The Mariana trench is located in the Pacific Ocean at a depth of about 11 000 m below the surface of the water. The density of seawater is 1025 kg/m3. (a) If an underwater vehicle were to explore such a depth, what force would the water exert on the vehicle's observation window (radius = 0.280 m)? (b) For comparison, determine the weight of a jetliner whose mass is 2.44 × 105 kg.

Answer:No it does not the chemical equation is unbalanced,this is because according to the law of conservation it states that mass cannot be created or destroyed ,the chemical equation N₂ + H₂ → NH₃ shows that energy has been created which is not possibly so the correct balanced chemical is equation is  1N₂ +3H₂ → 3NH₃

Explanation:

The impulse delivered to the pinball by the launcher is 1.8 kg•m/s to the east

Let us consider east as positive direction and west as negative direction.

Then, from the question, we get that the initial velocity of the pinball is

2m/s towards the west

or u = - 2 m/s

and the mass of the pinball is m = 150g = 0.15 kg

So, the initial momentum of the pinball is:

[tex]P_i=mu\\\\P_i=0.15\times(-2)\;kgm/s\\\\P_i=-0.3\;kgm/s[/tex]

Now, the final velocity of the pinball after being struck by the rod is 10  m/s towards the east,

or v = 10 m/s

So, the final momentum of the pinball is:

[tex]P_f=mu\\\\P_f=0.15\times(10)\;kgm/s\\\\P_f=1.5\;kgm/s[/tex]

Impulse is defined as the change in momentum, that is,

[tex]I=\Delta P\\\\I=P_f-P_i\\\\I=1.5-(-0.3)\\\\I=1.8\;kgm/s[/tex]

The impulse is 1.8 kgm/s towards the east.

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Answer:

[tex]\Delta s = 0.978\,m[/tex]

Explanation:

The pinball-spring system is modelled after the Principle of Energy Conservation:

[tex]U_{g,1} + U_{k,1} + K_{1} = U_{g,2} + U_{k,2} + K_{2}[/tex]

[tex]-(0.5\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right) \cdot \Delta h + \frac{1}{2}\cdot \left(120\,\frac{N}{m}\right)\cdot (-0.2\,m)^{2} = 0[/tex]

The height reached by the pinball is:

[tex]\Delta h = 0.489\,m[/tex]

The distance travelled by the pinball is:

[tex]\Delta s =\frac{0.489\,m}{\sin 30^{\circ}}[/tex]

[tex]\Delta s = 0.978\,m[/tex]

Answer:

Explanation:

the angular frequency ω    of the pendulum is given by the formula

ω  = [tex]\sqrt{\frac{k}{m} }[/tex]    , k is spring constant , m is mass attached .

= [tex]\sqrt{\frac{9}{.3} }[/tex]

= 5.48 rad /s

time period = 2π / ω

= 2 x 3.14 / 5.48

= 1.146 s

b )  formula for speed

v = ω[tex]\sqrt{(a^2-\ x^2)}[/tex]    , a is amplitude , x is displacement from equilibrium point.

for maximum speed x = 0

max speed = ωa

= 5.48 x 3.8 x 10⁻²  ( initial displacement becomes amplitude that is 3.8 cm )

= .208 m /s

20.8 cm / s

c )

when x = .02 m , velocity = ?

v = ω[tex]\sqrt{(a^2-\ x^2)}[/tex]  

=  5.48 [tex]\sqrt{(.038^2-\ .02^2)}[/tex]

= 5.48 x .0323109

= .177 m /s

17.7 cm /s .

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

Answer:

The correct option is C

Explanation:

From the question we are told that

  The initial speed of the rocket is  [tex]v_i = 100 m/s[/tex]

    The speed of the rocket engine sound is [tex]V[/tex]

    The final speed of the rocket is  [tex]v_f = 200 \ m/s[/tex]

The speed of the sound at [tex]v_f[/tex] would still remain V this because the speed of sound wave is constant and is not dependent on the speed of the observer(The mountain ) or the speed of the source (The rocket ).

