Identify two fields where physical quantities are used in motion calculations​

The force required to pull the two hemispheres is 46622.72N

( Note: 1 millibar=100 N/m2. One atmosphere is 1013 millibar = 1.013×105 N/m2 ]

The contact area between the hemispheres is (pi x 0.400^2) = 0.5024m^2.

Pressure difference = (940 - 12)

= 928 millibars.

(928 x 100)

= 92,800N/m^2.

Therefore, the required force to pull the two hemispheres is

(92800 x 0.5024)

= 46622.72N.

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Scientists assume the universe is a vast single system in which basic laws are consistent. I agree with this statement

The Big Bang Theory is known as one of the greatest consensus that was said to have occurred  among scientists, astronomers and also that of cosmologists.

Note that the theory is one that states the Universe as we all know it was formed as a result of a massive explosion that is not only formed by the majority of matter, but also with the use of the physical laws that works in our ever-expanding cosmos.

Therefore, In the case above, i do agree with the fact that Scientists assume the universe is a vast single system in which basic laws that are consistent.

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In a transverse wave, the disturbance are carried forward and the particles are undergoing to and fro motion. Hence in transverse waves, there is no horizontal motion.

A wave can be described as a disturbance that travels through a medium from one location to the other location without transporting any matter.

There are two types of matter,

(i) Transverse wave

(ii) Longitudinal wave

In a transverse wave, the particles of the medium move in a direction perpendicular to the direction of propagation of the wave. That means, the particles undergo to and fro motion about their mean position and moves in the direction of the wave propagation.

Examples are:

Whereas longitudinal waves undergo back and forth motion with respect to the mean position.

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Answer:

The red color in the sky at sunset (and sunrise) is due to an effect called Rayleigh scattering.

Explanation:

Answer:

11.54 meters

Explanation:

The owners take  10m / (1.3 + 1.3 m/s)  = 3.85 seconds to meet

The dog is running at constant speed of 3 m/s for these 3.85 seconds

   3 m/s * 3.85 sec = 11.54 meters

The ball orbit the Earth, when launched from the theoretical cannon of Newton, is option B. it is magnetically attracted.

Newton's cannonball was a hypothetical situation. Isaac Newton once proposed that gravity, which he believed to be a universal force, was the primary factor behind the planetary motion. In this experiment, Newton imagines projecting a stone or a cannonball onto the summit of a very tall mountain. The body should move away from Earth in the direction it was projected if there were no effects from gravity or air resistance.

Depending on the projectile's initial velocity and the gravitational force acting on it, the bullet will travel in a different direction. Low speeds result in a simple fallback to Earth. The Earth's surface causes the cannonball to deviate from its elliptical route.

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(1) The period of the oscillator is 1 second.

(2) The damping coefficient is 0.93.

The period of oscillation is the time taken to make one complete cycle.

From the graph, the time taken to make one complete oscillation is 1 second.

equation of the wave is given as;

y(t) = Ae^(-btx) cos(ωt)

3.5 = Ae^(-0) cos(0)

3.5 = A x 1

A = 3.5 cm

-3 = 3.5e^(-bx) cos(ω)

-3/3.5 = e^(-bx) cos(ω)

-0.857 = e^(-bx) cos(ω)

-0.857 / cos(ω) =  e^(-bx)

ln[-0.857 / cos(ω)] = -bx  

ln[-0.857 / cos(ω)] / b = - x  ---- (1)

-2 = 3.5e^(-2bx) cos(2ω)

-0.57 = e^(-2bx) cos(2ω)

ln[-0.57 / cos(2ω)] = -2bx  

ln[-0.57 / cos(2ω)] /2b = - x  ------(2)

solve (1) and (2)

ln[-0.57 / cos(2ω)]/2b = ln[-0.857 / cos(ω)] /b

-0.57 / cos(ω) = 2(-0.857 / cos(ω))

2(-0.857/cosω) = -0.57/cos2ω

-(2 x 0.857) / (-0.57) = cosω/cos 2ω

3 = cosω/cos 2ω

3(cos 2ω) =  cosω

3(2cos²ω - 1) = cos ω

6cos²ω - 6 = cosω

6cos²ω  - cosω - 6 = 0

let cosω  = y

6y² - y - 6 = 0

solve the quadratic equation;

y = 1.1 or -0.92

cosω = -0.92

ω  = arc cos(-0.92)

ω  = 2.74 rad/s

From equation (1)

ln[-0.857 / cos(ω)] / x = -b  ---- (1)

let x = 1

ln(-0.857/cos(2.74) = -b

-0.93 = -b

b = 0.93

Thus, the damping coefficient is 0.93.

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The length of the wire is 1.12m

What is Resistance ?

Resistance is opposition to the current flowing in the circuit.

Given ,

D= 0.6m

R= 0.6/2m

R= 28Ω

rho = 1Ωm

R= rho ×l/A

where R is resistance , l is length ,A is area and rho is resistivity

By applying value and Solving we get,

l= 1.12m

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The final speed of the electron is 4.64 * 10^5 m/s.

Given that the mass of the electron is obtained as 9.11 * 10^-31 Kg, we have that the charge of the electron is 1.6 * 10^-19 C.

E= F/q

F = Eq

F =  330 N/C * 1.6 * 10^-19 C

F = 5.28 * 10^-17 N

F = ma

a = F/m = 5.28 * 10^-17 N/ 9.11 * 10^-31 Kg

a = 5.8 * 10^13 m/s^2

Using

v = u + at

u = 0 m/s because the electron was initially at rest

v = at

v = 5.8 * 10^13 * 8.00 ✕ 10−⁹ s

v = 4.64 * 10^5 m/s

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The vertical and horizontal component of the maximum displacement are 0.15m and 0.85m.

As per conversation of momentum, mass of bullet × velocity= mass of pendulum × velocity

=> 0.029kg × 200 = 3.429 × velocity of pendulum

=> Velocity of pendulum= 5.8/3.429 = 1.7 m/s

1/2 × mass of pendulum × velocity²=1/2 × k × amplitude²

=> 10 = 13.44× amplitude²

=> Amplitude = √(10/13.44)

= 0.86m

Angle = tan inverse (amplitude/length of pendulum)

= tan inverse (0.86/2.5)

= 20°

Maximum displacement along vertical direction= L - L×cos20°

= 2.5 - 2.5×cos20°

= 0.15 m

Maximum displacement along horizontal direction= Lsin20°

= 2.5 × sin 20°

= 0.85m

Thus, we can conclude that the vertical and horizontal component of the maximum displacement are 0.15m and 0.85m.

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The constant angular acceleration (in rad/s2) of the centrifuge is 194.02 rad/s².

Apply the following kinematic equation;

ωf² = ωi² - 2αθ

where;

ωi = 3400 rev/min x 2π rad/rev x 1 min/60s = 356.05 rad/s

θ = 52 rev x 2π rad/rev = 326.7 rad

0 = ωi² - 2αθ

α = ωi²/2θ

α = ( 356.05²) / (2 x 326.7)

α = 194.02 rad/s²

Thus, the constant angular acceleration (in rad/s2) of the centrifuge is 194.02 rad/s².

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Based on the textbook, Felisha's argument is known to be unsound because not all the premises that were given was true.

2.  She stated that the “The Cowboys are popular and using only their huge fan base to tell is not right and if they are are better as she claim, why are they not the best but she stated that they are among the best.

3. A time when when i successfully argued a point of view was during my high school debate on the topic that doctors are better than lawyers and i gave a lot of valid reasons why the world cannot do without doctors.

The term fallacies are  seen as flawed or  deceptive and false kind of arguments that is known to be be proven wrong with the aid of reasoning.

Note that there are two key types of fallacies which are the formal fallacy and an informal fallacy.

Therefore, Based on the textbook, Felisha's argument is  seen as an unsound argument because not all the premises that were given was true and they were not standard enough.

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Answer:

D.-4.798m/s

Explanation:

Greetings !

Given values

[tex]u= 4ms \\ a = 0.21ms {}^{2} \\ t = 3.8sec[/tex]

Solve for V of the given expression

Firstly, recall the velocity-time equation

[tex]v = u + at[/tex]

plug in known values to the equation

[tex]v = (4) + (0.21)(3.8)[/tex]

solve for final velocity

[tex]v = 4.792ms[/tex]

Hope it helps!

A quantitative description of a system is important as it provides information about the carrying capacity of habitable worlds.

Quantitative description refers to the description of a system which is focused on the numerical value of the properties or components of the system.

For example, a quantitative description of the components of a given habitat will be focused on the number of the individual species in the habitat. It will also be focused on the numerical value of the non-living components of the system and how such values affect the living components of the system numerically.

This information will then be used by scientists to see how modifications of the various quantitative components of the habitat will help to improve the chances of survival of species found in the habitat. This leads to such concepts as the carrying capacity of a habitat which is the maximum number of species that the available resources is a habitat can easily sustain.

In conclusion, a quantitative description is important to in investigations about the habitable world.

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Answer:

The low energy X-ray calibration service is intended for thin-window plane-parallel chambers required for the dosimetry of superficial X-ray beams. Low-energy X-ray therapy is also known as superficial radiation therapy. This therapy uses beams of low-energy X-rays (radiation waves) to destroy skin cancer cells while not harming healthy tissue around or under the upper layers of skin.

-0.930 m/s is the velocity of the boat.

Mass of child and boat , [tex]m_1[/tex] = (25.0 + 38.4 )kg

                                               = 63.4 kg

Mass of the package, [tex]m_2[/tex] = 5.90 kg

Velocity of package thrown from boat , [tex]v_2[/tex] = 10.0m/s

[tex]v_1 =?[/tex]

Initial velocity v = 0 m/s

As the boat is at rest, [tex](m_1 + m_2) v=0[/tex]

According, to the law of conversation of momentum;

∴ Momentum before = Momentum after

   [tex]( m_1 + m_2 ) v = m_1v_1 + m_2v_2\\0 = m_1v_1 + m_2v_2\\0 = 63.4 v_1 + 5.90*10.0\\63.4 v_1 = - 5.9\\v_1 = - 0.930 m/s[/tex]

Negative direction shows the velocity in the direction opposite to the motion of the package.

Therefore, -0.930 m/s is the velocity of the boat.

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Explanation:

first rule of parallel circuits : the voltage is the same in the whole circuit, no matter where we measure.

so, we have 1.5V at every bulb.

second rule of parallel circuits : the total current is the sum of the individual branch currents.

so, if the battery really allows it, then we have 3 times 0.4 A = 3×0.4 = 1.2 A as current in the circuit.

The solution of this system is (i₁, i₂, i₃) = (- 10.852 A, 8.479 A, - 2.374 A). The negative signs indicate that real direction of the current is opposite than supposed.

In this question we must make use of Kirchhoff's laws to find the values of the missing currents in the circuit presented in the picture. There are two rules according to Kirchhoff's laws:

Based on the information given by the picture, we have the following system of linear equations that describes the entire circuit:

i₃ = i₁ + i₂    

- i₁ · R₁ + ε₁ - i₁ · r₁ - i₁ · R₅ + ε₂ - i₂ · (r₂ + R₂) = 0      

ε₂ - i₂ · (r₂ + R₂) - i₃ · r₄ - ε₄ - i₃ · r₃ + ε₃ - i₃ · R₃ = 0      

- i₁ - i₂ + i₃ = 0                                                      (1)

(R₁ + r₁ + R₅) · i₁ + (r₂ + R₂) · i₂ = ε₁ + ε₂                (2)

(r₂ + R₂) · i₂ + (r₄ + r₃ + R₃) · i₃ = ε₂ + ε₃ - ε₄        (3)

If we know that R₁ = 5 Ω, r₁ = 0.10 Ω, R₅ = 20 Ω, r₂ = 0.50 Ω, R₂ = 40 Ω, r₄ =  0.20 Ω, r₃ = 0.05 Ω, R₃ = 78 Ω, ε₁ = 22 V, ε₂ = 49 V, ε₃ = 10.5 V and ε₄ = 33 V, then the currents flowing in the circuit are:

- i₁ - i₂ + i₃ = 0

25.1 · i₁ + 40.5 · i₂ = 71

25.1 · i₂ + 78.25 · i₃ = 26.5

The solution of this system is (i₁, i₂, i₃) = (- 10.852 A, 8.479 A, - 2.374 A). The negative signs indicate that real direction of the current is opposite than supposed.

