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

The value of parameter D is -3.

Angular frequency is  the angular displacement of any element of the wave per unit time.

f(x) = Acos(x) + D

where;

The amplitude of a wave is the maximum displacement of the wave. It can also be described at the maximum upward displacement of a wave curve.

From the blue colored graph, y = -2 at x = 0

-2 = cos(0) + D

-2 = 1  +  D

-2 - 1 = D

-3 = D

Thus, the value of parameter D is -3.

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(a) The moment of inertia of the wheel  is 78.2 kgm².

(b) The mass (in kg) of the wheel is 1,436.2 kg.

(c) The angular speed (in rad/s) of the wheel at the end of this time period is 3.376 rad/s.

Apply principle of conservation of angular momentum;

Fr = Iα

where;

I = Fr/α

I = (200 x 0.33) / (0.844)

I = 78.2 kgm²

I = ¹/₂MR²

where;

2I = MR²

M = 2I/R²

M = (2 x 78.2) / (0.33²)

M = 1,436.2 kg

ω = αt

ω = 0.844 x 4

ω = 3.376 rad/s

Thus, the moment of inertia of the wheel  is 78.2 kgm².

The mass (in kg) of the wheel is 1,436.2 kg.

The angular speed (in rad/s) of the wheel at the end of this time period is 3.376 rad/s.

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The earth's orbital angular velocity around the Sun is 0.986° per day, and its axial angular velocity is 15.041° per hour.

Calculation:

The speeds, not the velocities, are what we require in this case, therefore that is how I will approach it.

Regular speed is calculated by dividing the distance traveled by the travel time.

So, speed = [tex]\frac{distance -travelled}{time}[/tex]

The angular velocity is essentially the same.

Hence, it's divided by the amount of time it took to turn the angle.

(a) Calculation of Earth's orbit around the Sun :

The Earth orbits the Sun in a circle that is 2 radians in diameter (360 degrees). It also takes a year, as we are aware (approx 365 days)

360°/365.25636 days = 0.986°/day ≈ 1° per day

(b) Next, we have to calculate Earth's angular velocity on its axis:

determining the Earth's angular velocity as it fulfills a complete revolution on its axis (a solar day)-

According to this, the angular speed of rotation = 360°/ 23h 56min 4.09s = 15.041° / hour.

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6.45 ×[tex]10^{-} ^{11}[/tex] m is far from the carbon atom and is the center of mass of the molecule.

The sum of the products of the masses of the two particles and their respective position vectors equals the product of the total mass of the system and the position vector of the center of mass.

The center of mass of CO molecule will be on the line joining C and O atoms. Let the CO molecule be along the X- axis, the C atm being at the origin (x=0). The center of mass relative to the C atom is given by

[tex]x_{cm} = \frac{m_{c}x_{c} + m_{o} x_{o} }{m_{c}+ m_{o} }[/tex]

Where, [tex]m_{c}[/tex] and [tex]m_{o}[/tex] are the respective masses of C and O atoms,  [tex]x_{c}[/tex] and  [tex]x_{o}[/tex] their distances relative to C atom.

Here, [tex]m_{c}[/tex] = 12 u , [tex]m_{o}[/tex] = 16 u and [tex]x_{o} =1.13[/tex]×[tex]10^-^{10}[/tex] m

[tex]x_{cm} = \frac{m_{c}x_{c} + m_{o} x_{o} }{m_{c}+ m_{o} }[/tex]

[tex]x_{cm} =[/tex] (12 × 0 + 16 × 1.13 × [tex]10^{-} ^{10}[/tex] ) / (12 + 16 )

[tex]x_{cm}[/tex] = 0.64571 × [tex]10^{-} ^{10}[/tex]

[tex]x_{cm}[/tex] = 6.45 ×[tex]10^{-} ^{11}[/tex] m

Therefore, 6.45 ×[tex]10^{-} ^{11}[/tex] m is far from the carbon atom is the center of mass of the molecule.

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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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(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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It will take you 0.23 seconds to hear your echo

Echo can be simply defined as the reflection of sound wave.