  A clear example when  lightning strikes you will first see (that is because it travels at  the speed of light which is greater than the speed of sound) but it would take some time before you hear the sound of the   lightning

Here we see that the speed of the lightning(speed of sound) does not affect the speed of the sound it generates  

Answer:

The angle of incidence is  [tex]\theta_i =42.6^o[/tex]

Explanation:

The diagram for this question is shown on the first uploaded image

From the question we are told that

   The distance between the mirror and the wall is  [tex]a = 33.7 cm[/tex]

    The height of the above the mirror is  [tex]b = 36.7 cm[/tex]

Generally the angle which the reflected ray make with the mirror is mathematically evaluated as

             [tex]\alpha =tan ^{-1} (\frac{b}{a})[/tex]

substituting values

             [tex]\alpha = tan ^{-1}( \frac{36.7}{33.7})[/tex]

            [tex]\alpha =47.4^o[/tex]

From the diagram we can deduce that  the angle of incidence is

             [tex]\theta_i = 90 - \alpha[/tex]

So          [tex]\theta_i = 90 - 47.4[/tex]

              [tex]\theta_i =42.6^o[/tex]

Answer:

It is somewhere in between

Explanation:

Wave length of sound from each of the  speakers = 340 / 1700 = 0.2 m = 20 cm

Distance between first speaker and the given point = 4 m.

Distance between second speaker and the given sound

D = √(4² + 2²)

D = √(16 + 4)

D = √20

D = 4.472 m

Path difference = 4.472 - 4 =

0.4722 m.

Path difference / wave length = 0.4772 / 0.2 = 2.386

This is a fractional integer which is neither an odd nor an  even multiple of half wavelength. Hence this point is of neither a perfect constructive nor a perfect destructive interference.

Final answer:

To maximize the power output of a circuit with a 5V battery and three resistors, the resistors should be placed in parallel because this arrangement supplies more current than a series combination, thereby maximizing power.

Explanation:

To maximize the power output for a circuit with a given voltage source, the resistors should be arranged to draw the highest current possible. For resistors connected in parallel, the total resistance of the circuit decreases, as opposed to when they are in series, where it increases. Since power (P) is calculated using the formula P = V^2 / R, where V is the voltage and R is the resistance, minimizing R will maximize P.

Because a parallel configuration guarantees that each resistor gets the full voltage of the battery, the correct answer is: The student should place all resistors in parallel because the battery supplies more current to a parallel combination than to a series combination. This is because in a parallel circuit, the voltage across each resistor is equal to the voltage of the source, which maximizes the current through each resistor due to Ohm's law (I = V/R), and therefore maximizes the total current in the circuit.

Answer option (a) is thus correct: The student should place all resistors in parallel because the battery supplies more current to a parallel combination than to a series combination.

Answer:

the  magnitude of the dipole moment is 3.5*10^-11Cm

Explanation:

The dipole moment is given by the following formula:

[tex]\mu=qr[/tex]

r: distance between the centers of the charges = 500mm

q: charges of the bowling balls = 0.70nC

By replacing you obtain:

[tex]\mu=(0.70*10^{-9}C)(500*10^{-4}m)=3.5*10^{-11}Cm[/tex]

hence, the magnitude of the dipole moment is 3.5*10^-11Cm

Answer:

Magnitude of magnetic field is 1.5 x 10^(-8) T in the positive y-direction

Explanation:

From maxwell's equations;

B = E/v

Where;

B is maximum magnitude of magnetic field

E is maximum electric field

v is speed of light which has a constant value of 3 x 10^(8) m/s

We are given, E = 4.5 V/m

Thus; B = 4.5/(3 x 10^(8))

B = 1.5 x 10^(-8) T

Now, for Electric field, vector E to be in the positive x-direction, the product of vector E and vector B will have to be in the positive z-direction when vector B is in the positive y-direction

Thus,

Magnitude of magnetic field is 1.5 x 10^(-8) T in the positive y-direction

Magnitude of magnetic field in the space at given instant in time is [tex]\bold { 1.5 x 10^{-8}\ T}[/tex]  in the positive y-direction.

From Maxwell's equations,

[tex]\bold {B = \dfrac Ev}[/tex]

Where;

B - maximum magnitude of magnetic field = ?

E- maximum electric field  = 4.5 V/m

v- speed of light =  3 x 10^(8) m/s

Put the values in the formula,

[tex]\bold {B = \dfrac {4.5}{3 x 10^8}}\\\\\bold {B = 1.5 x 10^{-8}\ T}[/tex]

When Electric field, is in the positive x-direction, vector B is in the positive y-direction and  the product of vector E and vector B will have to be in the positive z-direction.

Therefore, magnitude of magnetic field in the space at given instant in time is [tex]\bold { 1.5 x 10^{-8}\ T}[/tex]  in the positive y-direction.

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Answer:

Explanation:

moment of inertia of the disc I = 1/2 m r²

= .5 x 8 x .25²

= .25 kg m²

The wok done by force will be converted into rotational kinetic energy

F x d = 1/2 I ω²

F is force applied , d is displacement , I is moment of inertia of disc and ω

is angular velocity of disc

41 x .9 = 1/2 x .25 ω²

ω² = .25

ω = 17.18  rad / s

The angular speed should be 17.18  rad / s

Since

moment of inertia of the disc I = 1/2 m r²

= .5 x 8 x .25²

= .25 kg m²

Now the work done by force should be converted into the rotational kinetic energy

F x d = 1/2 I ω²

here,

F is the force applied,

d is displacement,

I is moment of inertia of disc

and ω is angular velocity of disc

So,

41 x .9 = 1/2 x .25 ω²

ω² = .25

ω = 17.18  rad / s

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Answer:

a .

15.68

m/s

(b).  

21.72

m

Explanation:

To determine if the arrow will hit the target, we can analyze the projectile motion. By calculating the time of flight and horizontal displacement, we can determine if the arrow will hit the target. The velocity and direction of the arrow at impact can also be determined using the horizontal and vertical components of velocity.

To determine if the arrow will hit the target, we need to analyze the projectile motion of the arrow. First, we need to calculate the time of flight using the equation t = 2 * (v * sin(theta))/g, where v is the initial velocity of the arrow and theta is the launch angle. Next, we can calculate the horizontal displacement using the equation x = v * cos(theta) * t, where x is the distance traveled horizontally. Comparing the calculated horizontal displacement with the distance to the target, we can determine if the arrow will hit the target.

Given the initial velocity of the arrow (230 km/hr), launch angle (3.5 degrees), and distance to the target (50 m), we can convert the initial velocity to m/s (230 km/hr * (1000 m/km) * (1 hr/3600 s)) and use the formulas to calculate time of flight and horizontal displacement. If the horizontal displacement is less than or equal to the distance to the target, the arrow will hit the target.

To find the velocity and direction of the arrow at impact, we can use the equations v_x = v * cos(theta) and v_y = v * sin(theta), where v_x is the horizontal component of velocity and v_y is the vertical component of velocity. Since the angle of impact is not specified, the direction of the arrow at impact will depend on the particular situation.

Answer:

0.0549 m

Explanation:

Given that

equation y(x,t)=Acos(kx−ωt)

speed  v = 8.5 m/s

amplitude A = 5.5*10^−2 m

wavelength λ   = 0.5 m

transverse displacement = ?

v = angular frequency / wave number

and

wave number = 2π/ λ

wave number =  2 * 3.142 / 0.5

wave number = 12.568

angular frequency = v k

angular frequency = 8.5 * 12.568

angular frequency = 106.828 rad/sec ~= 107 rad/sec

so

equation y(x,t)=Acos(kx−ωt)

y(x,t)= 5.5*10^−2 cos(12.568 x−107t)

when x =0 and and t = 0

maximum y(x,t)= 5.5*10^−2 cos(12.568 (0) − 107 (0))

maximum y(x,t)= 5.5*10^−2  m

and when x =  x = 1.52 m and t = 0.150 s

y(x,t)= 5.5*10^−2 cos(12.568 (1.52) −107(0.150) )

y(x,t)= 5.5*10^−2 × (0.9986)

y(x,t) = 0.0549 m

so the transverse displacement is  0.0549 m

Answer:

an orange fringe at 0°, yellow fringes at ±50° and red fringes farther out.

Explanation:

In the visible spectrum- red to violet, red has the highest wavelength.

The maximum internsity  for diffraction grating is given by,

Sinθ[tex](_{max} )[/tex] = mλ/d

It is concluded that the angle of diffraction increases with increase in wavelength'λ' .  So, red fringe will be farthest from the center, orange light will be at the center and yellow fringe will be at 50°.

Therefore, The new pattern consists of : an orange fringe at 0°, yellow fringes at ±50° and red fringes farther out.

Answer:

volume flow rate:2.68m³/s

diameter :58.27cm

Explanation:

flow rate = Q = πr² v  = amount per second that flows through the pipe

Given:

pipe's diameter= 0.581

r= 0.581/2=>0.2905m

speed 'v'= 10.1 m/s

Q= (3.142)(0.2905)²(10.1)

so

volume flow rate= 2.68m³/s

->if no oil has been added or subtracted or compressed then Q is the same everywhere

therefore,

Q= πr²v

2.68 = π(d/2)²(5.85)

d= (2.68x4) /(3.142 x 5.85)

d= 0.5827m =>58.27cm

First: CT Scan

Second: Fluoroscopy

Explanation:

Both correct

A CT scan can provide detailed images of a tumor and its surrounding area, which can be beneficial for a cancer patient. For ongoing throat issues, a Barium swallow study can be used, where the patient swallows a barium solution that is then visualized with an X-ray to identify abnormalities.

In the first scenario, the patient with an ongoing history of cancer might benefit most from a Computed Tomography (CT) scan. CT scans are capable of creating detailed pictures of organs, bones, and other tissues, making it an excellent tool for capturing the size and position of a tumor and surrounding blockages in the abdominal area. Furthermore, it can reveal whether the cancer has spread to other parts of the body.

In the second scenario, Joe's doctor chooses to use X-ray imaging to detect any abnormalities in his throat. The specific procedure used for this is known as a Barium swallow study. This involves swallowing a barium solution that coats the esophagus, enabling the X-ray to capture clear images of the region as the patient swallows.

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Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

The average distance calculated between a person's ears is 0.2 meters. The interaural time difference (ITD) is maximal when the sound source is directly to the side. For air (vs=338 m/s), Δ[tex]t_{max}[/tex] is 591.72 microseconds, and for seawater (vs= 1534 m/s), it is 130.38 microseconds. These calculations illustrate sound localization and the impact of medium on sound propagation.

Understanding Sound Localization and Interaural Time Difference

When a noise occurs directly to the right of a person, the sound waves reach the right ear before the left ear. We can calculate the maximum interaural time difference ( Δ[tex]t_{max}[/tex]) for sound reaching the ears using the given distance between the ears.

(a) The average distance between a person's ears was estimated as 0.2 meters (20 cm).

(b) To calculate Δ[tex]t_{max}[/tex] with the speed of sound in air (vs = 338 m/s), we can use the formula Δ[tex]t_{max}[/tex] = d / vs, where d is the distance between ears. Substituting the values, we get:

Δ[tex]t_{max}[/tex] = 0.2 m / 338 m/s = 0.0005917159763 seconds, or approximately 591.72 microseconds.

(c) Lastly, for sound traveling in seawater at room temperature where vs = 1534 m/s, we similarly get:

Δ[tex]t_{max}[/tex] = 0.2 m / 1534 m/s = 0.0001303763441 seconds, or approximately 130.38 microseconds.

This demonstrates the role of the medium in sound propagation and how it affects the interaural time difference.

