A block of mass m = 150 kg rests against a spring with a spring constant of k = 880 N/m on an inclined plane which makes an angle of θ degrees with the horizontal. Assume the spring has been compressed a distance d from its neutral position. Refer to the figure. show answer No Attempt 25%
Part (a) Set your coordinates to have the x-axis along the surface of the plane, with up the plane as positive, and the y-axis normal to the plane, with out of the plane as positive. Enter an expression for the normal force, FN, that the plane exerts on the block (in the y-direction) in terms of defined quantities and g. 25%
Part (b) Denoting the coefficient of static friction by μs, write an expression for the sum of the forces in the x-direction just before the block begins to slide up the inclined plane. Use defined quantities and g in your expression ΣFx = 25%
Part (c) Assuming the plane is frictionless, what will the angle of the plane be, in degrees, if the spring is compressed by gravity a distance 0.1 m? 25%
Part (d) Assuming θ = 45 degrees and the surface is frictionless, how far will the spring be compressed, d in meters?

Normal human body temperature is about 37°C, which is equivalent to 310.15 K. The peak wavelength emitted by a person with this temperature is 9.35 μm, which lies in the infrared part of the spectrum.

Normal human body temperature is about 37°C. To convert this temperature to kelvin, we can use the formula K = °C + 273.15. So, the temperature in kelvins would be:

K = 37 + 273.15 = 310.15 K.

The wavelength emitted by a person with this temperature can be determined using Wien's displacement law: λ = b/T, where b is Wien's constant and T is the absolute temperature. The peak wavelength (λmax) is the wavelength at which the emitted radiation is the greatest. For a temperature of 310.15 K, the peak wavelength can be calculated as: λmax = b/T = 2897.8/T = 9.35 μm.

The peak wavelength of 9.35 μm corresponds to the infrared part of the electromagnetic spectrum.

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Answer:

a)   A = 0.0221 m, b) T = 0.314 s

Explanation:

For this exercise we must separate the process into two parts, one when the body falls and another when it hits the tray

Let's look for the speed with which it reaches the tray, using energy conservation

Initial. Highest point

        Em₀ = U = m g h

Final. Tray Point

        [tex]Em_{f}[/tex] = K = ½ m v²

        Em₀ = Em_{f}

        m g h = ½ m v²

        v = √2gh

Now let's apply moment conservation

Initial. Just before the crash

          p₀ = m v

Final. After the crash

         p_{f} = (m + M) v_{f}

         p₀ = p_{f}

         m v = (m + M) v_{f}  

         v_{f}  = m / (m + M) √2gh

This is the turkey + plate system speed, which is the one that will oscillate

b) The angular velocity of the oscillation is

           w = √ k / m

The angular velocity is related to the frequency and period

         w = 2πf = 2π / T

          T = 2π √m / k

          T = 2π √ ((0.100 + 0.400) / 200)

          T = 0.314 s

a) To find the amplitude let's use the equation that describes the oscillatory motion

            y = A cos (wt + fi)

Speed ​​is

           v = dy / dt = - A w sin (wt + fi)

At the initial point the turkey + plate system has a maximum speed, in the previous equation the speed is maximum when cos (wt + fi) = ±1

               v = v_{f}  = A w

               m / (m + M) √ 2gh = A w

               A = m / (m + M) √2gh    1 / w

               w = 2π / T

Let's calculate

              A = 0.100 / (0.100 + 0.400) √(2 9.8 0.250)   0.314/2π

            A = 0.2   2.2136   0.049975

            A = 0.0221 m

Answer:

2019 light bulbs

Explanation:

So the electrical energy consumed by each 59-W light bulb within 7 hours is the product of the power and time duration

E = Pt = 59 * 7 = 413 Wh or 0.413 kWh

If each light bulb consumes 0.413 kWh, then the total number of light bulbs needed to consume 834 kWh would be

834 / 0.413 = 2019 light bulbs

Answer:

[tex]G=6750\ W.m^2[/tex] is the irradiation

[tex]\epsilon=0.4725[/tex]

No, it is not a grey body because we don't have its transmissivity is not zero.

Explanation:

Given that:

Temperature of air, [tex]T_{\infty}=310\ K[/tex]

temperature of plate, [tex]T_s=360\ K[/tex]

convective heat transfer coefficient, [tex]h=45\ W.m^{-2}.K^{-1}[/tex]

absorptivity of plate, [tex]\alpha=0.4[/tex]

radiosity of the plate, [tex]J=4500\ W.m^{-2}[/tex]

From the energy balance eq. :

[tex]E_{in}=E_{out}[/tex]

since two surfaces are involved

[tex]2G=2J+2q_{conv}[/tex]

[tex]G=J+q_{conv}[/tex]

where

[tex]q_{conv}=[/tex] heat transfer per unit area due to convection

[tex]G=[/tex] irradiation  per unit area

[tex]G=J+h.\Delta T[/tex]

[tex]G=4500+45\times (360-310)[/tex]

[tex]G=6750\ W.m^2[/tex] is the irradiation

Since radiosity includes transmissivity, reflectivity and emissivity.

[tex]J=G.\tau+G.\rho+E[/tex]

[tex]J=G.\tau+G.\rho+\epsilon\times \sigma\times T^{4}[/tex]

where:

[tex]\tau=[/tex] transmissivity

[tex]\rho=[/tex] reflectivity

[tex]\epsilon=[/tex] emissivity

Since the absorptivity of the plate is 0.4 and the plate is semitransparent so the transmissivity and reflectivity = 0.3 each.

