Logan observes a paramecium under a microscope. The eyepiece of the microscope has a horizontal scale marked in mm. The paramecium starts at the 65 mm mark and ends up at the 39 mm mark.
What is the paramecium's displacement? Express your answer with appropriate units.

The wave speed and wavelength will differ across strings of different thicknesses held at the same tension; the wave frequency remains the same.

When two strings of different thicknesses are made of the same material and held at the same tension, the wave speed and wavelength will differ, whereas the wave frequency will remain the same. This is because the speed of a wave on a string is determined by the tension (T) and the linear mass density (μ), where v = (T/μ)¹/2; as the tension is constant and the material is the same, it is the difference in thickness (hence, different densities) that causes a variance in wave speed. Since wave frequency (f) is related to the speed (v) and wavelength (λ) by the equation v = fλ, and the oscillator creates waves at a fixed frequency, a change in wave speed inherently impacts the wavelength.

Answer:

Vector C = -1.56i^ +1.07j^

This question requires that we use the properties of the scalar product of two vectors to find the required x and y component.

The dot product of two perpendicular vectors is equal to and the product of two vectors that are not parallel is equal to a nonzero value.

These are the properties that have been used in solving this problem alongside solving the simultaneous questions generator.

Explanation:

The full solution can be found in the attachment below.

Thank you for reading and I hope this post is helpful to you.

Answer:

m=57.65 kg

Explanation:

Given Data

Ricardo mass m₁=80 kg

Canoe mass m₂=30 kg

Canoe Length L= 3 m

Canoe moves x=40 cm

When Canoe was at rest the net total torque is zero.

Let the center of mass is at x distance from the canoe center and it will be towards the Ricardo cause. So the toque around the center of mass is given as

[tex]m_{1}(L/2-x)=m_{2}x+m_{2}(L/2-x)[/tex]

We have to find m₂.To find the value of m₂ first we need figure out the value of.As they changed their positions the center of mass moved to other side by distance 2x.

so

2x=40

x=40/2

x=20 cm

Substitute in the above equation we get

[tex]m_{x}=\frac{m_{1}(L/2-x)-m_{2}x }{L/2+x}\\m_{x}=\frac{80(\frac{3}{2}-0.2 )-30*0.2}{3/2+0.2}\\m_{x}=57.65 kg[/tex]

Answer:

(a) [tex]Q = 7.28\times 10^{14}[/tex]

(b) The charge inside the shell is placed at the center of the sphere and negatively charged.

Explanation:

Gauss’ Law can be used to determine the system.

[tex] \int{\vec{E}} \, d\vec{a} = \frac{Q_{enc}}{\epsilon_0}\\

E4\pi r^2 = \frac{Q_{enc}}{\epsilon_0}\\

(860)4\pi(0.77)^2 = \frac{Q_{enc}}{8.8\times 10^{-12}}\\

Q_enc = 7.28\times 10^{14}[/tex]

This is the net charge inside the sphere which causes the Electric field at the surface of the shell. Since the E-field is constant over the shell, then this charge is at the center and negatively charged because the E-field is radially inward.

The negative charge at the center attracts the same amount of positive charge at the surface of the shell.

Answer:

A) The total distance traveled when t = 8.3 s is 234 ft.

B) The average velocity of the particle is 4.1 ft/s.

C) The average speed at t = 8.3 s is 28 ft/s.

D) The instantaneous velocity at t = 8.3 s is 92 ft/s.

E) The acceleration of the particle at t = 8.3 s is 37.8 ft/s²

Explanation:

Hi there!

A) The position of the particle at a time "t", in feet, is given by the function "s":

s(t) = t³ - 6t² - 15t + 7

First, let´s find at which time the particle changes direction. The sign of the instantaneous velocity indicates the direction of the particle. We will consider the right direction as positive. The origin of the frame of reference is located at s = 0 and t = 0 so that the particle at t = 0 is located 7 ft to the right of the origin.

The instaneous velocity (v(t)) of the particle is the first derivative of s(t):

v(t) = ds/dt = 3t² - 12t - 15

The sign of v(t) indicates the direction of the particle. Notice that at t = 0,

v(0) = -15. So, initially, the particle is moving to the left.

So let´s find at which time v(t) is greater than zero:

v(t)>0

3t² - 12t - 15>0

Solving the quadratic equation with the quadratic formula:

For  every t > 5 s, v(t) > 0 (the other solution of the quadratic equation is -1. It is discarded because the time can´t be negative).

Then, the particle moves to the left until t = 5 s and, thereafter, it moves to the right.

To find the traveled distance at t= 8.3 s, we have to find how much distance the particle traveled to the left and how much distance it traveled to the right.

So, let´s find the position of the particle at t = 0, at t = 5 and at t = 8.3 s

s(t) = t³ - 6t² - 15t + 7

s(0) = 7 ft

s(5) = 5³ - 6 · 5² - 15 · 5 + 7 = -93 ft

s(8.3) = 8.3³ - 6 · 8.3² - 15 · 8.3 + 7 = 40.9 ft

So from t = 0 to t = 5, the particle traveled (93 + 7) 100 ft to the left, then from t = 5 to t = 8.3 the particle traveled (93 + 40.9) 134 ft to the right. Then, the total distance traveled when t = 8.3 s is (134 ft + 100 ft) 234 ft.

B) The average velocity (AV) is calculated as the displacement over time:

AV = Δs / Δt

Where:

Δs = displacement (final position - initial position).

Δt = elapsed time.

In this case:

final position = s(8.3) = 40.9 ft

initial position = s(0) = 7 ft

Δt = 8.3 s

So:

AV = (s(8.3) - s(0)) / 8.3 s

AV = (40.9 ft - 7 ft) / 8.3 s

AV = 4.1 ft/s

The average velocity of the particle is 4.1 ft/s (since it is positive, it is directed to the right).

C) The average speed is calculated as the traveled distance over time. The traveled distance at t = 8.3 s was already obtained in part A: 234 ft. Then, the average speed (as) will be:

as = distance / time

as = 234 ft / 8.3 s

as = 28 ft/s

The average speed at t = 8.3 s is 28 ft/s

D) The instantaneous velocity at any time t was obtained in part A:

v(t) = 3t² - 12t - 15

at t = 8.3 s

v(8.3) = 3(8.3)² - 12(8.3) - 15

v(8.3) = 92 ft/s

The instantaneous velocity at t = 8.3 s is 92 ft/s.

E) The particle acceleration at any time t, is obtained by derivating the velocity function:

v(t) = 3t² - 12t - 15

dv/dt = a(t) =  6t - 12

Then at t = 8.3 s

a(8.3) = 6(8.3) - 12

a(8.3) = 37.8 ft/s²

The acceleration of the particle at t = 8.3 s is 37.8 ft/s²

Given the missing information, we can only calculate the particle's instantaneous velocity and acceleration at 8.3 s, which can be obtained by taking derivatives of the position function.

This is a physics problem involving kinematics and calculus. Let's address each part of the question:

A. To find the total distance, we would need to know the initial position of the particle. Without this information, we cannot accurately calculate the total distance.

B. Average velocity is defined as the displacement divided by the time interval, which is irrelevant in this case because we do not know the displacement.

C. Similar to B, without displacement information, we cannot calculate average speed.

D. Instantaneous velocity is given by the first derivative of the position function. By taking the derivative of s, we can get the velocity function: v(t) = 3t^2 - 12t - 15. Plug in t = 8.3 s, we can then get the instantaneous velocity.

E. Acceleration is given by the first derivative of the velocity function or the second derivative of the position function. With our velocity function, v(t), we can take its derivative to find a(t) = 6t - 12. Plugging in t = 8.3 s will give the particle's acceleration at that time.

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To solve this problem we will start from the definition of energy of a spring mass system based on the simple harmonic movement. Using the relationship of equality and balance between both systems we will find the relationship of the amplitudes in terms of angular velocities. Using the equivalent expressions of angular velocity we will find the final ratio. This is,

The energy of the system having mass m is,

[tex]E_m = \frac{1}{2} m\omega_1^2A_1^2[/tex]

The energy of the system having mass 2m is,

[tex]E_{2m} = \frac{1}{2} (2m)\omega_1^2A_1^2[/tex]

For the two expressions mentioned above remember that the variables mean

m = mass

[tex]\omega =[/tex]Angular velocity

A = Amplitude

The energies of the two system are same then,

[tex]E_m = E_{2m}[/tex]

[tex]\frac{1}{2} m\omega_1^2A_1^2=\frac{1}{2} (2m)\omega_1^2A_1^2[/tex]

[tex]\frac{A_1^2}{A_2^2} = \frac{2\omega_2^2}{\omega_1^2}[/tex]

Remember that

[tex]k = m\omega^2 \rightarrow \omega^2 = k/m[/tex]

Replacing this value we have then

[tex]\frac{A_1}{A_2} = \sqrt{\frac{2(k/m_2)}{(k/m_1)^2}}[/tex]

[tex]\frac{A_1}{A_2} = \sqrt{2} \sqrt{\frac{m_1}{m_1}}[/tex]

But the value of the mass was previously given, then

[tex]\frac{A_1}{A_2} = \sqrt{2} \sqrt{\frac{m}{2m}}[/tex]

[tex]\frac{A_1}{A_2} = \sqrt{2} \sqrt{\frac{1}{2}}[/tex]

[tex]\frac{A_1}{A_2} = 1[/tex]

Therefore the ratio of the oscillation amplitudes it is the same.