Answer:

a) v = 786.93 m/s

b) v = 122.40 m/s

Explanation:

a) To find the average exhaust speed (v) of the engine we can use the following equation:

[tex] F = \frac{v\Delta m}{\Delta t} [/tex]

Where:

F: is the thrust by the engine = 5.26 N

Δm: is the mass of the fuel = 12.7 g

Δt: is the time of the burning of fuel = 1.90 s

[tex]v = \frac{F*\Delta t}{\Delta m} = \frac{5.26 N*1.90 s}{12.7 \cdot 10^{-3} kg} = 786.93 m/s[/tex]

b) To calculate the final velocity of the rocket we need to find the acceleration.

The acceleration (a) can be calculated as follows:

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

In the above equation, m is an average between the mass of the engine plus the rocket case mass and the mass of the engine plus the rocket case minus the fuel mass:

[tex]m = \frac{(m_{engine} + m_{rocket}) + (m_{engine} + m_{rocket} - m_{fuel})}{2} = \frac{2*m_{engine} + 2*m_{rocket} - m_{fuel}}{2} = \frac{2*25.0 g + 2*63.0 g - 12.7 g}{2} = 81.65 g[/tex]

Now, the acceleration is:

[tex] a = \frac{5.26 N}{81.65 \cdot 10^{-3} kg} = 64.42 m*s^{-2} [/tex]

Finally, the final velocity of the rocket can be calculated using the following kinematic equation:

[tex]v_{f} = v_{0} + at = 0 + 64.42 m*s^{-2}*1.90 s = 122.40 m/s[/tex]

I hope it helps you!

Final answer:

The average exhaust speed of the engine is calculated to be approximately 786.929 m/s, and the magnitude of the final velocity of the rocket if fired in outer space is roughly 94.916 m/s.

Explanation:

To calculate the average exhaust speed (v_ex), we can use the impulse-momentum theorem, which states that impulse is equal to the change in momentum of the system. Impulse is given by the product of the average force exerted by the engine (thrust) and the time interval during which the thrust is applied. If all the fuel is consumed, the change in mass (Δm) is the mass of the fuel.

Impulse = Thrust × Time = 5.26 N × 1.90 s = 9.994 N·s

The momentum change is equal to the mass of the fuel expelled times the average exhaust speed.

Δ(momentum) = Δm × v_ex

Substituting the impulse and solving for v_ex, we get:

v_ex = Impulse / Δm

v_ex = 9.994 N·s / 0.0127 kg = 786.929 m/s

Part B: Final Velocity of the Rocket in Space

The final velocity (V_final) of the rocket in space can be determined using the rocket equation, also known as Tsiolkovsky's rocket equation:

V_final = v_ex × ln(m_initial / m_final)

Where:

Calculating the final velocity:

V_final = 786.929 m/s × ln(0.088 kg / 0.0753 kg) ≈ 94.916 m/s

Answer:Identify

Explanation:

Answer:

b

Explanation:

[tex]w + h = 28[/tex]

[tex]l \times w \times h = v[/tex]

[tex]v = w \times h \times 24[/tex]

[tex]w = 28 - h[/tex]

[tex]h(28 - h) \times 24 = v[/tex]

[tex]24( - {h}^{2} + 28h) = v[/tex]

Answer:

the signs of heat and work are; -Q and -W

Explanation:

The first law of thermodynamics is given by; ΔU = Q − W

where;

ΔU is the change in internal energy of a system,

Q is the net heat transfer (the sum of all heat transfer into and out of the system)

W is the net work done (the sum of all work done on or by the system).

Now, The system in this case is the tire and since the air gets warmer, heat must have left the system. Therefore Q is negative (-Q).

Since work is done by the system, W remains negative.

Thus, the signs of heat and work are; -Q and - W

Answer:

The answer is A and C .

Explanation:

Reflection of object is reflected through the Normal (mirror) .

*incident angle = refracted angle

Answer:[tex]r_i=0.016119\ m\approx 16.119\ mm[/tex]

Explanation:

Given

mass of first particle is [tex]m_1=3.12\times 10^{-3}\ kg[/tex]

mass of second particle is [tex]m_2=7.1\times 10^{-3}\ kg[/tex]

Charge on both the particle [tex]q=8.8\times 10^{-6}\ C[/tex]

Now final speed of first particle is [tex]v_1=131\ m/s[/tex]

Final separation between particles is [tex]r=0.15\ m[/tex]

As there is no external force therefore linear momentum is conserved

[tex]0+0=m_1v_1+m_2v_2[/tex]

[tex]0=3.12\times 10^{-3}\times 131+7.1\times 10^{-3}\times v_2[/tex]

[tex]v_2=-\dfrac{3.12\times 10^{-3}}{7.1\times 10^{-3}}\times 131[/tex]

[tex]v_2=-57.56\ m/s[/tex]

Conserving total energy

Initial Kinetic energy +Initial  Potential energy=Final Kinetic energy +Final Potential energy

[tex]\Rightarrow 0+\frac{kq^2}{r_i}=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2+\frac{kq^2}{r_f}[/tex]

[tex]\Rightarrow \frac{9\times 10^9\times 8.8^2\times 10^{-12}}{r_i}=\frac{1}{2}\times 3.12\times 10^{-3}\times 131^2+\frac{1}{2}7.1\times 10^{-3}\times (57.56)^2+\frac{9\times 10^9\times 8.8^2\times 10^{-12}}{0.15}[/tex]

[tex]\Rightarrow \frac{0.696}{r_i}=26.771+11.76+4.646[/tex]

[tex]\Rightarrow \frac{0.696}{r_i}=43.177[/tex]

[tex]r_i=0.016119\ m\approx 16.119\ mm[/tex]

Using the law of conservation of linear momentum and conservation of total energy, the obtained initial separation between the particles is 0.01612 m.

Given that the masses of the two particles are;

[tex]m_1 = 3.12\times 10^{-3}\,kg\\m_2 = 7.1\times 10^{-3}\,kg[/tex]

Also, the charges of both the particles are equal.

[tex]q_1 = q_2 = +8.8\times 10^{-6} \,C[/tex]

The final separation between the particles is;

[tex]r_f = 0.15\,m[/tex]

Also, the final speed of the first particle is;

[tex](v_1)_{final} = 131\,m/s[/tex]

There is no external force applied here; so by the law of conservation of linear momentum, we have;

[tex](m_1v_1)_{initial} +(m_2v_2)_{initial}=(m_1v_1)_{final} +(m_2v_2)_{final}[/tex]

But initially, the particles are at rest, so the initial velocities are zero.

[tex]0+0=(3.12\times 10^{-3}\,kg \, \times 131\,m/s ) +(7.1\times 10^{-3}\,kg\,)\,(v_2)_{final}\\[/tex]

[tex]\implies (v_2)_{final}=-\frac{3.12\times 10^{-3}\,kg \, \times 131\,m/s}{7.1\times 10^{-3}\,kg} =-57.57\,m/s[/tex]

Now, applying the law of conservation of total energy, we get;

[tex](KE)_{\,initial}\, + \,(PE)_{\,initial} = (KE)_{\,final}\, + \,(PE)_{\,final}[/tex]

But initially, the particles are at rest; so they have no initial kinetic energy.

They have electrostatic potential alone initially.

[tex]0+k\frac{q^2}{r_i} = \frac{1}{2}(m_1 v_1)_{initial} + \frac{1}{2}(m_2 v_2)_{final} + k\frac{q^2}{r_f}[/tex]

Substituting the known values, we get;

[tex](9\times 10^9 \, Nm^2/C^2)\frac{(8.8\times 10^{-6} \,C)^2}{r_i} =[ \frac{1}{2}(3.12\times 10^{-3}\,kg )\times (131\,m/s)^2] + [\frac{1}{2}(7.1\times 10^{-3}\,kg) \times (-57.57\,m/s)^2 + (9\times 10^9 \, Nm^2/C^2)\frac{(8.8\times 10^{-6} \,C)^2}{r_f}[/tex]

[tex]\implies\frac{0.696\,Nm^2}{r_i} =26.77\,J+11.76\,J+ 4.64\,J[/tex]

