.5 seconds.

Explanation:

[tex]{ \boxed{ \bf{T {}^{2} = (\frac{4 {\pi}^{2} }{GM}) {r}^{3} }}}[/tex]

[tex]{ \tt{T {}^{2} = \frac{4 {(3.14)}^{2} }{(6.6 \times {10}^{ - 11} ) \times (5.98 \times {10}^{24} )} \times {((35880\times {10}^{3}) } + (6370 \times {10}^{3} )) {}^{3} }} \\ \\ { \tt{T {}^{2} = 7.54 \times {10}^{9} }} \\ { \tt{T = \sqrt{7.54 \times {10}^{9} } }} \\ { \tt{T = 86811.5 \: seconds}}[/tex]

Answer:

His angular velocity will increase.

Explanation:

According to the conservation of rotational momentum, the initial angular momentum of a system must be equal to the final angular momentum of the system.

The angular momentum of a system = [tex]I[/tex]'ω'

where

[tex]I[/tex]' is the initial rotational inertia

ω' is the initial angular velocity

the rotational inertia = [tex]mr'^{2}[/tex]

where m is the mass of the system

and r' is the initial radius of rotation

Note that the professor does not change his position about the axis of rotation, so we are working relative to the dumbbells.

we can see that with the mass of the dumbbells remaining constant, if we reduce the radius of rotation of the dumbbells to r, the rotational inertia will reduce to [tex]I[/tex].

From

[tex]I[/tex]'ω' = [tex]I[/tex]ω

since [tex]I[/tex] is now reduced, ω will be greater than ω'

therefore, the angular velocity increases.

Answer:

[tex]\displaystyle T = \sqrt{\frac{4\, \pi^{2} \, r^{3}}{G \cdot m}}[/tex].

The unit of both sides of this equation are [tex]\rm s[/tex].

Explanation:

The unit of the left-hand side is [tex]\rm s[/tex], same as the unit of [tex]T[/tex].

The following makes use of the fact that for any non-zero value [tex]x[/tex], the power [tex]x^{-1}[/tex] is equivalent to [tex]\displaystyle \frac{1}{x}[/tex].

On the right-hand side of this equation:

[tex]\begin{aligned}& \rm \sqrt{\frac{(m)^{3}}{(m^{3} \cdot kg^{-1} \cdot s^{-2}) \cdot (kg)}} \\ &= \rm \sqrt{\frac{m^{3}}{m^{3} \cdot s^{-2}}} = \sqrt{s^{2}} = s\end{aligned}[/tex].

Hence, the unit on the right-hand side of this equation is also [tex]\rm s[/tex].

Answer:

Explanation:

For image formation in objective lens

object distance u = 14 +1 = 15 mm

focal length f = 14 mm .

image distance v = ?

lens formula

[tex]\frac{1}{v} -\frac{1}{u} =\frac{1}{f}[/tex]

Putting the values

[tex]\frac{1}{v} +\frac{1}{15} =\frac{1}{14}[/tex]

v = 210 mm .

B )

magnification = v / u

= 210 / 15

= 14

size of image = 14 x 1.1 mm

= 15.4 mm

= 15 mm approx

C )

For final image to be at infinity , image produced by objective lens must fall at the focal point of eye piece . so objective lens's distance from the image formed by objective must be equal to focal length of eye piece that is 21 mm .

21 mm is the answer .

D )

overall magnification =

[tex]\frac{210}{15} \times \frac{D}{f_e}[/tex]

D = 25 cm , f_e = focal length of eye piece

= 14 x 250 / 21

= 166.67

= 170 ( in two significant figures )

(a) The distance of the image v=220mm

(b) SIze of the image 15 mm

(c) Distance of lens be from the image produced by the objective lens 21 mm

(d) overall magnification of the microscope 170

The objective lens of a microscope is the one at the bottom near the sample. At its simplest, it is a very high-powered magnifying glass, with very short focal length. This is brought very close to the specimen being examined so that the light from the specimen comes to a focus inside the microscope tube

For image formation in objective lens

object distance u = 14 +1 = 15 mm

focal length f = 14 mm .

image distance v = ?

By using lens formula

[tex]\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}[/tex]

Putting the values

[tex]\dfrac{1}{v}+\dfrac{1}{15}=\dfrac{1}{14}[/tex]

v = 210 mm .

B ) Magnification is the ratio of the size of the image to the size of the an object.