Answer:

D=1.54489 m

Explanation:

Given data

S=6.10 mm= 0.0061 m

To find

Depth of lake

Solution

To find the depth of lake first we need to find the initial time ball takes to hit the water.To get the value of time use below equation

[tex]S=v_{1}t+(1/2)gt^{2} \\ 0.0061m=(0m/s)t+(1/2)(9.8m/s^{2} )t^{2}\\ t^{2}=\frac{0.0061m}{4.9m/s^{2} }\\ t=\sqrt{1.245*10^{-3} }\\ t=0.035s[/tex]

So ball takes 0.035sec to hit the water

As we have found time Now we need to find the final velocity of ball when it enters the lake.So final velocity is given as

[tex]v_{f}=v_{i}+gt\\v_{f}=0+(9.8m/s^{2} )(0.035s)\\ v_{f}=0.346m/s[/tex]

Since there are (4.50-0.035) seconds left for (ball) it to reach the bottom of the lake

So the depth of lake given as:

[tex]D=|vt|\\D=|0.346m/s*4.465s|\\D=1.54489m[/tex]

Answer: d = 1.54m

The depth of the lake is 1.54m

Explanation:

The final velocity of the ball just before it hit the water can be derived using the equation below;

v^2 = u^2 + 2as ......1

Where ;

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance travelled.

Since the initial velocity is zero, and the acceleration is due to gravity, the equation becomes:

v^2 = 2gs

v = √2gs ......2

g = 9.8m/s^2

s = 6.10mm = 0.0061m

substituting into equation 2

v = √(2 × 9.8× 0.0061)

v = 0.346m/s

The time taken for the ball to hit water from the time of release can be given as:

d = ut + 0.5gt^2

Since u = 0

d = 0.5gt^2

Making t the subject of formula.

t = √(2d/g)

t = √( 2×0.0061/9.8)

t = 0.035s

The time taken for the ball to reach the bottom of the lake from the when it hits water is:

t2 = 4.5s - 0.035s = 4.465s

And since the ball falls for 4.465s to the bottom of the lake at the same velocity as v = 0.346m/s. The depth of the lake can be calculated as;

depth d = velocity × time = 0.346m/s × 4.465s

d = 1.54m

The depth of the lake is 1.54m

Final answer:

The horizontal plank on a pivot supported by a spring exhibits simple harmonic motion with angular frequency ω = √(3k/m). When the mass is 5.00 kg and the spring constant is 100 N/m, the frequency is approximately 1.95 Hz.

Explanation:

A horizontal plank of mass m and length L is pivoted at one end and supported by a spring of force constant k at the other. This setup is displaced by a small angle θ from its horizontal equilibrium position and released.

Part A: Derivation of Simple Harmonic Motion

To show the plank moves with simple harmonic motion (SHM) and to find the angular frequency ω, we analyze the forces acting on the plank when displaced. The restoring force F exerted by the spring is F = -kx, where x is the linear displacement of the spring. For small angles, θ, x ≈ Lθ. Thus, F ≈ -kLθ. Applying Newton's second law for rotational motion, τ = Iα, where τ is the torque, I is the moment of inertia of the plank, and α is the angular acceleration. The torque caused by the spring is τ = -kL²θ, and the moment of inertia of the plank about the pivot is I = (1/3)mL². From τ = Iα, we get α = -3k/mθ, indicating SHM with an angular frequency ω = √(3k/m).

Part B: Calculating the Frequency

Given m = 5.00 kg and k = 100 N/m, ω = √(3k/m) = √(3*100/5) = √60. Hence, the frequency f is f = ω/(2π) = √60/(2π) ≈ 1.95 Hz.

Final answer:

Yes, the Earth does gain momentum in the opposite direction due to the conservation of momentum principle when everyone jumps, but the effect is negligible given Earth's massive inertial mass. So the correct option is b.

Explanation:

The question posed is whether the Earth gains momentum in the opposite direction when the entire population of the world jumps and everyone is in the air. According to the conservation of momentum, the answer is yes, but the change in the Earth's momentum is incredibly small to the point of being imperceptible. This is because the inertial mass of Earth is so large compared to the combined mass of all the people that the result of this collective jump would be negligible when it comes to the Earth's momentum.

If we consider a closed system that includes both the Earth and the people jumping, then the total change of momentum for the system must be zero. When people jump, they exert a force on the Earth, and Earth exerts an equal and opposite force on them—this is Newton's third law. However, because of the Earth's substantially greater mass, it experiences an inconsequentially small acceleration in response to this force. While the Earth does indeed recoil much like when a force is applied through a goalpost from a football player hitting it, the Earth's recoil is immeasurably small.

Answer:

ω = 2.55 rad/sec

Explanation:

Assuming no other external forces acting in the horizontal plane, the only force keeping the  rider in a circular path of a radius equal to his distance to the center of rotation, is the centripetal force.

According to Newton's 2nd law, in the horizontal direction, we have:

F = Fc = m*a = m*ω²*r

We know that a = ac = 10*g = 98.0 m/s², and that r = 15.0 m.

Replacing these values in (1), and solving for ω, we get:

ω = √98.0m/s²/ 15.0 m = 2.55 rad/sec

The centrifugal force in a rocket.

The centrifugal force is the force that is related to the outwards or away from the body. The larger force is used to expose the aspiring astronomers to accelerations that are the same as those experienced by the rockets that are launched and the air reentries.

Thus the answer is  ω = 2.55 rad/sec

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Answer:

Air, water, rock and soil

Explanation:

The environment where the general properties of wave motion can be observed is the wave medium which is any substance or particle that carries the wave, or through which the wave travels.

To solve this problem we will apply the concept related to kinetic energy based on the ideal gas constant and temperature. From there and with the given values we will find the temperature of the system. As the temperature is the same it will be possible to apply the root mean square speed formula that is dependent on the element's molar mass, the ideal gas constant and the temperature, this would be:

[tex]KE = \frac{3}{2} RT[/tex]

Where,

KE = Average kinetic energy of an ideal gas

[tex]R = 8.314JK^{-1}mol^{-1}[/tex]= Ideal gas constant

T = Temperature

Replacing we have,

[tex]KE = \frac{3}{2} RT[/tex]

[tex]5930J/mol = \frac{3}{2}(8.314JK^{-1}mol^{-1})T[/tex]

[tex]T = 475.503K[/tex]

Therefore the temperature is 475.5K

RMS velocity of [tex]F_2[/tex] gas is

[tex]v_{rms} = \sqrt{\frac{3RT}{M}}[/tex]

Where,

M = Molar mass of [tex]F_2[/tex]

[tex]M = 38.00g/mol[/tex]

[tex]M = 38.00*10^{-3} kg/mol[/tex]

[tex]T = 475.5K[/tex]

[tex]R = 8.314JK^{-1}mol^{-1}[/tex]

Replacing we have,

[tex]v_{rms} = \sqrt{\frac{3RT}{M}}[/tex]

[tex]v_{rms} = \sqrt{\frac{3(8.314JK^{-1}mol^{-1})(475.5K )}{38.00*10^{-3} kg/mol}}[/tex]

[tex]v_{rms} = 558.662m/s[/tex]

Therefore, the RMS velocity of [tex]F_2[/tex] gas is 558.6m/s

To solve this problem we will start by calculating the distance traveled while relating the first acceleration in the given time. From that acceleration we will calculate its final speed with which we will calculate the distance traveled in the second segment. With this speed and the acceleration given, we will proceed to calculate the last leg of its route.

Expression for the first distance is

[tex]s_1 = ut +\frac{1}{2} at^2[/tex]

[tex]s_1 = 0+\frac{1}{2} (3.8)(6)^2[/tex]

[tex]s_1 = 68.4m[/tex]

The expression for the final speed is

[tex]v = v_0 +at[/tex]

[tex]v = 0+(3.8)(6)[/tex]

[tex]v = 22.8m/s[/tex]

Then the distance becomes as follows

[tex]s_2 = vt[/tex]

[tex]s_2 = (22.8)(1.6)[/tex]

[tex]s_2 = 36.48m[/tex]

The expression for the distance at last sop is

[tex]v_1^2=v_0^2 +2as_3[/tex]

[tex]22.8^2 = 0+2(3.3)s_3[/tex]

[tex]s_3 =78.7636m[/tex]

Therefore the required distance between the signs is,

[tex]S = s_1+s_2+s_3[/tex]

[tex]S = 68.4+36.48+78.76[/tex]

[tex]S = 183.64m[/tex]

Therefore the total distance between signs is 183.54m

Answer:

d = 0.247 mm

Explanation:

given,

λ = 633 nm

distance from the hole to the screen = L = 4 m

width of the central maximum = 2.5 cm

                                             2 y = 0.025 m

                                               y = 0.0125 m

For circular aperture

  [tex]sin \theta = 1.22\dfrac{\lambda}{d}[/tex]

using small angle approximation

  [tex]\theta = \dfrac{y}{D}[/tex]

now,

   [tex]\dfrac{y}{D} = 1.22\dfrac{\lambda}{d}[/tex]

   [tex]y = 1.22\dfrac{\lambda\ D}{d}[/tex]

   [tex]d = 1.22\dfrac{\lambda\ D}{y}[/tex]

   [tex]d = 1.22\dfrac{633\times 10^{-9}\times 4}{0.0125}[/tex]

         d =0.247 x 10⁻³ m

         d = 0.247 mm

the diameter of the hole is equal to 0.247 mm

The diameter of the hole is approximately 0.0201 mm.

The width of the central maximum in a diffraction pattern can be determined using the formula:

w = (2 * λ * D) / x

Where w is the width of the central maximum, λ is the wavelength of the light, D is the distance between the aperture and the screen, and x is the diameter of the hole. Rearranging the formula, we can solve for x:

x = (2 * λ * D) / w

Plugging in the given values, we get:

x = (2 * 633 * 10^-9 * 4.0) / 0.025

x ≈ 0.0201 mm

To solve this problem we will apply the concept of frequency in a string from the nodes, the tension, the linear density and the length of the string, that is,

[tex]f = \frac{n}{2L}(\sqrt{\frac{T}{\mu}})[/tex]

Here

n = Number of node

T = Tension

[tex]\mu[/tex] = Linear density

L = Length

Replacing the values in the frequency and value of n is one for fundamental overtone

[tex]f = \frac{n}{2L}(\sqrt{\frac{T}{\mu}})[/tex]

[tex]f = \frac{1}{2(62*10^{-2})}(\sqrt{\frac{500}{(\frac{3.7*10^{-3}}{90*10^{-2}})}})[/tex]

[tex]\mathbf{f = 281.2Hz}[/tex]

Similarly plug in 2 for n for first overtone and determine the value of frequency

[tex]f = \frac{n}{2L}(\sqrt{\frac{T}{\mu}})[/tex]

[tex]f = \frac{2}{2(62*10^{-2})}(\sqrt{\frac{500}{(\frac{3.7*10^{-3}}{90*10^{-2}})}})[/tex]

[tex]\mathbf{f = 562.4Hz}[/tex]

Similarly plug in 3 for n for first overtone and determine the value of frequency

[tex]f = \frac{n}{2L}(\sqrt{\frac{T}{\mu}})[/tex]

[tex]f = \frac{3}{2(62*10^{-2})}\bigg (\sqrt{\frac{500}{(\frac{3.7*10^{-3}}{90*10^{-2}})}} \bigg)[/tex]

[tex]\mathbf{f= 843.7Hz}[/tex]

Answer:

A. The wavelength doubles but the wave speed is unchanged

Explanation:

The relationship between the period and wavelength is direct. Doubling the period of the oscillator will correspondingly double the wavelength but the wave speed is unaffected

An oscillator creates periodic waves on a stretched string. If the period of the oscillator doubles, then the wavelength doubles but the wave speed is unchanged. So option A is correct here.

When the period of the oscillator doubles, it means that the time it takes for one complete oscillation or cycle of the wave doubles. The period of a wave is inversely proportional to its frequency. If the period doubles, the frequency is halved. The wavelength of a wave is the distance between two consecutive crests or troughs. The wavelength of a wave is inversely proportional to its frequency. When the frequency is halved, the wavelength doubles to maintain the relationship.

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Answer:4A

Explanation:

Given

Mass is displace x= A units from its mean position x=0'

When it is set to free it will oscillate about its mean position with maximum amplitude A i.e. from x=-A to x=A

One cycle is completed when block returns to its original position

so first block will go equilibrium position x=0 and then to x=-A

from x=-A it again moves back to x=0 and finally back to its starting position x=A

so it travels a distance of A+A+A+A=4A    

Answer: A) 141 m

Explanation:

Given that the boat travels at a speed of 1m/s due east in a river that flows 1m/s due south.

Let north represent positive y axis and east represent positive x axis.

Then we can resolve the resultant velocity of the boat to vector form.

Vr = i - j ( 1 m/s on x axis and -1m/s on y axis)

The time required to travel 100m from west to east at a speed of 1m/s is;

Time t = distance/speed = 100m/1m/s = 100s

Since the boat will use 100s to cross the river, We can now determine the resultant distance after 100s:

Distance = velocity × time = (i - j) × 100 = 100i - 100j

Distance = 100i - 100j (in vector form)

Magnitude of the Resultant distance can be given as:

dr = √(dx^2 + dy^2)

dr = √(100^2 + 100^2)

dr = √(20000)

dr = 141.42m

dr = 141m

A) The boat's overall displacement relative to its starting point is 141 m.

To solve this problem, we can break it down into two components: the magnitudes of the boat's eastward displacement and southward displacement. The time it takes for the boat to cross the river can be calculated using the width of the river and the boat's eastward speed. The distance the boat drifts downstream during this time can be calculated using the river's southward speed and the time taken. By combining these two displacements, we can determine the boat's overall displacement relative to its starting point.

