A particle moves so that its position (in meters) as a function of time (in seconds) is . Write expressions (in unit vector notation) for (a) its velocity and (b) its acceleration as functions of time.

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]

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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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:

[tex]6.945326087\times 10^{-11}\ C[/tex]

5434.78260873 N/C

2963369.48874 m/s

Explanation:

[tex]\epsilon_0[/tex] = Permittivity of free space = [tex]8.85\times 10^{-12}\ F/m[/tex]

q = Charge of electron = [tex]1.6\times 10^{-19}\ C[/tex]

V = Voltage = 25 V

Side = 3.8 cm

d = Separation = 4.6 mm

m = Mass of electron = [tex]9.11\times 10^{-31}\ kg[/tex]

Charge is given by

[tex]Q=\dfrac{VA\epsilon_0}{d}\\\Rightarrow Q=\dfrac{25\times (3.8\times 10^{-2})^2\times 8.85\times 10^{-12}}{4.6\times 10^{-3}}\\\Rightarrow Q=6.945326087\times 10^{-11}\ C[/tex]

Charge on each plate is [tex]6.945326087\times 10^{-11}\ C[/tex]

Electric field is given by

[tex]E=\dfrac{Q}{A\epsilon_0}\\\Rightarrow E=\dfrac{6.945326087\times 10^{-11}}{(3.8\times 10^{-2})^2\times 8.85\times 10^{-12}}\\\Rightarrow E=5434.78260873\ N/C[/tex]

The electric field strength is 5434.78260873 N/C

Acceleration is given by

[tex]a=\dfrac{qE}{m}\\\Rightarrow a=\dfrac{1.6\times 10^{-19}\times 5434.78260873}{9.11\times 10^{-31}}\\\Rightarrow a=9.5451725291\times 10^{14}\ m/s^2[/tex]

[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.5451725291\times 10^{14}\times 4.6\times 10^{-3}+0^2}\\\Rightarrow v=2963369.48874\ m/s[/tex]

The velocity is 2963369.48874 m/s

Answer:

Torque=6.261×10^-43Nm

Explanation:

Torque is given by÷

Torque=dqEsin(phi)

Where E is a constant=1.6022×10^-19

d= distance = 6.186×10^-30cm

Changing to metres=6.186×10^-28m

q= the charge=8500NC

Phi =42°

Torque= (6.186×10^-28)×8500×(1.6022×10^-19)sin42°

Torque= 8.425×10^-43 × 0.7431

Torque= 6.261 ×10^-43

The torque exerted by an 8500 N/C electric field exerts on the dipole is 6.261 ×10⁻⁴³ Nm.

The force which causes the object to rotate about any axis is called perpendicular distance.

Torque is a twisting or turning force that frequently results in rotation around an axis, which may be a fixed point or the center of mass. The ability of something rotating, such as a gear or a shaft, to overcome turning resistance is another way to think of torque.

Given:

A single water molecule was oriented such that its dipole moment of magnitude 6.186 × 10 − 30 Cm is along a z-axis,

The electric field  = 8500 N/C,

∅ = 42°

Calculate the torque as shown below,

Torque = d × q ×  E × sin(∅)

Where E is a constant = 1.6022×10⁻¹⁹

distance = 6.186×10⁻³⁰ cm =  6.186×10⁻²⁸ m

Torque = 6.186×10⁻²⁸ × 8500 × (1.6022 ×  10⁻¹⁹) sin42°

Torque = 8.425×10⁻⁴³ × 0.7431

Torque = 6.261 ×10⁻⁴³

Thus, the torque is 6.261 ×10⁻⁴³ Nm.

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Answer:

(a) E = 3.065 * 10^21 N/C

(b) The direction is radially outward.

Explanation:

Parameters given:

Number of protons = 94

Charge of proton = 1.6023 * 10^-19 C

Radius of nucleus = 6.65 fm = 6.65 * 10^-15

At the surface of the nucleus, Electric field, E is given as:

E = kq/r^2

Where k = Coulombs constant

r = radius of nucleus

=> E = (9 * 10^9 * 94* 1.6023 * 10^-19)/(6.65 * 10^21)

E = 3.065 * 10^21 N/C

The direction of the field is radially outward.

The magnitude of the electric field at the surface of a plutonium-239 nucleus is roughly 7.94 x 10^21 NC^-1. The direction of the electric field, created by positively charged protons, is radially outward.

The question pertains to the electric field created by the protons of a plutonium-239 atom. The electric field resulting from a spherical charge distribution can be determined by using Coulomb's law transformed to a volumetric charge distribution. The formula is E = kQ/r2, where k is the Coulomb constant (8.99 x 109 Nm2/C2), Q is the total charge, and r is the distance from the center of the field (which in this case is the radius of the nucleus).

(a) Magnitude of the electric field: Here, the total charge Q is the charge of a single proton (1.60 x 10-19 C) multiplied by the total number of protons which is 94. Therefore, upon inserting these values into the formula, the calculated electric field at the surface of the nucleus is around 7.94 x 1021 NC-1.

(b) Direction of the electric field: The direction of the electric field is always from positive to negative. Since protons are positively charged and they are creating the field, the electric field direction will be radially outward.

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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

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]

1)

first you find the maxium force that the car can produce.

f=ma

Fmax=(1100kg)(6m/s^2)

then use f = ma again to find the accel with the passengers

Fmax=(1100kg +1650kg)(a)

=> a = (1100kg)(6m/s^2)/( 1100kg +1650kg)

= 2.4 m/s^2

The maximum magnitude of the acceleration of the two‑car system is  [tex]2.44 \;\rm m/s^{2}[/tex].

Given data:

The mass of car is, m = 1200 kg.

The maximum acceleration is, [tex]a =6.00 \;\rm m/s^{2}[/tex].

The mass of stall is, m' = 1750 kg.

As per the Newton's Second law of motion, the applied force on the car is equal to the product of mass and acceleration of car. So, the maximum force to push the car is given as,

F = ma

[tex]F = 1200 \times 6.00\\\\F= 7200\;\rm N[/tex]

Now for two-car system, the total mass is,

M = m + m'

M = 1200 +1750 = 2950 kg.

So, the acceleration for the two-car system is,

[tex]F = M \times a'\\\\7200 =2950 \times a'\\\\a'=2.44 \;\rm m/s^{2}[/tex]

Thus, the maximum magnitude of the acceleration a of the two‑car system is  [tex]2.44 \;\rm m/s^{2}[/tex].

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Answer: Water on Earth was transported here by comets. The correct option is A.

Explanation:

Comets are made up of water with ice, rock and minerals.

Alot of research and hypotheses has been made to prove the origin of water on planet earth. Extraplanetary source such as comets, trans-Neptunian objects, and water-rich meteoroids (protoplanets) are believed to have delivered water to Earth.

Final answer:

Water on Earth primarily originated from the (b) solar nebula during planetary accretion, with minor contributions from comets and volcanic activity. Hence, (b) is the correct option.

Explanation:

As Earth coalesced from the protoplanetary disk around the young Sun, volatile compounds, including water, were incorporated into the developing planet. Comets may have contributed some water, but the predominant source was the solar nebula.

While volcanic activity released water vapor, it played a minor role compared to the water acquired during accretion. Chemical reactions involving hydrogen and oxygen likely occurred after Earth's formation, but the bulk of the water was established during the early stages of planetary accretion from the solar nebula.

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: 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

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)   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:

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:

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

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

Answer:

a. -7.11+5.7

Explanation:

Given:

Now the acceleration:

[tex]a=\frac{v_w}{t} +\frac{v_n}{t}[/tex]

[tex]a=\frac{-25\hat i}{3.5} +\frac{20\hat j}{3.5}[/tex]

[tex]a=\frac{-50}{7} \hat i+\frac{40}{7} \hat j[/tex]

[tex]a=-7.1428\hat i+5.7142\hat j[/tex]

[tex]a=\sqrt{(\frac{50}{7} )^2+(\frac{40}{7} )^2}[/tex]

[tex]a=9.1473\ m.s^{-2}[/tex]

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.

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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Answer:

[tex]\frac{4}{3}\rho\pi(r_2^3 - r_1^3)[/tex]

Explanation:

The volume of the shell is the difference between the volume of the outer sphere and the volume of the inner sphere

[tex]V = V_o - V_i = \frac{4}{3}\pi r_2^3 - \frac{4}{3}\pi r_1^3[/tex]

[tex]V = \frac{4}{3}\pi(r_2^3 - r_1^3)[/tex]

The mass of the shell is the product of the volume and its density

[tex]m = V\rho = \frac{4}{3}\rho\pi(r_2^3 - r_1^3)[/tex]

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:

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:

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

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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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:

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.

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:

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