[tex]4500=0.3\times 6750+0.3\times 6750+\epsilon\times 5.67\times 10^{-8}\times 360^4[/tex]

[tex]\epsilon=0.4725[/tex]

No, it is not a grey body because we don't have its transmissivity is not zero.

The Stefan-Boltzmann law of radiation can be used to find the irradiation and emissivity of a plate. The irradiation can be found using the net rate of heat transfer equation, and the emissivity can be found by rearranging the equation. The plate is diffuse-gray if its emissivity is between 0 and 1.

The irradiation and emissivity of the plate can be found using the Stefan-Boltzmann law of radiation. The rate of heat transfer by emitted radiation is given by P = 6 AeT^4, where o = 5.67 × 10^-8 J/s. m². K^4 is the Stefan-Boltzmann constant, A is the surface area of the object, and T is its temperature in kelvins.

To find the irradiation, we can use the equation Qnet = σe A (T₂^4 – T₁^4), where Qnet is the net rate of heat transfer, σ is the Stefan-Boltzmann constant, e is the emissivity of the body, A is the surface area of the object, T₂ is the temperature of the surrounding air, and T₁ is the temperature of the plate.

To find the emissivity of the plate, we can rearrange the equation Qnet = σe A (T₂^4 – T₁^4) to solve for e. The plate is diffuse-gray if its emissivity is between 0 and 1. If the emissivity is 0, the plate is a perfect reflector, and if the emissivity is 1, the plate is a perfect absorber and emitter of radiation.

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Answer:

a) [tex]v_{max}=0.4\ m.s^{-1}[/tex]

b) [tex]a_{max}=1.6\ m.s^{-2}[/tex]

c) [tex]v_x=0.32\ m.s^{-1}[/tex]

d) [tex]a_x=0.96\ m.s^{-1}[/tex]

e) [tex]\Delta t=0.232\ s[/tex]

Explanation:

Given:

mass of the object attached to the spring, [tex]m=0.5\ kg[/tex]

spring constant of the given spring, [tex]k=8\ N.m^{-1}[/tex]

amplitude of vibration, [tex]A=0.1\ m[/tex]

a)

Now, maximum velocity is obtained at the maximum Kinetic energy and the maximum kinetic energy is obtained when the whole spring potential energy is transformed.

Max. spring potential energy:

[tex]PE_s=\frac{1}{2} .k.A^2[/tex]

[tex]PE_s=0.5\times 8\times 0.1^2[/tex]

[tex]PE_s=0.04\ J[/tex]

When this whole spring potential is converted into kinetic energy:

[tex]KE_{max}=0.04\ J[/tex]

[tex]\frac{1}{2}.m.v_{max}^2=0.04[/tex]

[tex]0.5\times 0.5\times v_{max}^2=0.04[/tex]

[tex]v_{max}=0.4\ m.s^{-1}[/tex]

b)

Max. Force of spring on the mass:

[tex]F_{max}=k.A[/tex]

[tex]F_{max}=8\times 0.1[/tex]

[tex]F_{max}=0.8\ N[/tex]

Now acceleration:

[tex]a_{max}=\frac{F_{max}}{m}[/tex]

[tex]a_{max}=\frac{0.8}{0.5}[/tex]

[tex]a_{max}=1.6\ m.s^{-2}[/tex]

c)

Kinetic energy when the displacement is, [tex]\Delta x=0.06\ m[/tex]:

[tex]KE_x=PE_s-PE_x[/tex]

[tex]\frac{1}{2} .m.v_x^2=PE_s-\frac{1}{2} .k.\Delta x^2[/tex]

[tex]\frac{1}{2}\times 0.5\times v_x^2=0.04-\frac{1}{2} \times 8\times 0.06^2[/tex]

[tex]v_x=0.32\ m.s^{-1}[/tex]

d)

Spring force on the mass at the given position, [tex]\Delta x=0.06\ m[/tex]:

[tex]F=k.\Delta x[/tex]

[tex]F=8\times 0.06[/tex]

[tex]F=0.48\ N[/tex]

therefore acceleration:

[tex]a_x=\frac{F}{m}[/tex]

[tex]a_x=\frac{0.48}{0.5}[/tex]

[tex]a_x=0.96\ m.s^{-1}[/tex]

e)

Frequency of oscillation:

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

[tex]\omega=\sqrt{\frac{8}{0.5} }[/tex]

[tex]\omega=4\ rad.s^{-1}[/tex]

So the wave equation is:

[tex]x=A.\sin\ (\omega.t)[/tex]

where x = position of the oscillating mass

put x=0

[tex]0=0.1\times \sin\ (4t)[/tex]

[tex]t=0\ s[/tex]

Now put x=0.08

[tex]0.08=0.1\times \sin\ (4t)[/tex]

[tex]t=0.232\ s[/tex]

So, the time taken in going from point x = 0 cm to x = 8 cm is:

[tex]\Delta t=0.232\ s[/tex]

Final answer:

The problem involves calculating the maximum speed, maximum acceleration, speed and acceleration at a certain distance from equilibrium, and the time interval for an object in simple harmonic motion. Using formulas for SHM including maximum speed (v_max = ωA), maximum acceleration (a_max = ω^2A), and expressions for speed and acceleration at a given position, one can determine these values for the object on the spring.

Explanation:

The student has asked us to calculate various properties of an object undergoing simple harmonic motion (SHM) when attached to a spring with a known force constant and amplitude. To answer this question, one needs to use equations that describe SHM.

Maximum Speed (v_max) Calculation:

The maximum speed (v_max) of an object in SHM occurs when it passes through the equilibrium point and can be calculated using the formula v_max = ωA, where ω is the angular frequency (ω = sqrt(k/m)) and A is the amplitude of the motion.

Maximum Acceleration (a_max) Calculation:

The maximum acceleration (a_max) occurs at the maximum displacement and is given by a_max = ω^2A.

Speed at 6 cm from Equilibrium:

To find the speed at a certain position x, we use the formula v = ω sqrt(A^2 - x^2).

Acceleration at 6 cm from Equilibrium:

Acceleration at any position x is a = -ω^2x.

Time Interval to Move from 0 to 8 cm:

The time interval to move from x = 0 to a certain position x can be found using the formula for time in SHM as a function of position.

Answer:

v_b = 10.628 m /s (13.605 degrees east of south)

Explanation:

Given:

- Velocity of mia v_m = + 6 j m/s

- Velocity of ball wrt mia v_bm = 5.0 m/s     30 degree due east of south

Find:

What are the magnitude and direction of the velocity of the ball relative to the ground? v_b

Solution:

- The relation of velocity in two different frame is given:

                             v_b  - v_m = v_bm

- Components along the direction of v_b,m:

                             v_b*cos(Q) - v_m*cos(30) = 5

                             v_b*cos(Q) = 5 + 6 sqrt(3) / 2

                             v_b*cos(Q) = 5 + 3sqrt(3)  

- Components orthogonal the direction of v_b,m:

                             -v_m*sin(30) = v_b*sin(Q)

                             -6*0.5 = v_b*sin(Q)

                              -3 = v_b*sin(Q)

- Divide two equations:

                               tan(Q) = - 3 / 5 + 3sqrt(3)

                               Q = arctan(- 3 / 5 + 3sqrt(3)

                               Q = -16.395 degrees

                               v_b =  -3 / sin(-16.395)

                               v_b = 10.63 m/s

The magnitude of the soccer ball's velocity relative to the ground is 2.89 m/s, and its direction is approximately 56.31° north of east. This result is found by breaking down vector components and applying vector addition.

You're asking about the velocity of the ball relative to the ground when soccer players Mia and Alice are running, and Alice passes the ball to Mia. To solve this, we'll use vector addition. Mia is running due north at 6.00 m/s, and the velocity of the ball relative to Mia is 5.00 m/s at 30° east of south. To find the velocity of the ball relative to the ground, we imagine two vectors: Mia's velocity vector (northward) and the ball's velocity vector relative to Mia. The latter will have an east and a south component due to the 30° angle.

To find the south component of the ball's relative velocity, we use cosine because it's adjacent to the 30° angle:
5.00 m/s * cos(30°) = 4.33 m/s. The east component is found using sine:
5.00 m/s * sin(30°) = 2.50 m/s.

Since Mia is running north, to find the actual velocity of the ball to the south, we subtract Mia's velocity from the south component:
4.33 m/s - 6.00 m/s = -1.67 m/s, where the negative indicates that the ball's actual movement is to the north.

Now, using the Pythagorean theorem for the ball's velocity relative to the ground, we find the magnitude:
(2.50 m/s)² + (-1.67 m/s)² = √(6.25 + 2.7889)
√8.3389 m²/s² = 2.89 m/s.

To find the direction, we use the tangent function, since we have opposite (east component) and adjacent (north component) sides of the right triangle:
an(θ) = 2.50 / 1.67; θ = tan(⁻¹)(2.50 / 1.67);
θ ≈ 56.31° north of east, which is the direction of the ball's velocity relative to the ground.

Answer:

find answers below

Explanation:

a)

S1 560 e^(j acos 0.707 ) ⋅kVA            S1 = ( ) 395.92 396.04j  kW+ ⋅

S2 := 132 kW⋅                                      S2 = 132 kW⋅

Sd S1 +S2 :=

Sd = 527.92 396.04j  kW

Sd = 660 kVA ⋅

arg Sd = 36.877 deg ⋅

Vd 2.2 e^(− j⋅0⋅deg)⋅kV

current in the line =(Sd/3 Vd) ⋅

⎯

:=

I Line =  79.988 60.006j A

Line = 99.994 A

arg I( ) Line = −36.877⋅deg

r Line := 0.4⋅Ω                          resistance in the line xLine := 2.7⋅Ω

Vsan Vd+ (r Line+ j xLine) ILine

Vsan = ( ) kV 2.394 0.192j + ⋅ Vsan = 2.402 kV⋅ arg V( ) san = 4.584 deg ⋅

Vsab 3 e

j 30 ⋅ ⋅deg ⋅ Vsan := ⋅

Vsab = ( ) kV 3.425 2.361j + ⋅ Vsab = 4.16 kV⋅ arg V( ) sab = 34.584 deg

b)

PLine 3 I ( ) Line

2 ⋅ r

Line := ⋅

PLine = 12 kW⋅

QLine 3 I ( ) Line

2 ⋅ xLine := ⋅ QLine = 80.99 kVAR ⋅

c)

Ss 3 Vsan ⋅ I

Line := ⋅

⎯ Ss = ( ) kVA 539.919 477.03j + ⋅

Ss = 720.5 kVA ⋅ arg S( )s = 41.461 deg ⋅

Ps Re S( )s := Ps = 540 kW⋅

Qs Im S( )s := Qs = 477 kVAR ⋅

S1 S2 + PLine + j QLine + ⋅ = ( ) kVA 539.919 477.03j + ⋅ Check

Answer:

M_c = 100.8 Nm

Explanation:

Given:

F_a = 2.5 KN

Find:

Determine the moment of this force about C for the two cases shown.

Solution:

- Draw horizontal and vertical vectors at point A.

- Take moments about point C as follows:

                        M_c = F_a*( 42 / 150 ) *144

                        M_c = 2.5*( 42 / 150 ) *144

                        M_c = 100.8 Nm

- We see that the vertical component of force at point A passes through C.

Hence, its moment about C is zero.

A) 7.297 N is the x component Fx of the resultant force.

Part B) - 3.654 N is the y component Fy of the resultant force.

Part C) 8.16N is the magnitude F of the resultant force.

D) 26.55° from the negative x-axis in a clockwise direction.

A) x-component of forces is 7.297 N

For x-component FX;

FX = F1 cos 65 + F2 cos 53.9

FX = 9.20 cos 65 + 5.8 cos 53.9

NB both forces are positive if resolved along the x-axis.

FX = 3.88 + 3.417 = 7.297 N

B) 6-component of forces is -3.654 N

For y-component FY

FY = -F1 sin 65 + F2 sin 53. 9

FY = - 9.2 sin 65 + 5.8 sin 53.9

NB F1 is negative along y-axis

FY = - 8.34 + 4.686 = - 3.654 N

C) The magnitude of the resultant force is 8.16 N

Magnitude :

[tex](FX^2 + FY^2)^{0.5}\\(7.297^2 + (-3.654)^2)^{0.5} \\= 8.16 \text N[/tex]

D) Angle made with negative x-axis is 26.56°

[tex]\text{Tan}^{ -1[/tex] of FY/FX gives the angle of the resultant force.

= [tex]\text{tan}^{ -1[/tex] of -3.654/7.297

=[tex]\text{tan}^{ -1[/tex] of -0.5

= -26.56°

i.e. 26.55° from the negative x-axis in a clockwise direction.

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(A) The x-component of the resultant force is approximately 7.039 N.

(B) The y-component of the resultant force is approximately 3.654 N.

(C) The magnitude of the resultant force is approximately 7.95 N.

(D) The angle θ with the negative x-axis is approximately 28.6 degrees (measured clockwise).

To find the x and y components of the resultant force, as well as its magnitude and angle with the negative x-axis, we can use vector addition. Given the forces F₁ and F₂:

F₁ = 9.20 N (65.0° above the negative x-axis)

F₂ = 5.80 N (53.9° below the negative x-axis)

(A) Finding the x-component (Fₓ) of the resultant force:

We'll first find the x-component of each force:

For F₁:

F₁ₓ = F₁ * cos(65.0°)

For F₂:

F₂ₓ = F₂ * cos(53.9°)

Now, calculate the net x-component:

Fₓ = F₁ₓ + F₂ₓ

Calculate Fₓ:

F₁ₓ = 9.20 N * cos(65.0°) ≈ 4.016 N

F₂ₓ = 5.80 N * cos(53.9°) ≈ 3.023 N

Fₓ = 4.016 N + 3.023 N ≈ 7.039 N

(B) Finding the y-component (Fᵧ) of the resultant force:

Similarly, we'll find the y-component of each force:

For F₁:

F₁ᵧ = F₁ * sin(65.0°)

For F₂:

F₂ᵧ = -F₂ * sin(53.9°) [negative because it's below the x-axis]

Now, calculate the net y-component:

Fᵧ = F₁ᵧ + F₂ᵧ

Calculate Fᵧ:

F₁ᵧ = 9.20 N * sin(65.0°) ≈ 8.111 N

F₂ᵧ = -5.80 N * sin(53.9°) ≈ -4.457 N

Fᵧ = 8.111 N - 4.457 N ≈ 3.654 N

(C) Finding the magnitude (F) of the resultant force:

The magnitude of the resultant force is given by the Pythagorean theorem:

F = √(Fₓ² + Fᵧ²)

Calculate F:

F = √(Fₓ² + Fᵧ²) ≈ √(7.039 N)² + (3.654 N)² ≈ √(49.53 N² + 13.35 N²) ≈ √62.88 N² ≈ 7.95 N

(D) Finding the angle (θ) with the negative x-axis:

The angle θ with the negative x-axis can be found using the inverse tangent function:

θ = arctan(Fᵧ / Fₓ)

Calculate θ:

θ = arctan(Fᵧ / Fₓ) ≈ arctan(3.654 N / 7.039 N) ≈ arctan(0.519) ≈ 28.6°

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Explanation:

a. 1 liter (L) is equal to 61.0237 cubic inches ([tex]in^3[/tex]), So:

[tex]1L*\frac{61.0237in^3}{1L}=61.0237in^3[/tex]

b. 1 J is equal to [tex]9.47817*10^{-4}[/tex] BTU. Thus:

[tex]650J*\frac{9.47817*10^{-4}BTU}{1J}=6.16081*10^{-1}BTU[/tex]

c. 1 kW is equal to 737.56 [tex]\frac{ft\cdot lbf}{s}[/tex]. So:

[tex]0.135kW*\frac{737.56\frac{ft\cdot lbf}{s}}{1kW}=99.57\frac{ft\cdot lbf}{s}[/tex]

d. 1 [tex]\frac{g}{s}[/tex] is equal to 0.1323 [tex]\frac{lb}{min}[/tex]. Therefore:

[tex]378\frac{g}{s}*\frac{0.1323\frac{lb}{min}}{1\frac{g}{s}}=50.01\frac{lb}{min}[/tex]

e. 1 kPa is equal to 0.145[tex]\frac{lbf}{in^2}[/tex]. Thus:

[tex]304kPa*\frac{0.145\frac{lbf}{in^2}}{1kPa}=44.08\frac{lbf}{in^2}[/tex]

f. 1 [tex]\frac{m^3}{h}[/tex] is equal to [tex]9.81*10^{-3}\frac{ft^3}{s}[/tex]. So:

[tex]55\frac{m^3}{h}*\frac{9.81*10^{-3}\frac{ft^3}{s}}{1\frac{m^3}{h}}=5.40*10^{-1}\frac{ft^3}{s}[/tex]

g. 1 [tex]\frac{km}{h}[/tex] is equal to 0.911 [tex]\frac{ft}{s}[/tex]. Therefore:

[tex]50\frac{km}{h}*\frac{0.911\frac{ft}{s}}{1\frac{km}{h}}=45.55\frac{ft}{s}[/tex]

h. 1 N is equal to [tex]1.1*10^{-4}[/tex] ton. Thus:

[tex]8896N*\frac{1.1*10^{-4}ton}{1N}=0.98ton[/tex]

Answer:

F_e = +/- 0.013133 N

Explanation:

Given:

- charge on each sphere q = -56 nC

- separation of spheres r = 3.0 cm

- charge constant k = 8.99*10^9

Find:

- Magnitude of Electric Force F_e on each sphere.

Solution:

The magnitude of electric Force F_e of one sphere on other is:

                                F_e = k*q^2 / r^2

- Plugging the given values :

                                F_e = (8.99*10^9) * (56*10^-19)^2 / (0.03)^2

                                F_e = 0.013133 N

- An equal and opposite force is experienced on the other sphere. Hence, F_e = + / - 0.013133 N

Answer:

4.has gained two electrons

Explanation:

There exist electrovalent bonding the compound MgS . In electrovalent bonding, there is a transfer of electrons from the metal to non-metal.

Magnesium atom has an atomic number 12 and its electron configuration is 2,8,2

Sulfur atom , a non-metal has atomic number of 16 and its electron configuration = 2,8,6

This means that magnesium as a metal needs to loose two electrons from its valence shell to attain its stable structure.Also sulfur requires two more electron to achieve its octet structure.

Hence a transfer of electrons will take place from magnesium atom to sulfur atom, sulfur gaining two electrons.

Final answer:

In MgS, the sulfide ion has gained two electrons, resulting in a charge of -2, represented as S²-. Magnesium loses two electrons to form Mg²+, balancing the electron transfer to create a stable ionic compound.

Explanation:

In the compound MgS, the sulfide ion has gained two electrons to achieve a stable electron configuration. When sulfur (S), which has an atomic number of 16, gains two electrons, it results in an ion with 18 electrons and 16 protons. This gives the ion a charge of -2, since there are two more negative electrons than positive protons. Therefore, the sulfide ion is represented as S²-. The magnesium atom loses its two valence electrons to become a Mg²+ cation, as magnesium is in Group 2A of the periodic table and tends to lose two electrons to achieve a noble gas electron configuration. This electron transfer process is balanced, meaning the number of electrons lost by magnesium is equal to the number of electrons gained by sulfur, forming a stable ionic compound.

Answer:

Aristotle

Explanation:

Aristotelian theory of the Universe

For two millennia, the philosophical tradition considered that the universe was eternal and did not change. The wise Aristotle said so, with total clarity and his ideas dominated Western thought for more than two thousand years.

This distinguished philosopher believed that the stars are made of an imperishable matter and that the landscapes of the sky are immutable.

From the time of Aristotle until the beginning of the twentieth century, the idea that the universe was static, that the cosmos had been eternally equal, was admitted.

In those years, the origin of the universe was not really considered in a scientific way, since it was based on the basis that the gods had created everything that exists, at the time they wanted it, according to their omnimous power.

So that all the efforts of the wise men of the time focused on discovering the existing organization in the universe created by the gods.

According to Aristotle and the thinkers of the fourth century B.C. what is below the Moon is a changing world, what is beyond the Moon is an immutable world.

Answer:

Explanation:

When the distance is decreasing, the frequency of the received wave form will be higher than the source wave form. Besides sound and radio waves, the Doppler effect also affects the light emitted by other bodies in space. If a body in space is "blue shifted," its light waves are compacted and it is coming towards us.

The Doppler effect is a phenomenon that describes how the perceived frequency and wavelength of waves change due to relative motion. It can result in a blue shift or a red shift, depending on the direction of motion. The Doppler effect has applications in astronomy and medicine, among other fields.

The Doppler effect is a phenomenon in physics that describes how the perceived frequency and wavelength of waves change when there is relative motion between the source of the waves and the observer. This effect can alter the way in which we perceive radiation, such as light or sound.

When an object emitting waves moves towards an observer, the waves are compressed, resulting in a higher frequency and shorter wavelength. This is known as a blue shift. On the other hand, when an object moves away from an observer, the waves are stretched, resulting in a lower frequency and longer wavelength, which is called a red shift.

The Doppler effect has several applications in various fields. For example, it is used in astronomy to determine the speed and direction of celestial objects based on their red or blue shifts. In medicine, it is used in ultrasound technology to measure blood flow velocity. Overall, the Doppler effect allows us to understand and interpret the motion of objects emitting waves.

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Answer:

F = 88.88[lb-f]

Explanation:

To solve this problem, we must first look at the attached image.

Another important data that is needed to solve the problem is the distance from the Centre O to points C and D. The distance is 0.9 [ft].

Then we do a sum of moments around Point O and it equals zero.

[tex]SM_{O} =0\\F*0.9+F*0.9-160=0\\2*F*0.9=160\\F=88.88[lb-f][/tex]

Therefrore the couple of forces needed are 88.88[lb-f]

Answer:

P_2 = 62.69 psi

Explanation:

given,

P₁ = 70 psia               T₁ = 55° F    = (55 + 459.67) R

P₂ = ?                         T₂ = 115° F    = (115 + 459.67) R

we know,

p = ρ RT

ρ is the density which is constant

R is also constant

now,

[tex]\dfrac{P_2}{P_1} =\dfrac{T_2}{T_1}[/tex]

[tex]\dfrac{P_2}{70} =\dfrac{55+459.67}{115+459.67}[/tex]

    P_2 = 62.69 psi

Hence, the increase in Pressure is equal to P_2 = 62.69 psi

Answer:

Pressure of gas A = 544.324 kPa

Pressure of gas B = 2217.784 kPa

Explanation:

Data provided in the question:

Total volume of the cylinder, V = 5.55 L = 0.00555 m³     [1 m³ = 1000 L]

Moles on gas A, [tex]n_a[/tex] = 1.21 mol

Moles on gas A, [tex]n_b[/tex] = 4.93 mol

Temperature, T = 27.3°C = 27.3 + 273 = 300.3 K

Now,

Pressure = [tex]\frac{nRT}{V}[/tex]

here,

R is the ideal gas constant = 8.314 J/mol.K

Therefore,

Pressure of gas A = [tex]\frac{n_aRT}{V}[/tex]

= [tex]\frac{1.21\times8.314\times300.3}{0.00555}[/tex]

= 544324.32 Pa

= 544.324 kPa

Pressure of gas B = [tex]\frac{n_bRT}{V}[/tex]

= [tex]\frac{4.93\times8.314\times300.3}{0.00555}[/tex]

= 2217784.22 Pa

= 2217.784 kPa

Answer: The molar mass of metal (M) is 47.86 g/mol

Explanation:

Let the atomic mass of metal (M) be x

Atomic mass of [tex](MO_2)=(x+32)g/mol[/tex]

To calculate the mass of metal, we use the equation:

[tex]\text{Mass percent of metal}=\frac{\text{Mass of metal}}{\text{Mass of metal oxide}}\times 100[/tex]

Mass percent of metal = 59.93 %

Mass of metal = x g/mol

Mass of metal oxide = (x + 32) g/mol

Putting values in above equation, we get:

[tex]59.93=\frac{x}{(x+32)}\times 100\\\\59.93(x+32)=100x\\\\x=47.86g/mol[/tex]

Hence, the molar mass of metal (M) is 47.86 g/mol

Final answer:

The initial momentum of the bullet before the collision is equal to the momentum of the bullet and block at the instant they reach the maximum height h. Option A

Explanation:

The correct statement of the collision in this scenario is A) The initial momentum of the bullet before the collision is equal to the momentum of the bullet and block at the instant they reach the maximum height h.

According to the law of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision, as long as there are no external forces acting on the system. In this case, the bullet and block form a closed system, and therefore, the initial momentum of the bullet is equal to the momentum of the bullet and block at the maximum height.

To understand why this statement is true, we can consider the momentum of the bullet and block system at different points in the motion. Initially, the bullet has momentum in the positive x-direction, and this momentum is transferred to the bullet and block as they rise to the maximum height. Therefore, the statement A is correct.

Answer:

q=3.5*10^-4

Explanation:

concept:

The force acting on both charges is given by the coulomb law:

F=kq1q2/r^2

the centripetal force is given by:

Fc=mv^2/r

The kinetic energy is given by:

KE=1/2mv^2

The tension force:

when the plane is uncharged

T=mv^2/r

T=2(K.E)/r

T=2(50 J)/r

T=100/r

when the plane is charged

T+k*|q|^2/r^2=2(K.E)charged/r

100/r+k*|q|^2/r^2=2(53.5 J)/r

q=√(2r[53.5 J-50 J]/k)                                          √= square root on whole

q=√2(2)(53.5 J-50 J)/8.99*10^9

q=3.5*10^-4

Since the Units presented are not in the International System we will proceed to convert them. We know that,

[tex]1 mi/h = 0.447 m/s[/tex]

So the speed in SI would be

[tex]V=95mi/h(\frac{0.447m/s}{1mi/h})[/tex]

[tex]V=42.465 m/s[/tex]

The change in frequency when the wave is reflected is

[tex]f'=f(1+\frac{V}{c})[/tex]

Or we can rearrange the equation as

[tex]f' = f + f\frac{V}{c}[/tex]

f' = Apparent frequency

f = Original Frequency

c = Speed of light

[tex]f'-f = f\frac{V}{c}[/tex]

[tex]\Delta f = f\frac{V}{c}[/tex]

Replacing,

[tex]\Delta f = (10.525*10^9)(\frac{42.465}{3*10^8})[/tex]

[tex]\Delta f =1489.8 Hz[/tex]

Since the waves are reflected, hence the change in frequency at the gun is equal to twice the change in frequency

[tex]\Delta f_T = 2 \Delta f[/tex]

[tex]\Delta f_T = 2(1489.8Hz)[/tex]

[tex]\Delta f_T = 2979.63Hz[/tex]

Therefore the increase in frequency is 2979.63Hz

The increase in frequency of these waves is equal to 2979.6 Hertz.

Given the following data:

Conversion:

1 mi/h = 0.447 m/s.

95.0 mi/h = [tex]95 \times 0.447[/tex] = 42.465 m/s.