The eastward displacement of the boat can be found using the formula: eastward displacement = eastward speed x time.

Plugging in the given values, we get: eastward displacement = 1 m/s x (100 m / 1 m/s) = 100 m.

The southward displacement of the boat can be found using the formula: southward displacement = southward speed x time.

Plugging in the given values, we get: southward displacement = 1 m/s x (100 m / 1 m/s) = 100 m.

Therefore, the boat's overall displacement relative to its starting point, which is the combination of the eastward and southward displacements, is equal to the square root of (eastward displacement squared + southward displacement squared).

Plugging in the calculated values, we get overall displacement = sqrt((100 m)^2 + (100 m)^2) = sqrt(2) x 100 m = 141 m.

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Answer: Electron affinity of F equals

275.8kJ/mol

Explanation: Electron affinity is the energy change when an atom gains an electron.

Let's first calculate the energy required -E(r)to dissociate KF into ions not neutral atom which is given.

E(r) = {z1*z2*e²}/{4π*permitivity of space*r}

z1 is -1 for flourine

z2 is +1 for potassium

e is magnitude of charge 1.602*EXP{-9}C

r is ionic bond length of KF(is a constant for KF 0.217nm)

permitivity of free space 8.854*EXP{-12}.

Now let's solve

E(r)= {(-1)*(1)*(1.602*EXP{-9})²} /

{4*3.142*8.854*EXP(-12)*0.217*EXP(-9)

E(r) = - 1.063*EXP{-18}J

But the energy is released out that is exothermic so we find - E(r)

Which is +1.603*EXP{-18}J

Let's now convert this into kJ/mol

By multiplying by Avogadro constant 6.022*EXP(23) for the mole and diving by 1000 for the kilo

So we have,

1.603*EXP(-18) *6.022*EXP(23)/1000

-E(r) = 640.2kJ/mol.

Now let's obtain our electron affinity for F

We use this equation

Energy of dissociation (nuetral atom)= electron affinity of F +(-E(r)) + ionization energy of K.

498kJ/mol

=e affinity of F + 640.2kJ/mol

+(-418kJ/mol)

(Notice the negative sign in ionization energy for K. since it ionize by losing an electron)

Making electron affinity of F subject of formula we have

Electron affinity (F)=498+418-640.2

=275.8kJ/mol.

Bohr's model of atom postulated that the electrons revolves around the nucleus only in those orbits which have fixed energy and do not lose energy while revolving in them.

According to Bohr's model, the energy at infinite distance is taken to be zero and as it approaches the atom, it starts becoming more negative.

The [tex]n^{th}[/tex] shell of electrons is calculated by

[tex]n^2 = \frac{kZ^2}{E_n}[/tex]

Here

E_n = Energy at [tex]n^{th}[/tex] level

k = Constant

n = Number of shell

Z = Atomic number of the element

Replacing we have that

[tex]n^2 = \frac{-(2.179*10^{-18}J)(1)^2}{-8.72*10^{-20}J}[/tex]

[tex]n = 24.98[/tex]

[tex]n \approx 25[/tex]

Thus

[tex]n = \pm 5[/tex]

Since number of shell cannot be negative we have that n = 5

Now the distance of electron from nucleus is given according to relation

[tex]r = (0.529)(n^2)[/tex]

[tex]r = (0.529)(5^2)[/tex]

[tex]r = 13.225 \AA[/tex]

Therefore the distance of electron from nucleus is 13.225A

The electron in a hydrogen atom is approximately 2.116 angstroms away from the nucleus.

To determine the distance of an electron from the nucleus in a hydrogen atom, we can use the equation for the energy of an electron in a hydrogen atom: E = -13.6eV / n^2, where n is the principal quantum number. We can convert the energy from joules to electron volts (eV) by using the conversion factor 1eV = 1.6x10^-19J. Substituting the given energy, we have:

-8.72x10^-20J = -13.6eV / n^2.

By rearranging the equation and solving for n, we find that n≈2. Thus, the electron is in the second energy level. The distance from the nucleus in angstroms (Å) can be calculated using the formula r = 0.529n^2Å, where r is the distance from the nucleus. Substituting n = 2, we get:

r = 0.529 x 2^2 = 2.116Å.

Therefore, the electron is approximately 2.116 angstroms away from the nucleus.

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Answer:

Answer:

The reason why the B-2 "stealth" bomber has no vertical fins (or stabilizers) at all, is because by design, the vertical fins are not required to provide stability; the stability for yaw movement is provided through a computer that controls the B-2 stealth bomber; hence the B-2 stealth bomber does not need a vertical stabilizer in order to fly.