[tex]\rm magnification = \dfrac{v} { u}[/tex]

[tex]M= \dfrac{210} { 15}[/tex]

M= 14

size of image = 14 x 1.1 mm

= 15.4 mm

= 15 mm approx

C )

For final image to be at infinity , image produced by objective lens must fall at the focal point of eye piece . so objective lens's distance from the image formed by objective must be equal to focal length of eye piece that is 21 mm .

21 mm is the answer .

D )

overall magnification =

[tex]\dfrac{210}{15}\times \dfrac{D}{f_e}[/tex]

D = 25 cm , f_e = focal length of eye piece

[tex]= 14 \times \dfrac{ 250} { 21}[/tex]

= 166.67

= 170 ( in two significant figures )

Hence all the answers are:

(a) The distance of the image v=220mm

(b) SIze of the image 15 mm

(c) Distance of lens be from the image produced by the objective lens 21 mm

(d) overall magnification of the microscope 170

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Answer:

the net work is zero

Explanation:

Work is defined by the expression

        W = F. ds

Bold type indicates vectors

In this problem, the friction force does not decrease, therefore it will be zero.

Consequently for work on a closed path it is zero.

The work in going from the initial point (0, 0, 0) to the end of each segment is positive and when it returns from the point of origin the angle is 180º, therefore the work is negative, consequently the net work is zero

Answer:

Net force = 2.3686 × 10^(20) N

Explanation:

To solve this, we have to find the force of the earth acting on the moon and the force of the sun acting on the moon and find the difference.

Now, from standards;

Mass of earth;M_e = 5.98 × 10^(24) kg

Mass of moon;M_m = 7.36 × 10^(22) kg

Mass of sun;M_s = 1.99 × 10^(30) kg

Distance between the sun and earth;d_se = 1.5 × 10^(11) m

Distance between moon and earth;d_em = 3.84 × 10^(8) m

Distance between sun and moon;d_sm = (1.5 × 10^(11)) - (3.84 × 10^(8)) = 1496.96 × 10^(8) m

Gravitational constant;G = 6.67 × 10^(-11) Nm²/kg²

Now formula for gravitational force between the earth and the moon is;

F_em = (G × M_e × M_m)/(d_em)²

Plugging in relevant values, we have;

F_em = (6.67 × 10^(-11) × 5.98 × 10^(24) × 7.36 × 10^(22))/(3.84 × 10^(8))²

F_em = 1.9909 × 10^(20) N

Similarly, formula for gravitational force between the sun and moon is;

F_sm = (G × M_s × M_m)/(d_sm)²

Plugging in relevant values, we have;

F_se = (6.67 × 10^(-11) × 1.99 × 10^(30) ×

7.36 × 10^(22))/(1496.96 × 10^(8))²

F_se = 4.3595 × 10^(20) N

Thus, net force = F_se - F_em

Net force = (4.3595 × 10^(20) N) - (1.9909 × 10^(20) N) = 2.3686 × 10^(20) N

Answer:

The initial kinetic energy of the bullet is closest to 491.87 J

Explanation:

Given;

mass of bullet, m₁ = 18g = 0.018kg

mass of block, m₂ = 10kg

height moved by the block, h = 9 mm = 0.009 m

time taken for the bullet to travel through the block, t = 0.001s

let the initial velocity of the bullet = v₁

let the final velocity of the bullet = v₂

Apply the principle of conservation of linear momentum;

initial momentum = final momentum

0.018v₁ = v₂(0.018 + 10)

0.018v₁ = 10.018v₂ -----equation (1)

Apply the law of conservation of energy when the bullet lifts the block through 9mm

mgh = ¹/₂mv₂²

gh = ¹/₂v₂²

v₂² = 2gh

v₂ = √2gh

v₂ = √(2 x 9.8 x 0.009)

v₂ = 0.42 m/s

Substitute in v₂ in equation 1, to determine the initial velocity of the bullet;

0.018v₁ = 10.018v₂

0.018v₁  =  10.018(0.42)

0.018v₁  = 4.208

v₁ = 4.208 / 0.018

v₁ = 233.78 m/s

Now, determine the initial kinetic energy of the bullet;

K.E₁ = ¹/₂m₁v₁²

K.E₁ = ¹/₂(0.018)(233.78)²

K.E₁ = 491.87 J

Therefore, the initial kinetic energy of the bullet is closest to 491.87 J

Answer:

3.9E-8

Explanation:

We know that

Mv²/r = Bqv

So

r= mv/Bq

But E is 1/2mv² which is 5.53eV

m²v² =2m x 5.53eV

mv = √( 2 x 9.1E-31)

So

r= √( 2 x9.1E-31 x 5.53)/ 7.83x10^-4 x1.6E-19

= 3.9x10-8cm

​

Explanation: