A person is being pulled by gravity with a force of 500 N. What is the force with which the person pulls Earth?
1,000 N
O100 N
500 N
0 250 N

Answers

Answer 1

Answer:

The correct answer is 500 N

Explanation:

This is an exercise in Newton's third law or law of action and reaction

The Earth exerts a force on the person, which we call a weight of 500 N directed downwards, we can call this action and the person exerts a force on the Earth of equal magnitude 500N and in the opposite direction, that is directed upwards.

Which force we call action does not matter, the analysis and conclusions are the same

The correct answer is 500N


Related Questions

If you were to come back to our solar system in 6 billion years, what might you expect to find?

A) a red giant star

B) a rapidly spinning pulsar

C) a white dwarf

D) a black hole

E) Everything will be essentially the same as it is now

Answers

Answer:

A)a red giant star

what effect does decreasing the field current below its nominal value have on the speed versus voltage characteristic of a separately excited dc motor

Answers

Answer

The effect is that it Decreases the field current IF and increases slope K1

A spherical balloon has a radius of 6.95m and is filled with helium. The density of helium is 0.179 kg/m3, and the density of air is 1.29 kg/m3. The skin and structure of the balloon has a mass of 960kg . Neglect the buoyant force on the cargo volume itself. Determine the largest mass of cargo the balloon can lift.

Answers

Answer:

602.27 kg

Explanation:

The computation of the largest mass of cargo the balloon can lift is shown below:-

Volume of helium inside the ballon= (4 ÷ 3) × π × r^3

= (4 ÷ 3) × 3.14 × 6.953

= 1406.19 m3

Mass the balloon can carry = volume × (density of air-density of helium)

= 1406.19 × (1.29-0.179)

= 1562.27 kg

Mass of cargo it can carry = Mass it can carry - Mass of structure

= 1562.27 - 960

= 602.27 kg

What happens when two polarizers are placed in a straight line, one behind the other? A. They allow light to pass only if they are polarized in exactly the same direction. B. They block all light if they are polarized in exactly the same direction. C. They allow light to pass only if their directions of polarizations are exactly 90° apart. D. They block all light if their directions of polarizations are exactly 90° apart. E. They block all light if their directions of polarizations are either exactly the same or exactly 90° apart.

Answers

Answer:

C

They allow light to pass only if their directions of polarizations are exactly 90° apart.

Which best identifies the requirements for work to be performed? an object that has a force acting on it an object that is moving and has no net force a force acting on a motionless object a force that moves an object

Answers

Answer:

a force that moves an object

Explanation:

the formula for work is force * distance

This question involves the concepts of work, force, and displacement.

The statement that best identifies the requirements for work to be performed is "a force that moves an object".

Work is defined as the product of force applied on an object and the distance moved by the object. Mathematically,

Work = (Force)(Displacement)

Hence, both the applied force and the displacement of the object as a result of the application of the force is necessary for the work to be done. If any one of these values becomes zero, the work automatically becomes zero, which means no work is performed.

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An inductor is hooked up to an AC voltage source. The voltage source has EMF V0 and frequency f. The current amplitude in the inductor is I0.
Part A
What is the reactance XL of the inductor?
Express your answer in terms of V0 and I0.
Part B
What is the inductance L of the inductor?
Express your answer in terms of V0, f, and I0.

Answers

Answer:

a. The reactance of the inductor is XL = V₀/I₀

b. The inductance of the inductor is L = V₀/2πfI₀

Explanation:

PART A

Since the voltage across the inductor V₀ = I₀XL where V₀ = e.m.f of voltage source, I₀ = current amplitude and XL = reactance of the inductor,

XL = V₀/I₀

So, the reactance of the inductor is XL = V₀/I₀

PART B

The inductance of the inductor is gotten from XL = 2πfL where f = frequency of voltage source and L = inductance of inductor

Since XL = V₀/I₀ = 2πfL

V₀/I₀ = 2πfL

L = V₀/2πfI₀

So the inductance of the inductor is L = V₀/2πfI₀

A) The reactance XL of the inductor :  [tex]\frac{V_{0} }{I_{0} }[/tex]  

B) The Inductance L of the inductor : [tex]\frac{V_{0} }{2\pi fl_{0} }[/tex]  

A) Expressing the Reactance of the inductor

Voltage across the Inductor = V₀ = I₀XL   ---- ( 1 )

Where :  V₀ = emf voltage ,  I₀ = current

from equation ( 1 )

∴ XL ( reactance ) = [tex]\frac{V_{0} }{I_{0} }[/tex]  

B ) Expressing the Inductance of the Inductor

Inductance of an inductor is expressed as : XL = 2πfL

from part A

XL = [tex]\frac{V_{0} }{I_{0} }[/tex] = 2πfL

∴ The inductance L of the Inductor expressed in terms of V₀, F and I₀

L = [tex]\frac{V_{0} }{2\pi fl_{0} }[/tex]

Hence we can conclude that The reactance XL of the inductor :  [tex]\frac{V_{0} }{I_{0} }[/tex]  and The Inductance L of the inductor : [tex]\frac{V_{0} }{2\pi fl_{0} }[/tex]  .

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Currents in DC transmission lines can be 100 A or higher. Some people are concerned that the electromagnetic fields from such lines near their homes could pose health dangers.
A. For a line that has current 150 A and a height of 8.0 m above the ground, what magnetic field does the line produce at ground level? Express your answer in teslas.
B. What magnetic field does the line produce at ground level as a percent of earth's magnetic field which is 0.50 G?
C. Is this value of magnetic field cause for worry? Choose your answer below.
i. Yes. Since this field does not differ a lot from the earth's magnetic field, it would be expected to have almost the same effect as the earth's field.
ii. No. Since this field is much lesser than the earth's magnetic field, it would be expected to have less effect than the earth's field.
iii. Yes. Since this field is much greater than the earth's magnetic field, it would be expected to have more effect than the earth's field.
iv. No. Since this field does not differ a lot from the earth's magnetic field, it would be expected to have almost the same effect as the earth's field.

Answers

Answer:

Explanation:

magnetic field due to an infinite current carrying conductor

B = k x 2I / r where k = 10⁻⁷  , I is current in conductor and r is distance from wire

putting the given data

B = 10⁻⁷ x 2 x 100 / 8

= 25 x 10⁻⁷ T .

B )

earth's magnetic field = .5 gauss

= .5 x 10⁻⁴ T

= 5 x 10⁻⁵ T

percent required = (25 x 10⁻⁷ / 5 x 10⁻⁵) x 100

= 5 %

C )

ii.  No. Since this field is much lesser than the earth's magnetic field, it would be expected to have less effect than the earth's field.

What is the power P of the eye when viewing an object 61.0 cm away? Assume the lens-to-retina distance is 2.00 cm , and express the answer in diopters.

Answers

Answer:

The power of the eye is 51.64 diopters

Explanation:

The power of the eye is given by;

[tex]P = \frac{1}{f} = \frac{1}{d_o} +\frac{1}{d_i}[/tex]

where;

P is the power of the eye in diopter

f is the focal length of the eye

[tex]d_o[/tex] is the distance between the eye and the object

[tex]d_i[/tex] is the distance between the eye and the image

Given;

[tex]d_o[/tex] = 61.0 cm = 0.61 m

[tex]d_i[/tex] = 2.0 cm = 0.02 m

[tex]P = \frac{1}{d_o} +\frac{1}{d_i} \\\\P = \frac{1}{0.61} + \frac{1}{0.02} \\\\P = 51.64 \ D[/tex]

Therefore, the power of the eye is 51.64 diopters.

The power P of the eye when viewing an object 61.0 cm away is 51.639D

The power of a lens is a reciprocal of its focal length and it is expressed as:

[tex]P=\frac{1}{f}[/tex]

According to the mirror formula

[tex]\frac{1}{f} =\frac{1}{d_i} +\frac{1}{d_0}[/tex]

where

[tex]d_i[/tex] is the distance from the lens to the image = 61.0cm = 0.61m

[tex]d_0[/tex] is the distance from the lens to the object = 2.00cm = 0.02m

[tex]P=\frac{1}{f} =\frac{1}{0.02} +\frac{1}{0.61}\\P=50+1.639\\P=51.639D[/tex]

Hence the power P of the eye when viewing an object 61.0 cm away is 51.639D

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Two automobiles are equipped with the same singlefrequency horn. When one is at rest and the other is moving toward the first at 20 m/s , the driver at rest hears a beat frequency of 9.0 Hz.

Requried:
What is the frequency the horns emit?

Answers

Answer: f ≈ 8.5Hz

Explanation: The phenomenon known as Doppler Shift is characterized as a change in frequency when one observer is stationary and the source emitting the frequency is moving or when both observer and source are moving.

For a source moving and a stationary observer, to determine the frequency:

[tex]f_{0} = f_{s}.\frac{c}{c-v_{s}}[/tex]

where:

[tex]f_{0}[/tex] is frequency of observer;

[tex]f_{s}[/tex] is frequency of source;

c is the constant speed of sound c = 340m/s;

[tex]v_{s}[/tex] is velocity of source;

Rearraging for frequency of source:

[tex]f_{0} = f_{s}.\frac{c}{c-v_{s}}[/tex]

[tex]f_{s} = f_{0}.\frac{c-v_{s}}{c}[/tex]

Replacing and calculating:

[tex]f_{s} = 9.(\frac{340-20}{340})[/tex]

[tex]f_{s} = 9.(0.9412)[/tex]

[tex]f_{s} =[/tex] 8.5

Frequency the horns emit is 8.5Hz.

A 10kg block with an initial velocity of 10 m/s slides 1o m across a horizontal surface and comes to rest. it takes the block 2 seconds to stop. The stopping force acting on the block is about

Answers

Answer:

-50N

Explanation:

F=ma=m(Vf-Vi)/t

m=10kgVf=0m/sVi=10m/st=2s

F=(10)(-10)/(2)=-50N

So the force acting on the block is -50N, where the negative sign simply tells us that the force is opposite to the direction of movement.

In a physics laboratory experiment, a coil with 250 turns enclosing an area of 14 cm2 is rotated in a time interval of 0.030 s from a position where its plane is perpendicular to the earth's magnetic field to a position where its plane is parallel to the field. The earth's magnetic field at the lab location is 5.0×10^−5 T.Required:a. What is the total magnetic flux through the coil before it is rotated? After it is rotated? b. What is the average emf induced in the coil?

Answers

Explanation:

Consider a loop of wire, which has an area of [tex]A=14 \mathrm{cm}^{2}[/tex] and [tex]N=250[/tex] turns, it is initially placed perpendicularly in the earth magnetic field. Then it is rotated from this position to a position where its plane is parallel to the field as shown in the following figure in [tex]\Delta t=0.030[/tex] s. Given that the earth's magnetic field at the position of the loop is [tex]B=5.0 \times 10^{-5} \mathrm{T}[/tex], the flux through the loop before it is rotated is,

[tex]\Phi_{B, i} &=B A \cos \left(\phi_{i}\right)=B A \cos \left(0^{\circ}\right[/tex]

[tex]=\left(5.0 \times 10^{-5} \mathrm{T}\right)\left(14 \times 10^{-4} \mathrm{m}^{2}\right)(1)[/tex]

[tex]=7.0 \times 10^{-8} \mathrm{Wb}[/tex]

[tex]\quad\left[\Phi_{B, i}=7.0 \times 10^{-8} \mathrm{Wb}\right[/tex]

after it is rotated, the angle between the area and the magnetic field is [tex]\phi=90^{\circ}[/tex] thus,

[tex]\Phi_{B, f}=B A \cos \left(\phi_{f}\right)=B A \cos \left(90^{\circ}\right)=0[/tex]

[tex]\qquad \Phi_{B, f}=0[/tex]

(b) The average magnitude of the emf induced in the coil equals the change in the flux divided by the time of this change, and multiplied by the number of turns, that is,

[tex]{\left|\mathcal{E}_{\mathrm{av}}\right|=N\left|\frac{\Phi_{B, f}-\Phi_{B, i}}{\Delta t}\right|}{=} & \frac{1.40 \times 10^{-5} \mathrm{Wb}}{0.030 \mathrm{s}}[/tex]

[tex]& 3.6 \times 10^{-4} \mathrm{V}=0.36 \mathrm{mV}[/tex]

[tex]\mathbb{E}=0.36 \mathrm{mV}[/tex]

(a) The initial and final flux through the coil is 1.75 × 10⁻⁵ Wb and 0 Wb

(b) The induced EMF in the coil is 0.583 mV

Flux and induced EMF:

Given that the coil has N = 250 turns

and an area of A = 14cm² = 1.4×10⁻³m².

It is rotated for a time period of Δt = 0.030s such that it is parallel with the earth's magnetic field that is B = 5×10⁻⁵T

(a) The flux passing through the coil is given by:

Ф = NBAcosθ

where θ is the angle between area vector and the magnetic field

The area vector is perpendicular to the plane of the coil.

So, initially, θ = 0°, as area vector and earth's magnetic field both are perpendicular to the plane of the coil

So the initial flux is:

Φ = NABcos0° = NAB

Ф = 250×1.4×10⁻³×5×10⁻⁵ Wb

Ф = 1.75 × 10⁻⁵ Wb

Finally, θ = 90°, and since cos90°, the final flux through the coil is 0

(b) The EMF induced is given by:

E = -ΔФ/Δt

E = -(0 - 1.75 × 10⁻⁵)/0.030

E = 0.583 × 10⁻³ V

E = 0.583 mV

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A particle moves along line segments from the origin to the points (2, 0, 0), (2, 3, 1), (0, 3, 1), and back to the origin under the influence of the force field F(x, y, z).

Required:
Find the work done.

Answers

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

The entropy of any substance at any temperature above absolute zero is called the: Select the correct answer below:
a. absolute entropy
b. Third Law entropy
c. standard entropy
d. free entropy
e. none of the above

Answers

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.

5. The speed of a transverse wave on a string is 170 m/s when the string tension is 120 ????. To what value must the tension be changed to raise the wave speed to 180 m/s?

Answers

Answer:

The tension on string when the speed was raised is 134.53 N

Explanation:

Given;

Tension on the string, T = 120 N

initial speed of the transverse wave, v₁ = 170 m/s

final speed of the transverse wave, v₂ = 180 m/s

The speed of the wave is given as;

[tex]v = \sqrt{\frac{T}{\mu} }[/tex]

where;

μ is mass per unit length

[tex]v^2 = \frac{T}{\mu} \\\\\mu = \frac{T}{v^2} \\\\\frac{T_1}{v_1^2} = \frac{T_2}{v_2^2}[/tex]

The final tension T₂ will be calculated as;

[tex]T_2 = \frac{T_1 v_2^2}{v_1^2} \\\\T_2 = \frac{120*180^2}{170^2} \\\\T_2 = 134.53 \ N[/tex]

Therefore, the tension on string when the speed was raised is 134.53 N

You have a lightweight spring whose unstretched length is 4.0 cm. First, you attach one end of the spring to the ceiling and hang a 1.8 g mass from it. This stretches the spring to a length of 5.1 cm . You then attach two small plastic beads to the opposite ends of the spring, lay the spring on a frictionless table, and give each plastic bead the same charge. This stretches the spring to a length of 4.3 cm .

Requried:
What is the magnitude of the charge (in nC) on each bead?

Answers

Answer:

2.2nC

Explanation:

Call the amount by which the spring’s unstretched length L,

the amount it stretches while hanging x1

and the amount it stretches while on the table x2.

Combining Hooke’s law with Newton’s second law, given that the stretched spring is not accelerating,

we have mg−kx1 =0, or k = mg /x1 , where k is the spring constant. On the other hand,

applying Coulomb’s law to the second part tells us ke q2/ (L+x2)2 − kx2 = 0 or q2 = kx2(L+x2)2/ke,

where ke is the Coulomb constant. Combining these,

we get q = √(mgx2(L+x2)²/x1ke =2.2nC

A spring with spring constant 15 N/m hangs from the ceiling. A ball is attached to the spring and allowed to come to rest. It is then pulled down 6.0 cm and released. If the ball makes 30 oscillations in 20 s, what are its (a) mass and (b) maximum speed?

Answers

Answer:

a

   [tex]m = 0.169 \ kg[/tex]

b

  [tex]|v_{max} |= 0.5653 \ m/s[/tex]

Explanation:

From the question we are told that

    The  spring constant is  [tex]k = 14 \ N/m[/tex]

     The  maximum extension of the spring is  [tex]A = 6.0 \ cm = 0.06 \ m[/tex]

     The number of oscillation is  [tex]n = 30[/tex]

      The  time taken is  [tex]t = 20 \ s[/tex]

Generally the the angular speed of this oscillations is mathematically represented as

           [tex]w = \frac{2 \pi}{T}[/tex]

where T is the period which is mathematically represented as

     [tex]T = \frac{t}{n}[/tex]

substituting values

     [tex]T = \frac{20}{30 }[/tex]

     [tex]T = 0.667 \ s[/tex]

Thus  

       [tex]w = \frac{2 * 3.142 }{ 0.667}[/tex]

       [tex]w = 9.421 \ rad/s[/tex]

this angular speed can also be represented mathematically as

       [tex]w = \sqrt{\frac{k}{m} }[/tex]

=>   [tex]m =\frac{k }{w^2}[/tex]

substituting values

      [tex]m =\frac{ 15 }{(9.421)^2}[/tex]

      [tex]m = 0.169 \ kg[/tex]

In SHM (simple harmonic motion )the equation for velocity is  mathematically represented as

        [tex]v = - Awsin (wt)[/tex]

The  velocity is maximum when  [tex]wt = \(90^o) \ or \ 1.5708\ rad[/tex]

     [tex]v_{max} = - A* w[/tex]

=>   [tex]|v_{max} |= A* w[/tex]

=>    [tex]|v_{max} |= 0.06 * 9.421[/tex]

=>   [tex]|v_{max} |= 0.5653 \ m/s[/tex]

Tech A says parallel circuits are like links in a chain. Tech B says total current in a parallel circuit equals the sum of the current flowing in each branch of the circuit. Who is correct?

Answers

Answer: Only Tech B is correct.

Explanation:

First, tech A is wrong.

The circuits that can be compared with links in a chain are the series circuit, and it can be related to the links in a chain because if one of the elements breaks, the current can not flow furthermore (because the elements in the circuit are connected in series) while in a parallel circuit if one of the branches breaks, the current still can flow by other branches.

Also in a parallel circuit, the sum of the currents of each path is equal to the current that comes from the source, so Tech B is correct, the total current is equal to the sum of the currents flowing in each branch of the circuit.

A soccer ball of mass 0.4 kg is moving horizontally with a speed of 20 m/s when it is kicked by a player. The kicking force is so large that the ball flies up at an angle of 30 degrees above the ground. The player however claims (s)he aimed her/his foot at a 40 degree angle above the ground. Calculate the average kicking force magnitude and the final speed of the ball, if you are given that the foot was in contact with the ball for one hundredth of a second.

Answers

Answer:

v_{f} = 74 m/s, F = 230 N

Explanation:

We can work on this exercise using the relationship between momentum and moment

        I = ∫ F dt = Δp

bold indicates vectors

we can write this equations in its components

X axis

       Fₓ t = m ( -v_{xo})

Y axis  

        t = m (v_{yf} - v_{yo})

in this case with the ball it travels horizontally v_{yo} = 0

Let's use trigonometry to write the final velocities and the force

        sin 30 = v_{yf} / vf

        cos 30 = v_{xf} / vf

        v_{yf} = vf sin 30

        v_{xf} = vf cos 30

         sin40 = F_{y} / F

         F_{y} = F sin 40

         cos 40 = Fₓ / F

         Fₓ = F cos 40

let's substitute

      F cos 40 t = m ( cos 30 - vₓ₀)

      F sin 40 t = m (v_{f} sin 30-0)

we have two equations and two unknowns, so the system can be solved

        F cos 40 0.1 = 0.4 (v_{f} cos 30 - 20)

        F sin 40 0.1 = 0.4 v_{f} sin 30

we clear fen the second equation and subtitles in the first

         F = 4 sin30 /sin40     v_{f}

         F = 3.111 v_{f}

        (3,111 v_{f}) cos 40 = 4 v_{f} cos 30 - 80

        v_{f} (3,111 cos 40 -4 cos30) = - 80

        v_{f} (- 1.0812) = - 80

        v_{f} = 73.99

        v_{f} = 74 m/s

now we can calculate the force

          F = 3.111 73.99

          F = 230 N

An electron moves on a circular orbit in a uniform magnetic field of 7.83×10-4 T. The kinetic energy of the electron is 55.3 eV. What is the diameter of the orbit?

Answers

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:

A uniform bar has two small balls glued to its ends. The bar is 2.10 m long and with mass 3.70 kg , while the balls each have mass 0.700 kg and can be treated as point masses.

Required:
Find the moment of inertia of this combination about an axis
a. perpendicular to the bar through its center.
b. perpendicular to the bar through one of the balls.
c. parallel to the bar through both balls.
d. parallel to the bar and 0.500 m from it.

Answers

Answer:

Explanation:

a )

moment of inertia in the first case will be sum of moment of inertia of two balls + moment of inertia of bar

= 2 x .700 x (2.1 / 2 )² + 3.7 x 2.1² / 12

= 1.5435 + 1.35975

= 2.90325 kg m²

b )

moment of inertia required

= moment of inertia of bar + moment of inertia of the other ball

= 3.70 x (2.1² / 3 )  + .7 x 2.1²

= 5.439 + 3.087

= 8.526 kg m²

c )

In this case moment of inertia of the combination = 0 as distance of masses from given axis is zero .

d )

masses = 3.7 + .7 = 4.4 kg

distance from axis = .5 m  

moment of inertia about given axis

= 4.4 x .5²

= 1.1 kg m².

The ability of a water strider to move along the surface of water without breaking the surface is due to:

Answers

Answer:

The ability of a water strider to move along the surface of water without breaking the surface is due to:

SURFACE TENSION

Explanation:

this is because Water molecules are more attracted to each other than they are to other materials, so they generate a force to stay together called surface tension. Which allows the strider to move without breaking the surface

In order to waken a sleeping child, the volume on an alarm clock is doubled. Under this new scenario, how much more energy will be striking the child's ear drums each second?

Answers

Answer:4 times more energy will be striking the childbearing

Explanation:

Because Volume is directly proportional to amplitude of sound. Energy is proportional to amplitude squared. If you triple the amplitude, you multiply the energy by 4

What is the angle between a wire carrying an 8.40 A current and the 1.20 T field it is in, if 50.0 cm of the wire experiences a magnetic force of 2.55 N? ° (b) What is the force (in N) on the wire if it is rotated to make an angle of 90° with the field? N

Answers

Answer:

A. 30.38°

B 5.04N

Explanation:

Using

F= ILBsin theta

2 .55N= 8.4Ax 0.5mx 1.2T x sintheta

Theta = 30.38°

B. If theta is 90°

Then

F= 8.4Ax 0.5mx 1.2x sin 90°

F= 5.04N

Which is produced around a wire when an electrical current is in the wire? magnetic field solenoid electron flow electromagnet

Answers

Answer:

A. magnetic field

Explanation:

The magnetic field is produced around a wire when an electrical current is in the wire because of the magnetic effect of the electric current therefore the correct answer is option A .

What is a magnetic field ?

A magnetic field could be understood as an area around a magnet, magnetic material, or an electric charge in which magnetic force is exerted.

As given in the problem statement we have to find out what is produced around a wire when an electrical current is in the wire.

The magnetic field is produced as a result when an electrical current is passed through the conducting wire .

Option A is the appropriate response because a wire's magnetic field is created when an electrical current flows through it due to the magnetic influence of the electric current .

Learn more about the magnetic fields here, refer to the link given below;

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Give an example of a fad diet that is not healthy and one that is healthy. Explain how you know the difference.

Answers

Answer:

 Good Diet: ! gallon of water a day, Fruits, Vegetables, White meats(Chicken), Don't eat past 3 PM.

Bad Diet: Pizza, Red meat, Baked goods, Eating at late hours.

Explanation: I know the difference because, When you drink water first thing in the morning it gets your metabolism running. Than means you can digest foods better, you want to feed your body good foods but you should not eat until you feel stuffed. You should eat until you are no longer starving. Than you should drink a cup of water in between meals. I know you should not eat past 3 pm because your body needs time to digest foods because you should never go to sleep with a full stomach. I know the difference between good food and bad food because when you eat healthy food and a balanced diet, your body will have more energy and you wont feel tired afterwards. Eating bad foods and food with artificial sugars will clump up in your kidneys, and your body will have small bursts of energy but you will feel lazy afterwards...Your body is supposed to stay energized from a healthy meal in order to give you the energy your body needs to exercise. If you feel droopy all the time and you don't want to do anything, than you are unhealthy.

Answer:

A vegetarian diet is an example of a good fad diet if you do it correctly. It can help you get lots of veggies and good nutrients from them while still following the non-meat diet you want. This can be effective and good for weight loss becasue you are still eating and getting all the good nutrients and calories from less fatty foods. 

Vegan diet (some can be successful but many people fail and do not do good that is why I choose this) The problem with this fad diet is that it can cause nutritional deficiencies and lead to a host of additional health problems, including negatively impacting hormonal health and metabolism. Many people also struggle to find healthy vegan food and end up eating bad and fatty foods instead. 