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(1) The INCREASE in the sound level from the ambient work environment level (in dB) is 1 dB.

(2) The factor that sound intensity exceed the 1.5- Hours/day intensity limits from the graph is 19.6 %.

The INCREASE in the sound level from the ambient work environment level (in dB) is calculated as follows;

Increase in sound level = final sound level - original sound level

Increase in sound level = 86 dB - 85 dB = 1 dB

from the graph at 1.5 hours/day, sound level = 97 dB

Increase in sound intensity = final sound level - original sound level

Increase in sound intensity = 116 dB - 97 dB = 19 dB

Factor increase = 19/97 = 0.196 = 19.6%

Thus, the factor that sound intensity exceed the 1.5- Hours/day intensity limits from the graph is 19.6 %.

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Answer:

(a)If two objects collide and one is initially at rest, is it possible for both to be at rest after the collision?

No. Because if you have initial momentum P⃗ ≠0 , if both of the objects were at rest after the collision the total momentum of the system would be P⃗ =0 , which violates conservation of momentum

(b)Is it possible for only one to be at rest after the collision?

Yes, that is perfectly possible. It characteristically, happens when both objects are of the same mass. When two objects of the same mass collide and Kinetic energy is conserved (Perfectly Elastic collision) then the two objects interchange velocities.

Answer:

long-term

Explanation:

usually a short term goal is able to be accomplished in a week or two. the question gives a 3 month time frame for the person to build up to the end goal.

(a) The kinetic energy of the projectile when it reaches the highest point in its trajectory is 900 J.

(b) The work done  in firing the projectile is 2,500 J.

The kinetic energy of the projectile when it reaches the highest point in its trajectory is calculated as follows;

K.E = ¹/₂mv₀ₓ²

where;

Note: At maximum height the final vertical velocity is zero and the final horizontal velocity is equal to the initial horizontal velocity.

K.E = (0.5)(2)(30²)

K.E = 900 J

Based on the principle of conservation of energy, the work done in firing the projectile is equal to the initial kinetic energy of the projectile.

W = K.E(i) = ¹/₂mv²

where;

v = √(30² + 40²)

v = 50 m/s

W = (0.5)(2)(50²)

W = 2,500 J

Thus, the kinetic energy of the projectile when it reaches the highest point in its trajectory is 900 J.

The work done  in firing the projectile is 2,500 J.

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2.89watts.

How much power is radiated as a sound from a band whose intensity is 1.6x10-3 w/m2 at a distance of 12m:

Formula: [tex]I\frac{P}{4\pi r^{2} }[/tex]

I=1.6x10-3 w/m2

r=12m

[tex]I\frac{P}{4\pi r^{2} }[/tex]

[tex]P=I4\pi r^{2}[/tex]

   [tex]=(1.6x10-3\frac{W}{m^{2} } ) (4\pi (12m)^{2} )[/tex]

The power is radiated as a sound from a band whose intensity is 1.6x10-3 w/m2 at a distance of 12m  [tex]=2.89watts.[/tex]

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Answer:

potential energy = (mass × height × acceleration due to gravity.)

750 = mass × 3 × 10

hence, mass = 750/30 = 25 kg.

Given the following data:

First of all, we would calculate the electric field due to B₁ to B₃ and the electric field due to B₂ to B₃ respectively.