Given that you are 38.6 m away from a rock wall, you yell. To know how much time in seconds it will take you to hear your echo to two significant digits, You must make sure to account for the travel from you to the wall and from the wall back to you.

Speed V is the distance per time

V = D / T

Substitute all the parameters into the formula

340 = 77.2 / T

Make T the subject of formula

T = 77.2 / 340

T = 0.2270

T = 0.23 s to two significant digits

Therefore, it will take you 0.23 seconds to hear your echo.

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Answer:

Impulse is defined as change in momentum of an object divided by time interval.

at t= 0s

initial velocity = u , initial momentum = mu

at some time t .

final velocity = V, final momentum= mv.

now, change in momentum= ( final - initial) = ( mv-mu)

time interval = (t-0) = t

impulse force = (mv-mu)/ ( t)

Ft = (mv-mu) proved .

this law is known as Newton's second law.

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 center of mass of the car with all people seated will be 3.20 m far from the front of the car.

Taking the front of the car as reference:

m₁x₁ + m₂x₂ + m₃x₃ = mx

where,

m₁ = mass of empty car = 1300 kg

x₁ = distance of center of mass of empty car from front = 2.45 m

m₂ = mass of 2 people sitting in front = 2 x 60 kg = 120 kg

x₂ = distance of center of mass of 2 people sitting in front from the front    = 2.7 m

m₃ = mass of 3 people sitting in back seat = 3 x 60 kg = 180 kg

x₃ = distance of center of mass of 3 people sitting in the back seat from the front = 3.65 m

m = total mass of car with all people = 1300 kg + 120 kg + 180 kg = 1600kg

x = distance of center of mass of the car from the front when all are seated

Therefore,

(1300 kg)(2.45 m) + (120 kg)(2.7 m) + (180kg)(3.65 m) = (1300 kg)x

x = 4166 kg.m/1300 kg

x = 3.20 m

Therefore, the center of mass of the car with all people seated will be 3.20 m far from the front of the car.

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E =+823.12N/C is the electric field at point P

Each point in space has an electric field associated with it when a charge of any kind is present. The value of E, often known as the electric field strength, electric field intensity, or just the electric field, expresses the strength and direction of the electric field. in the diagram, q1 = +6.39*10^-9 C and

q2 = +3.22*10^-9 C hence E = 823.12 N/C

A region of space surrounding andd P Include a + or - sign toindicate the direction.P910.424 m-0.636 m- electrically charged particle or object known as an electric field is one in which an electric charge would experience force. A vector quantity called an electric field can be represented by arrows pointing in the direction of or away from charges. The force per unit charge exerted on a positive test charge that is at rest at a given position is the force per unit charge that is used to define the electric field analytically.

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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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a) The original momentum of the fullback is 331.2 kgm/s.

b) The impulse exerted on the fullback is - 331.2 kgm/s

c) The impulse exerted on the tackler is 331.2 kgm/s

d) The average force exerted on the tackler is 389.64 N

Mass of the fullback, m = 92 kg

Initial velocity of the fullback , u = 3.6 m/s

Time of the motion of the fullback , t = 0.85s

The original momentum of the fullback ;

[tex]P_{i} = mv\\P_{i} = (92) (3.6)\\P_{i} = 331.2 kg m/s[/tex]

The impulse exerted on the full-back;

J = ΔP = [tex]m v_f - m v_i[/tex]

J = [tex]m ( v_f - v_i)[/tex]

J = 92 ( 0- 3.6)

J = - 331.2 kgm/s

The impulse exerted on the tackler;

[tex]J _1 = - J _2\\J_2 = - J_1\\J_2 = -( - 331.2)\\J_2 = 331.2 kgm/s[/tex]

The average force exerted on the tackler;

F= Δmv/t

[tex]F =\frac{331.2}{0.85} \\F = 389.64 N[/tex]

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(a) The magnitude of the force is 1.32 x 10⁻⁵ N and direction of the electric force on the sphere towards the right.

(b) The work done on the sphere by the electric force as it moves from A to B is 2.5 x 10⁻⁶ J.

(c) The change of the electric potential energy as the sphere moves from A to B is 2.5 x 10⁻⁶ J.