To determine the increase in frequency:

Mathematically, the change in frequency of a wave is given by this formula:

[tex]F' = F(1+\frac{V}{c} )\\\\F' = F+F\frac{V}{c}\\\\F' - F=F\frac{V}{c}\\\\\Delta F = F\frac{V}{c}[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]\Delta F = 10.525 \times 10^9 \times \frac{42.465}{3 \times 10^8}\\\\\Delta F = \frac{446.944 \times 10^9}{3 \times 10^8}\\\\\Delta F = 1489.8\;Hertz[/tex]

Since the waves were reflected, the increase in frequency toward the gun is double (twice) the change in frequency. Thus, we have;

[tex]F_2 = 2\Delta F\\\\F_2 = 2\times 1489.8\\\\F_2 = 2979.6\;Hertz[/tex]

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Answer:a)1.11×10^-21Nm

b) 1.16×10^-3m

Explanation:see attachment

1) Current: 4.5 A

2) Time taken: 4.7 s

Explanation:

1)

The electric current intensity is defined as the rate at which charge flows in a conductor; mathematically:

[tex]I=\frac{q}{t}[/tex]

where

I is the current

q is the amount of charge passing a given point in a time t

For the wire in this problem, we have

q = 9.0 C is the amount of charge

t = 2.0 s is the time interval

Solving for I, we find the current:

[tex]I=\frac{9.0}{2.0}=4.5 A[/tex]

2)

To solve this problem, we can use again the same formula

[tex]I=\frac{q}{t}[/tex]

where

I is the current

q is the amount of charge passing a given point in a time t

In this problem, we have:

I = 3.0 A (current)

q = 14.0 C (charge)

Therefore, the time taken for the charge to move past a particular spot in the wire is

[tex]t=\frac{q}{I}=\frac{14.0}{3.0}=4.7 s[/tex]

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The electric current in the wire in the first scenario is 4.5 A, and it will take 4.7 seconds for 14.0 C of charge to move past a particular spot on the wire with a constant current of 3.0 A in the second scenario.

The amount of electric current in a circuit is defined by the amount of electric charge that passes through it in a given amount of time. This is represented by the formula I = Q / t, where I is the current, Q is the charge and t is the time.

In the first part of your question, we are given that Q = 9.0 C and t = 2.0 s. We can substitute these values into the formula to find the current: I = 9.0 C / 2.0 s = 4.5 A.

In the second part, we are given that I = 3.0 A and Q = 14.0 C. This time, we need to rearrange the formula to find t: t = Q / I = 14.0 C / 3.0 A = 4.7 s.

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The entrance of certain frequencies through the atmosphere varies depending on the sector. This concept of permittivity by which the rays can enter the earth is called the Radio Window and has a spectrum that ranges from 5MHz to 30GHz. Of the radiation produced by astronomical objects, one part is 'filtered' and the other has the ability to be absorbed by the earth due to its opacity effect.

Ultraviolet rays, X-rays and even gamma rays are completely affected by the ozone layer in Earth's atmosphere.

This makes it difficult to use X-ray telescopes located in Antarctica.

The correct option is D.

Earth's atmosphere limits X-ray observations, making Antarctica unsuitable due to extreme cold.

Earth's atmosphere acts as a barrier for X-ray observations, as it blocks most radiation at wavelengths shorter than visible light. Telescopes located in Antarctica would not work well due to the extreme cold which can affect their performance.

Answer:

The first possible value of d is 0.85 m

The second possible value of d is 2.55 m

The third possible value of d is 4.25 m

Explanation:

Given that,

Distance =d

Frequency of sound wave= 200 Hz

We need to calculate the wavelength

Using formula of wavelength

[tex]\lambda=\dfrac{v}{f}[/tex]

Put the value into the formula

[tex]\lambda=\dfrac{340}{200}[/tex]

[tex]\lambda=1.7\ m[/tex]

The separation between the speakers in the destructive interference is

[tex]\Delta x= d[/tex]

The equation for destructive interference

[tex]2\pi\times\dfrac{\Delta x}{\lambda}-\Delta\phi_{0}=(m+\dfrac{1}{2})2\pi[/tex]

The loudspeakers are in phase

So, [tex]\Delta\phi_{0}=0[/tex]

The equation for destructive interference is

[tex]2\pi\times\dfrac{d}{\lambda}=(m+\dfrac{1}{2})2\pi[/tex]....(I)

Here, m = 0,1,2,3.....

We need to calculate the first possible value of d

For, m = 0

Put the value in the equation (I)

[tex]2\pi\times\dfrac{d_{1}}{1.7}=(0+\dfrac{1}{2})2\pi[/tex]

[tex]d_{1}=\dfrac{1.7}{2}[/tex]

[tex]d_{1}=0.85\ m[/tex]

We need to calculate the second possible value of d

For, m = 1

Put the value in the equation (I)

[tex]2\pi\times\dfrac{d_{2}}{1.7}=(1+\dfrac{1}{2})2\pi[/tex]

[tex]d_{2}=\dfrac{1.7\times3}{2}[/tex]

[tex]d_{2}=2.55\ m[/tex]

We need to calculate the third possible value of d

For, m = 1

Put the value in the equation (I)

[tex]2\pi\times\dfrac{d_{3}}{1.7}=(2+\dfrac{1}{2})2\pi[/tex]

[tex]d_{3}=\dfrac{1.7\times5}{2}[/tex]

[tex]d_{3}=4.25\ m[/tex]

Hence, The first possible value of d is 0.85 m

The second possible value of d is 2.55 m

The third possible value of d is 4.25 m

the radial acceleration of the blade tip is approximately 1210.50 times g .