To solve this problem we will proceed to find the density from the specific gravity. Later we will find the specific volume as the inverse of the density. Finally with the data obtained we will find the total weight in the bottle.

a) [tex]\rho = \gamma * 1000[/tex]

Here,

[tex]\rho[/tex] = Density

[tex]\gamma[/tex] = Specific gravity

[tex]\rho = 1.1 * 1000[/tex]

[tex]\rho = 1100 kg/m3[/tex]

b)

[tex]\text{Specific volume}= \frac{1}{\rho}[/tex]

[tex]\upsilon = \frac{1}{1100}[/tex]

[tex]\upsilon = 0.00090909 m^3/kg[/tex]

From the equivalences of meters to feet and kilograms to pounds, we have to

[tex]1m = 3.280839895 ft[/tex]

[tex]1 kg = 2.2046 lbm[/tex]

Converting the previous value to British units:

[tex]\upsilon = 0.00090909 m^3/kg (\frac{3.280839895^3 ft^3}{1m^3} )(\frac{1kg}{2.2046 lbm})[/tex]

[tex]\upsilon= 0.0145757 ft^3 / lbm[/tex]

c)

[tex]V = 2*10^{-3} m^3[/tex]

Mass of the liquid in bottle is

[tex]m = V\rho[/tex]

[tex]m= (2*10^{-3} m^3 )(1100kg/m^3)[/tex]

[tex]m = 2.2kg = 2200g[/tex]

Therefore the Total weight

[tex]W= 157 + 2200 = 2357 g[/tex]

Answer:

Tbc = 230.69 N ; Fac = 172.31 N

Explanation:

Sum of forces in y direction:

[tex]T_{BC} * sin (35) = 300*sin (70) + F_{AC}*sin (60) .... Eq 1\\[/tex]

Sum of forces in x direction:

[tex]T_{BC} * cos (35) + F_{AC}*cos (60)= 300*cos (70) .... Eq 1\\[/tex]

Solving Eq 1 and Eq 2 simultaneously:

[tex]T_{BC} = 281.9077862 + \sqrt{3} / 2 * F_{AC}\\\\F_{AC} (1.736868124) = 300*cos (70) - 491.4912266*cos (35)\\\\F_{AC} = - \frac{300}{1.736868124}\\\\F_{AC} = - 172.73 N\\\\T_{BC} = 230.69 N[/tex]

Answer: Tbc = 230.69 N ; Fac = 172.31 N

(a) The tension in the cable at AC is -200.67 N.

(b) The  tension in the cable at BC is 328.99 N.

The sum of the forces in y-direction is calculated as follows;

T(BC)sin(35) = 400 x sin(65) + F(AC) sin(60) --- (1)

T(BC)cos(35) + F(AC) cos(60) = 400 x cos(65)  ---(2)

From equation(1);

[tex]T_{BC} = \frac{400 \times sin(65) \ + \ F_{AC} sin(60)}{sin(35)} \\\\T_{BC} = 632 + 1.51F_{AC}[/tex]

From equation (2);

0.82T(BC) + 0.5F(AC) = 169.1

[tex]0.82(632 + 1.51F_A_C) + 0.5F_A_C= 169.1\\\\518.24 + 1.24F_A_C + 0.5F_A_C = 169.1\\\\F_A_C = \frac{-349.14}{1.74} \\\\F_A_C = -200.67 \ N[/tex]

T(BC) = 632 + 1.51(-200.67)

T(BC) = 328.99 N

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The question is incomplete. This is the complete question: A spring has an unstretched length of 22 cm. A 150 g mass hanging from the spring stretches it to an equilibrium length of 30 cm. Suppose the mass is pulled down to where the spring's length is 38 cm. When it is released, it begins to oscillate. What is the amplitude of the oscillation?

Answer:

The amplitude of the oscillation is 8 cm.

Explanation:

The amplitude of the oscillation, which is the maximum displacement of the stretched spring from equilibrium or rest, can be calculated by subtracting the spring’s length at equilibrium (when being stretched by 150g mass) from the spring’s length when it was pulled down.

Amplitude = A = the spring’s length when it was pulled down before oscillating (i.e., 38cm) — the spring’s length at equilibrium (i.e., 30cm)

Therefore, A = 38cm — 30cm = 8cm.

Answer:

d) F

Explanation:

According to columb's law:

"The magnitude of electrostatic force between two charges is directly proportional to the product of magnitude of two charges and inversly proportional to separation between them."

If q₁ and q₂ are magnitude of two charges, d is distance between them and k is dielectric constant, then force F is given by

                                              [tex]F=\frac{kq_{1}q_{2}}{d^2}[/tex]

According to this force exerted on point charge Q is same as that of 3Q, so force point 3Q charge experience is also F

The magnitude of the electric force of attraction between the iron nucleus and its innermost electron is approximately 2.645 × 10^-3 Newtons.