The electric field due to B₁ to B₃ is given by:

E₁ = kq₁/r₁²

E₁ = (9 × 10⁹ × 17 × 10⁻⁶)/(0.05² + 0.1²)

E₁ = 153 × 10³/0.0125

Electric field, E₁ = 12.24 × 10⁶ N/C.

The electric field due to B₂ to B₃ is given by:

E₂ = kq₂/r₂²

E₂ = (9 × 10⁹ × 5 × 10⁻⁶)/(0.1²)

E₂ = 45 × 10³/0.01

Electric field, E₂ = 4.5 × 10⁶ N/C.

Also, the magnitude of the angle formed is given by tan trigonometry:

Tanθ = 5/10

Tanθ = 0.5

θ = tan⁻¹(0.5)

θ = 26.56°.

Next, we would calculate the x-component of the electric field at the observation location (Bee number 3) due to B₁:

E₁x = E₁cosθ - E₂

E₁x = 12.24 × 10⁶ × cos(26.56) - 4.5 × 10⁶

E₁x = 10.95 × 10⁶ - 4.5 × 10⁶

E₁x = 6.5 × 10⁶ N/C.

Similarly, we would calculate the y-component of the electric field at the observation location (Bee number 3) due to B₁:

E₁y = -E₁sinθ

E₁y = -12.24 × 10⁶ × sin(26.56)

E₁y = -12.24 × 10⁶ × 0.4471

E₁y = -5.5 × 10⁶ N/C.

Also, the magnitude of the electric field is given by:

Exy = √(E₁x² + E₁y²)

Exy = √(6.5 × 10⁶)² + (-5.5 × 10⁶)²)

Exy = √4.225 × 10¹³ + 3.025 × 10¹³)

Exy = √(7.25 × 10¹³)

Exy = 8.5 × 10⁶ N/C.

We would calculate the x-component of the electric field at the observation location (Bee number 3) due to B₂:

E₂x = -E₂sinθ

E₂x = -4.5 × 10⁶ × sin(90)

E₂x = -4.5 × 10⁶ N/C.

Similarly, we would calculate the y-component of the electric field at the observation location (Bee number 3) due to B₁:

E₂y = E₂cosθ

E₂y = -4.5 × 10⁶ × cos(90)

E₂y = 0 N/C.

The magnitude of the net electric field at B₃ is given by:

Eₙ = √(E₁x + E₂x)² + E₁y²)

Eₙ = √(6.5 × 10⁶ + (-4.5 × 10⁶))² + (-5.5 × 10⁶)²)

Eₙ = √(2.5 × 10⁶)² + (3.025 × 10¹³)

Eₙ = √(6.25 × 10¹²) + (3.025 × 10¹³)

Eₙ = √(3.65 × 10¹³)

Eₙ = 6.04 × 10⁶ N/C.

For the direction, we have:

Tanθ = x/y

Tanθ = 6.5/-5.5

Tanθ = -1.1818

θ = tan⁻¹(-1.1818)

θ = 49.76°.

The magnitude of the force on B₃ due to this net electric field is given by:

F = B₃ × Eₙ

F = 26 × 10⁻⁶ × 6.04 × 10⁶

F = 157.04 Newton.

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Answer:

Approximately [tex]3.86\; {\rm N}[/tex] (given that the magnitude of this charge is [tex]-7.35\; {\rm \mu C}[/tex].)

Explanation:

If a charge of magnitude [tex]q[/tex] is placed in an electric field of magnitude [tex]E[/tex], the magnitude of the electrostatic force on that charge would be [tex]F = E\, q[/tex].

The magnitude of this charge is [tex]q = 7.35\; {\rm \mu C}[/tex]. Apply the unit conversion [tex]1\; {\rm \mu C} = 10^{-6}\; {\rm C}[/tex]:

[tex]\begin{aligned} q &= 7.35\; {\mu C} \times \frac{10^{-6}\; {\rm C}}{1\; {\mu C}} = 7.35\times 10^{-6}\; {\rm C}\end{aligned}[/tex].