(d) The potential difference between A and B is 56.7 V.

The electric force on the sphere is calculated as follows;

F = Eq

where;

F = 300 x (44 x 10⁻⁹)

F = 1.32 x 10⁻⁵ N

The direction of the force is towards the right.

W = Fd

W = 1.32 x 10⁻⁵ N  x 0.189 m

W = 2.5 x 10⁻⁶ J

The change in the electric potential energy (in J) as the sphere moves from A to B is equal to work done in moving the charge = 2.5 x 10⁻⁶ J.

VB − VA =  Ed

VB − VA =  300 N/C  x   0.189 m

VB − VA =  56.7 V

Thus, the magnitude of the force is 1.32 x 10⁻⁵ N and direction of the electric force on the sphere towards the right.

The work done on the sphere by the electric force as it moves from A to B is 2.5 x 10⁻⁶ J.

The change of the electric potential energy as the sphere moves from A to B is 2.5 x 10⁻⁶ J.

The potential difference between A and B is 56.7 V.

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Answer:

4

Explanation:

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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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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.

(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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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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The magnitude of the acceleration of the electron is 5.187 x 10¹³ m/s².

The speed of the electron at the given time is 4.41 x 10⁵ m/s.

The magnitude of the acceleration of the electron is calculated as follows;

Force on the electron, F = ma = Eq

ma = Eq

a = Eq/m

where;

a = (295 x 1.6 x 10⁻¹⁹) / (9.1 x 10⁻³¹)

a = 5.187 x 10¹³ m/s²

The speed of the electron at the given time is calculated as follows;

v = at

v = ( 5.187 x 10¹³)(8.5 x 10⁻⁹)

v = 4.41 x 10⁵ m/s

Thus, the magnitude of the acceleration of the electron is 5.187 x 10¹³ m/s².

The speed of the electron at the given time is 4.41 x 10⁵ m/s.

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Answer:

A

Explanation:

Work = force * distance

          = 100 n  *  0     = 0   work done

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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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

They pick up speed and move away from the usual.

Using meter sticks, a group of pupils makes a straight line (shoulder to shoulder) and connects to their closest neighbor. Two media are separated in the room by a strip of masking tape. Students can be seen moving normally in one of the media (on one side of the tape). The kids in the other medium (or on the opposite side of the tape) move very slowly and in baby stages. The class moves in the union in a straight line in the direction of the diagonal piece of masking tape. As they get close to the masking tape, the pupils stay in line. Each pupil immediately changes the speed of her or his walk as she or he reaches the tape.

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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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The rate at which the person is now losing thermal energy is 99 J/s.

The thickness of the person's skin is calculated as follows;

Q = k(ΔT)h

where;

h = Q/k(ΔT)

h = (120) / (0.22 x 303)

h = 1.8 m

Q = k(ΔT)h

where;

Q = (0.18)(306)(1.8)

Q = 99 J/s

Thus, the rate at which the person is now losing thermal energy is 99 J/s.

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1.3 second of time will be required for reflected sunlight to travel from the Moon to Earth if the distance between Earth and the Moon is 3.85 × 105 km

Speed is the distance travelled per time taken. It is a scalar quantity. And the S.I unit is meter per second. That is, m/s

In the given question, we want to find how much time is required for reflected sunlight to travel from the Moon to Earth if the distance between Earth and the Moon is 3.85 × 10^5 km.

What are the parameters to consider ?

The parameters are;

Speed = distance S ÷ Time t

Convert kilometer to meter by multiplying it by 1000

C = S/t

3 × [tex]10^{8}[/tex] =  3.85 × [tex]10^{8}[/tex] / t

Make t the subject of formula

t = 3.85 × [tex]10^{8}[/tex] / 3 × [tex]10^{8}[/tex]

t = 1.2833

t = 1.3 s

Therefore, 1.3 second of time will be required for reflected sunlight to travel from the Moon to Earth if the distance between Earth and the Moon is 3.85 × 105 km

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Answer:

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

750 = mass × 3 × 10

hence, mass = 750/30 = 25 kg.