To find the linear speed of the blade tip and the radial acceleration of the blade tip, we can use the following formulas:

(a) Linear speed of the blade tip:

[tex]\[ \text{Linear speed} = \text{Angular speed} \times \text{Radius} \]\[ v = \omega \times r \][/tex]

(b) Radial acceleration of the blade tip:

[tex]\[ \text{Radial acceleration} = \omega^2 \times r \]\[ a_r = \omega^2 \times r \][/tex]

Given:

- Angular speed [tex](\( \omega \))[/tex] = 550 rev/min

- Radius [tex](\( r \))[/tex] = 3.40 m

- Acceleration due to gravity[tex](\( g \))[/tex] = 9.81 m/s²

First, let's convert the angular speed from rev/min to rad/s:

[tex]\[ \omega = \frac{550 \text{ rev/min} \times 2\pi \text{ rad/rev}}{60 \text{ s/min}} \]\[ \omega = \frac{550 \times 2\pi}{60} \text{ rad/s} \]\[ \omega \approx 57.91 \text{ rad/s} \][/tex]

(a) Linear speed of the blade tip:

[tex]\[ v = \omega \times r \]\[ v = 57.91 \times 3.40 \]\[ v \approx 197.04 \text{ m/s} \][/tex]

So, the linear speed of the blade tip is approximately 197.04 m/s.

(b) Radial acceleration of the blade tip:

[tex]\[ a_r = \omega^2 \times r \]\[ a_r = (57.91)^2 \times 3.40 \]\[ a_r \approx 11854.76 \text{ m/s}^2 \][/tex]

Now, express the radial acceleration as a multiple of ( g ):

[tex]\[ \text{Multiple of } g = \frac{a_r}{g} \]\[ \text{Multiple of } g = \frac{11854.76}{9.81} \]\[ \text{Multiple of } g \approx 1210.50 \][/tex]

So, the radial acceleration of the blade tip is approximately 1210.50 times ( g ).

Answer:

[tex]v = 3.81\ m/s[/tex]

Explanation:

given,

Radius of the ball, r₁ = 53 cm

Radius of another ball, r₂ = 26 cm

center to center distance between the balls = 223 cm

time, t = 18.9 s

surface to surface distance between them

S = 223 - (53+26)

S = 144 cm

Speed of the ball = ?

[tex]relative\ speed = \dfrac{distance}{time}[/tex]

[tex]2 v = \dfrac{144}{18.9}[/tex]

both the balls are moving towards each other so, speed doubles.

[tex]v= \dfrac{7.62}{2}[/tex]

[tex]v = 3.81\ m/s[/tex]

Speed of the balls is equal to 3.81 m/s

Answer:

Therefore the magnitude of the net electric filed at the center of the triangle is [tex]=\frac{6K}{A^2}[/tex] N/C

Explanation:

Given,A = side of the triangle. q = 1 C

The center of a triangle is the centroid of the triangle.

The distance between centroid to any vertices of a equilateral triangle is

[tex]=\frac{2}{3}[/tex] of the height of the equilateral triangle

[tex]=\frac{2}{3}\times \frac{\sqrt{3} }{2} \times A[/tex]

[tex]=\frac{1}{\sqrt{3} } A[/tex]

Electric field= [tex]\frac{Kq}{d^2}[/tex]       K=8.99×10⁹ Nm²/C², q=charge  and d = distance

Therefore the magnitude of the net electric filed at the center of the triangle is

=2[tex]\frac{Kq}{d^2}[/tex]         [both charge are at same distance from the centroid]

=[tex]\frac{2k}{(\frac{1}{\sqrt{3} }A)^2 }[/tex]

[tex]=\frac{6K}{A^2}[/tex] N /C      [K=8.99×10⁹ Nm²/C²]

Answer:

[tex]{P_2}=146.53\ kPa[/tex]

Explanation:

Volume ,V = 3 L

Initial temperature ,T₁ = 273 K

Initial pressure ,P₁ = 105 kPa

Final temperature ,T₂ = 381 K

Lets take final pressure =P₂

We know that ,If the volume of the gas is constant ,then we can say that

[tex]\dfrac{P_2}{P_1}=\dfrac{T_2}{T_1}[/tex]

[tex]{P_2}=P_1\times \dfrac{T_2}{T_1}[/tex]

Now by putting the values in the above equation we get

[tex]{P_2}=105\times \dfrac{381}{273}\ kPa[/tex]

[tex]{P_2}=146.53\ kPa[/tex]

Therefore the final pressure will be 146.53 kPa.

The final pressure of the cylinder after being heated is 75.57 kPa.

To solve this problem, we can use the equation for Charles's Law, which states that at constant pressure, the volume of a gas is directly proportional to its temperature. In this case, the initial volume is 3.0 L and the initial temperature is 273 K. The final temperature is 381 K. Now we can set up a proportion:

(Initial volume) / (Initial temperature) = (Final volume) / (Final temperature)

Plugging in the numbers, we get:

(3.0 L) / (273 K) = (Final volume) / (381 K)

Solving for the final volume gives us:

Final volume = [(3.0 L) / (273 K)] x (381 K) =  4.1732 L

Since the volume remains the same, the pressure is inversely proportional to the new volume. So, if the initial pressure is 105 kPa, the final pressure can be calculated using the following equation:

(Initial pressure) x (Initial volume) = (Final pressure) x (Final volume)

Plugging in the numbers, we get:

(105 kPa) x (3.0 L) = (Final pressure) x (4.1732 L)

Solving for the final pressure gives us:

Final pressure = [(105 kPa) x (3.0 L)] / (4.1732 L) = 75.57 kPa

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Answer:

true

Explanation:

The statement "the energy of a photon is inversely proportional to the wavelength of the radiation" is definitely true.

Wavelength may be characterized as the space or distance between two successive crests of a wave that especially includes the points in a sound wave or electromagnetic wave. The wavelength may be represented by a letter known as lambda (λ).

When the energy of the radiation increases, the waves travel faster which directs the small gap between two successive crests. This means that the wavelength decreases.

On contrary, when the energy of the radiation decreases, the waves travel slower which leads to a huge gap between two successive crests. This means that the wavelength increases.

Therefore, the statement "the energy of a photon is inversely proportional to the wavelength of the radiation" is definitely true.

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