An electric field of magnitude [tex]E = 5.25\times 10^{5}\; {\rm N \cdot C^{-1}}[/tex] would exert on this charge a force with a magnitude of:

[tex]\begin{aligned}F &= E\, q \\ &= 5.25 \times 10^{5}\; {\rm N \cdot C^{-1}} \times (-7.35\times 10^{-6}\; {\rm C}) \\ &\approx 3.86\; {\rm N}\end{aligned}[/tex].

Note that the electric charge in this question is negative. Hence, electrostatic force on this charge would be opposite in direction to the the electric field. Since the electric field points due south, the electrostatic force on this charge would point due north.

Answer:

Thermal conductivity formula

Explanation:

The object as point particles is [tex]r_{cm} =0.5909 m, 0.6364 m.[/tex]

a)

The center of mass of a three-particle system is expressed as

[tex]r_{cm}=[/tex] [tex]$$r_{c m}=\frac{\sum_{i=1}^{3} m_{i} r_{i}}{\sum_{i=1}^{3} m_{i}} \Rightarrow \frac{m_{1} r_{1}+m_{2} r_{2}+m_{3} r_{3}}{m_{1}+m_{2}+m_{3}}$$[/tex]

When the system is only on [tex]$\mathrm{x}$[/tex]-axis (i.e. [tex]$\mathrm{y}=0$[/tex] )

[tex]$$\begin{aligned}x_{c m} &=\frac{m_{1} x_{1}+m_{2} x_{2}+m_{3} x_{3}}{m_{1}+m_{2}+m_{3}} \\x_{c m} &=\frac{(5.0 \times 0.5)+(2.0 \times 0)+(4.0 \times 1.0)}{5.0+2.0+4.0} \\x_{c m} &=0.5909 \mathrm{~m} \\\text { Therefore } r_{c m}=(0.5909 \mathrm{~m}, 0)\end{aligned}$$[/tex]

b)

When the two particles are shifted

[tex]\begin{aligned}&x_{c m}=\frac{m_{1} x_{1}+m_{2} x_{2}+m_{3} x_{3}}{m_{1}+m_{2}+m_{3}} \\&x_{c m}=\frac{(5.0 \times 0.5)+(2.0 \times 0)+(4.0 \times 1.0)}{5.0+2.0+4.0} \\&x_{c m}=0.5909 \mathrm{~m} \\&y_{c m}=\frac{m_{1} y_{1}+m_{2} y_{2}+m_{3} y_{3}}{m_{1}+m_{2}+m_{3}} \\&y_{c m}=\frac{(5.0 \times 1.0)+(2.0 \times 0)+(4.0 \times 0.5)}{5.0+2.0+4.0} \\&y_{c m}=0.6364 \mathrm{~m}\end{aligned}[/tex]

Therefore [tex]r_{cm} =0.5909 m, 0.6364 m.[/tex]

The object as point particles is [tex]r_{cm} =0.5909 m, 0.6364 m.[/tex]

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Answer:

The term significant figures refers to the number of important single digits (0 through 9 inclusive) in the coefficient of an expression in scientific notation . The number of significant figures in an expression indicates the confidence or precision with which an engineer or scientist states a quantity.

Answer:

each of the digits of a number that are used to express it to the required degree of accuracy, starting from the first nonzero digit.

Explanation:

The wavelength of the wave is 1.16m and the velocity is 23.64m/s.

To find the answer, we have to know more about the Transverse waves.

                [tex]y(x,t)=0.048sin(5.40x-128t)[/tex]

               [tex]amplitude, a=0.048m\\wave vector, k=5.40\\angular frequency,w=128Hz[/tex]

                [tex]k=\frac{2\pi }{wavelength} \\\\wavelength=\frac{2\pi }{k} =\frac{2*3.14}{5.40} \\\\wavelength=1.16m[/tex]

                         [tex]v=frequency*wavelength\\2\pi f=w, thus,\\f=\frac{w}{2\pi } =\frac{128}{2*3.14}=20.38s^{-1}\\\\v=1.16*20.38=23.64m/s[/tex]

Thus, we can conclude that, the wavelength of the wave is 1.16m and the velocity is 23.64m/s

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