Explanation:

Got a 100

From a static hot air balloon, a 10kg projectile is launched at a speed of 10m / s upwards. If the balloon has a mass of 90kg. What is the final velocity of the latter? Select one:

a. 0.57m / s down
b. 2.56m / s down
c. 1.11m / s down
d. 2.03m / s down
e. 3.15m / s down

Answers

Answer:

c. 1.11 m/s down

Explanation:

Momentum is conserved.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

Assuming the balloon and projectile are originally at rest:

(90 kg) (0 m/s) + (10 kg) (0 m/s) = (90 kg) v + (10 kg) (10 m/s)

0 kg m/s = (90 kg) v + 100 kg m/s

v = -1.11 m/s

An interference pattern is produced by light with a wavelength 520 nm from a distant source incident on two identical parallel slits separated by a distance (between centers) of 0.440 mm.
1. If the slits are very narrow, what would be the angular position of the first-order, two-slit, interference maxima?
2. What would be the angular position of the second-order, two-slit, interference maxima in this case?
3. Let the slits have a width 0.310 mm . In terms of the intensity I0 at the center of the central maximum, what is the intensity at the angular position of θ1?
4. What is the intensity at the angular position of θ2?

Answers

Answer:

1) θ = 0.00118 rad, 2)  θ = 0.00236 rad , 3) I / I₀ = 0.1738, 4)  I / Io = 0.216

Explanation:

In the double-slit interference phenomenon it is explained for constructive interference by the equation

          d sin θ = m λ

1) the first order maximum occurs for m = 1

           sin θ = λ  / d

           θ = sin⁻¹ λ  / d

let's reduce the magnitudes to the SI system

           λ  = 520 nm = 520 10⁻⁹  θ = 0.00118 radm

           d = 0.440 mm = 0.440 10⁻³ m ³

let's calculate

           θ = sin⁻¹ (520 10⁻⁹ / 0.44 10⁻³)

            θ = sin⁻¹ (1.18 10⁻³)

            θ = 0.00118 rad

2) the second order maximum occurs for m = 2

            θ = sin⁻¹ (m λ  / d)

            θ = sin⁻¹ (2 5¹20 10⁻⁹ / 0.44 10⁻³)

            θ = 0.00236 rad

3) To calculate the intensity of the interference spectrum, the diffraction phenomenon must be included, so the equation remains

          I = I₀ cos² (π d sin θ /λ ) sinc² (pi b sin θ /λ )

where the function sinc = sin x / x

and b is the width of the slits

we caption the values

             x = π 0.310 10⁻³ sin 0.00118 / 520 10⁻⁹)

             x = 2.21

            I / I₀ = cos² (π 0.44 10⁻³ sin 0.00118 / 520 10⁻⁹) (sin (2.21) /2.21)²

remember angles are in radians

            I / I₀ = cos² (3.0945) [0.363] 2

            I / I₀ = 0.9978 0.1318

            I / I₀ = 0.1738

4) the maximum second intensity is

            I / I₀ = cos² (π d sinθ / λ) sinc² (πb sin θ /λ)

            x =π 0.310 10⁻³ sin 0.00236 / 520 10⁻⁹)

            x = 4.41

            I / Io = cos² (π 0.44 10⁻³ sin 0.00236 / 520 10⁻⁹) (sin 4.41 / 4.41)²

            I / Io = cos² 6.273    0.216

            I / Io = 0.216

.

Categorize each ray tracing statement as relating to ray 1, ray 2, or ray 3.
A. Drawn from the top of the object so that it passes through the center of the lens at the optical axis.
B. Drawn from the top of the object so that it passes through the focal point on the same side of the lens as the object.
C. Drawn parallel to the optical axis from the top of the object.
D. Ray bends parallel to the optical axis.
E. Ray bends so that it passes through the focal point on the opposite side of the lens as the object.
F. Ray does not bend.

Answers

Answer:

statement 1 with answer C

statement 2 with answer F

statement 3 with answer B

Statement 1 with E

Statement 2 with A

Statement 3 with D

Explanation:

In this exercise you are asked to relate each with the answers

In general, in the optics diagram,

* Ray 1 is a horizontal ray that after stopping by the optical system goes to the focal point

* Ray 2 is a ray that passes through the intercept point between the optical axis and the system and does not deviate

* Ray 3 is a ray that passes through the focal length and after passing the optical system, it comes out horizontally.

With these statements, let's review the answers

statement 1 with answer C

statement 2 with answer F

statement 3 with answer B

Statement 1 with E

Statement 2 with A

Statement 3 with D

Now the friends are ready to tackle a homework problem. A pulse is sent traveling along a rope under a tension of 29 N whose mass per unit length abruptly changes, from 19 kg/m to 45 kg/m. The length of the rope is 2.5 m for the first section and 2.8 m for the second, and the second rope is rigidly fixed to a wall. Two pulses will eventually be detected at the origin: the pulse that was reflected from the medium discontinuity and the pulse that was originally transmitted, which hits the wall and is reflected back and transmitted through the first rope. What is the time difference, Δt, between the two pulses detected at the origin? s

Answers

Answer:

The time difference is 2.97 sec.