[tex]\implies r_i =\frac{0.696\,Nm^2}{43.17\,J}=0.01612\,m[/tex]

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Answer:

T

Explanation:

Answer:

true

Explanation:

Answer:

Explanation:

Given a particle of mass

M = 1.7 × 10^-3 kg

Given a potential as a function of x

U(x) = -17 J Cos[x/0.35 m]

U(x) = -17 Cos(x/0.35)

Angular frequency at x = 0

Let find the force at x = 0

F = dU/dx

F = -17 × -Sin(x/0.35) / 0.35

F = 48.57 Sin(x/0.35)

At x = 0

Sin(0) =0

Then,

F = 0 N

So, from hooke's law

F = -kx

Then,

0 = -kx

This shows that k = 0

Then, angular frequency can be calculated using

ω = √(k/m)

So, since k = 0 at x = 0

Then,

ω = √0/m

ω = √0

ω = 0 rad/s

So, the angular frequency is 0 rad/s

Answer:

Explanation:

This is a problem based on interference pf light waves.

wavelength of light λ = 553 nm

slit separation d = .134 x 10⁻³ m

screen distance D = 4.5 m

for fourth order bright fringe, path- length  difference = 4 x λ

= 4 x 553

= 2212 nm .

= 2.212 μm

Answer:

Explanation:

λ

given λ = 700 nm

a) for first maxima, d*sin(θ) =  λ

sin(theta) = λ/d

= 700*10^-9/(0.025*10^-3)

= 0.028

theta θ = sin^-1(0.028)

= 1.60 degrees

b) given R = 1 m,

delta_y = λ*R/d

= 700*10^-9*1/(0.025*10^-3)

= 0.028 m or 2.8 cm

c) for first maxima, d*sin(θ) = λ

sin(θ) = λ/d

= 700*10^-9/(2.5*10^-3)

= 0.00028

theta = sin^-1(0.00028)

= 0.0160 degrees

d) R = 25 mm = 0.025 m

δ_y = λ*R/d

= 700*10^-9*0.025/(2.5*10^-3)

= 7*10^-6 m or 7 micro m

e) No. But the position of maxima and minima will be shifted.

Answer:[tex]d=7.94\times 10^5\ m[/tex]

Explanation:

Given

Speed of Primary wave [tex]v_1=8\ km/s[/tex]

Speed of secondary wave [tex]v_2=4.5\ km/s[/tex]

difference in timing of two waves are [tex]77.2\ s[/tex]

Suppose both travel a distance of d km then

[tex]t_1=\frac{d}{8}\quad \ldots (i)[/tex]

[tex]t_2=\frac{d}{4.5}\quad \ldots (ii)[/tex]

Subtract (ii) from (i)

[tex]\frac{d}{4.5}-\frac{d}{8}=77.2[/tex]

[tex]d[\frac{1}{4.5}-\frac{1}{8}]=77.2[/tex]

[tex]d[0.0972]=77.2[/tex]

[tex]d=794.23\ km[/tex]

[tex]d=7.94\times 10^5\ m[/tex]

Final answer:

To calculate the distance to the earthquake's epicenter, we use the time difference between the arrival of P-waves and S-waves and their speeds. By setting up an equation and solving for the distance, the seismograph is found to be approximately 618,940 meters from the epicenter.

Explanation:

When an earthquake occurs, two types of waves are generated: P-waves (primary waves) and S-waves (secondary waves), each with distinctive speeds. To determine the distance to the epicenter of the earthquake, we use the formula d = v × t, where d is distance, v is velocity, and t is time. Given that the P-wave has a speed of 8.0 km/s and the S-wave has a speed of 4.5 km/s, and the time difference of arrival between the two waves is 77.2 seconds, we can calculate the distance from the seismograph to the earthquake's epicenter.

Let the distance be d, then:

To find the distance d, we solve the equation:

Now, to convert kilometers to meters:

Therefore, the seismograph is approximately 618,940 meters from the earthquake's epicenter.