To calculate the magnitude of the electric force of attraction between an iron nucleus and its innermost electron, we can use Coulomb's law:

Given:

Charge of the iron nucleus, q1 = +26e

Charge of the electron, q2 = -e (where e is the elementary charge, 1.6 × 10^-19 C)

Distance between them, r = 1.5 × 10^-12 m

Convert the charge of the nucleus from elementary charges to coulombs:

q1 = +26e × 1.6 × 10^-19 C/e = +4.16 × 10^-18 C

Calculate the electric force using Coulomb's law:

F = (8.9875 × 10^9 N m^2/C^2) × |(+4.16 × 10^-18 C) × (-1.6 × 10^-19 C)| / (1.5 × 10^-12 m)^2

F ≈ (8.9875 × 10^9) × | -6.656 × 10^-37 | / (2.25 × 10^-24)

F ≈ (8.9875 × 10^9) × 2.947 × 10^-13

F ≈ 2.645 × 10^-3 N

So, the magnitude of the electric force of attraction between the iron nucleus and its innermost electron is approximately 2.645 × 10^-3 Newtons.

Answer:

0.0986 h or 5 minutes 55 seconds.

Explanation:

Speed: This can be defined as the rate of change of distance of a body. The S.I unit of speed is m/s. Speed is a scalar quantity, because it can only be represented by magnitude alone.

Mathematically,

Speed = distance/time.

S = d/t ........................... Equation 1

making t  the subject  of the equation

t = d/S ......................... Equation 2

Form the question,

Time taken for the faster runner to reach the finish line

t₁ = d/S₁................... Equation 3

Where t₁ = time taken for the faster runner to reach the finish line, d = distance, S₁ = speed of the faster runner.

Given: d = 5.0 km, S₁ = 14.7 km/h.

Substituting into equation 3

t₁ = 5/14.7

t₁ = 0.340 h

Also,

t₂ = d/S₂................... Equation 4

Where t₂ = time taken for the slower runner to reached the finished line, d = distance, S₂ = speed of the slower runner.

Given: d = 5 km, S₂ = 11.4 km/h.

Substitute into equation 4,

t₂ = 5/11.4

t₂ = 0.4386 h.

The time the faster runner have to wait at the finish line to see the slower runner cross = t₂ - t₁ = 0.4386-0.340

The time the faster runner have to wait at the finish line to see the slower runner cross = 0.0986 h = 5 mins 55 s.

Answer:

The final speed of the car in the first 10 seconds will be 20 m/s.

Explanation:

Given that,

Initial speed of the car, u = 0

Acceleration of the car, [tex]a=2\ m/s^2[/tex]

Time, t = 10 s

We need to find the speed of the car in the first 10 seconds. It can be calculated using first equation of motion. It is given by :

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

v is the final speed of the car

[tex]v=at[/tex]

[tex]v=2\ m/s^2\times 10\ s[/tex]

v = 20 m/s

So, the final speed of the car in the first 10 seconds will be 20 m/s. Hence, this is the required solution.

Answer:

[tex]x=2L\pm \sqrt{\frac{(-2L)^2-4\times 1\times(-L^2)}{2\times 1} }[/tex]

Explanation:

Given:

Now the third charge must be placed  on the line joining the two charges at a distance where the intensity of electric field is same for both the charges that point will not lie between the two charges because they are opposite in nature.

[tex]E_1=E_2[/tex]

[tex]\frac{1}{4\pi\epsilon_0} \times \frac{q_1}{x^2} =\frac{1}{4\pi\epsilon_0} \times \frac{q_2}{(L+x)^2}[/tex]

[tex]\frac{2.37}{x^2} =\frac{4.74}{L^2+x^2+2xL}[/tex]

[tex]2x^2=L^2+x^2+2xL[/tex]

[tex]x^2-2L.x-L^2=0[/tex]

[tex]x=2L\pm \sqrt{\frac{(-2L)^2-4\times 1\times(-L^2)}{2\times 1} }[/tex]

Answer:b

Explanation:

Given

Force of attraction is F when charges are d distance apart.

Electrostatic force is given by

[tex]F=\frac{kq_1q_2}{d^2}---1[/tex]

where k=constant

[tex]q_1[/tex] and [tex]q_2[/tex] are charges

d=distance between them

In order to double the force i.e. 2F

[tex]2F=\frac{kq_1q_2}{d'^2}----2[/tex]

divide 1 and 2 we get

[tex]\frac{F}{2F}=\frac{d'^2}{d^2}[/tex]

[tex]d'=\frac{d}{\sqrt{2}}[/tex]