Explanation:

Given that,

Tension = 29 N

Mass per unit length [tex]\mu_{1}=19\ kg/m[/tex]

Mass per unit length [tex]\mu_{2}=45\ kg/m[/tex]

Length of first section = 2.5 m

Length of second section = 2.8 m

We need to total distance of first pulse

Using formula for distance

[tex]d=2.5+2.5[/tex]

[tex]d_{1}=5.0\ m[/tex]

We need to total distance of second pulse

Using formula for distance

[tex]d=2.8+2.8[/tex]

[tex]d_{2}=5.6\ m[/tex]

We need to calculate the speed of pulse in the first string

Using formula of speed

[tex]v_{1}=\sqrt{\dfrac{T}{\mu_{1}}}[/tex]

Put the value into the formula

[tex]v_{1}=\sqrt{\dfrac{29}{19}}[/tex]

[tex]v_{1}=1.24\ m/s[/tex]

We need to calculate the speed of pulse in the second string

Using formula of speed

[tex]v_{2}=\sqrt{\dfrac{T}}{\mu_{2}}}[/tex]

Put the value into the formula

[tex]v_{2}=\sqrt{\dfrac{29}{45}}[/tex]

[tex]v_{2}=0.80\ m/s[/tex]

We need to calculate the time for first pulse

Using formula of time

[tex]t_{1}=\dfrac{d_{1}}{v_{1}}[/tex]

Put the value into the formula

[tex]t_{1}=\dfrac{5.0}{1.24}[/tex]

[tex]t_{1}=4.03\ sec[/tex]

We need to calculate the time for second pulse

Using formula of time

[tex]t_{2}=\dfrac{d_{1}}{v_{1}}[/tex]

Put the value into the formula

[tex]t_{2}=\dfrac{5.6}{0.80}[/tex]

[tex]t_{2}=7\ sec[/tex]

We need to calculate the time difference

Using formula of time difference

[tex]\Delta t=t_{2}-t_{1}[/tex]

Put the value into the formula

[tex]\Delta t=7-4.03[/tex]

[tex]\Delta t=2.97\ sec[/tex]

Hence, The time difference is 2.97 sec.

A bar magnet is dropped from above and falls through the loop of wire. The north pole of the bar magnet points downward towards the page as it falls. Which statement is correct?a. The current in the loop always flows in a clockwise direction. b·The current in the loop always flows in a counterclockwise direction. c. The current in the loop flows first in a clockwise, then in a counterclockwise direction. d. The current in the loop flows first in a counterclockwise, then in a clockwise direction. e. No current flows in the loop because both ends of the magnet move through the loop.

Answers

Answer:

b. The current in the loop always flows in a counterclockwise direction.

Explanation:

When a magnet falls through a loop of wire, it induces an induced current on the loop of wire. This induced current is due to the motion of the magnet through the loop, which cause a change in the flux linkage of the magnet. According to Lenz law, the induced current acts in such a way as to repel the force or action that produces it. For this magnet, the only opposition possible is to stop its fall by inducing a like pole on the wire loop to repel its motion down. An induced current that flows counterclockwise in the wire loop has a polarity that is equivalent to a north pole on a magnet, and this will try to repel the motion of the magnet through the coil. Also, when the magnet goes pass the wire loop, this induced north pole will try to attract the south end of the magnet, all in a bid to stop its motion downwards.

The current in the loop always flows in a counterclockwise direction. Hence, option (b) is correct.

The given problem is based on the concept and fundamentals of magnetic bars. When a magnet falls through a loop of wire, it induces an induced current on the loop of wire. There is some magnitude of current induced in the wire.

This induced current is due to the motion of the magnet through the loop, which cause a change in the flux linkage of the magnet. According to Lenz law, the induced current acts in such a way as to repel the force or action that produces it. For this magnet, the only opposition possible is to stop its fall by inducing a like pole on the wire loop to repel its motion down. An induced current that flows counterclockwise in the wire loop has a polarity that is equivalent to a north pole on a magnet, and this will try to repel the motion of the magnet through the coil. Also, when the magnet goes pass the wire loop, this induced north pole will try to attract the south end of the magnet, all in a bid to stop its motion downwards.

Thus, we can say that the current in the loop always flows in a counterclockwise direction. Hence, option (b) is correct.

Learn more about the magnetic field here:

https://brainly.com/question/14848188

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