The 500-kg cylindrical drum is supported by a metal cable (A-B) and a rigid plate (B-C). If the contact between the drum and all surfaces is frictionless and its radius is 500 mm, compute the elongation of the cable (A-B). The elastic modulus of A-B is 200 GPa. A, B, and C are pinned joints. Note that the cross section of Cable A-B is 5 mm2 .

Answer:

The boat is approaching the dock at a rate of 2.5 ft/s.

Explanation:

Let the rope length be 'l' at any time 't', the distance of boat from dock be 'b' at any time 't'.

Given:

The height of dock above water (h) = 6 feet

Rate of pull of rope or rate of change of rope is, [tex]\frac{dl}{dt}=2\ ft/s[/tex]

As clear from the question, the height is fixed and only the length 'l' and distance 'b' varies with time 't'.

Now, the above situation represents a right angled triangle as shown below.

Using Pythagoras Theorem, we have:

[tex]l^2=h^2+b^2\\\\l^2=6^2+b^2\\\\l^2=36+b^2----------(1)[/tex]

Now, differentiating the above equation with time 't', we get:

[tex]2l\frac{dl}{dt}=0+2b\frac{db}{dt}\\\\l\frac{dl}{dt}=b\frac{db}{dt}\\\\\frac{db}{dt}=\frac{l}{b}\frac{dl}{dt}------(2)[/tex]

Now, the distance 'b' can be calculated using 'l=10 ft' in equation (1). This gives,

[tex]b^2=10^2-36\\\\b=\sqrt{64}=8\ ft[/tex]

Now, substituting all the given values in equation (2) and solve for [tex]\frac{db}{dt}[/tex]. This gives,

[tex]\frac{db}{dt}=\frac{10}{8}\times 2\\\\\frac{db}{dt}=2.5\ ft/s[/tex]

Therefore, the boat is approaching the dock at a rate of 2.5 ft/s.

1 hour = 3600 seconds

2 hrs = 7200 sec

(24 frame/sec) x (72 sec) =

172,800 frames.

If each frame of a motion picture film is 35 mm high, and 24 frames go by in a second, then the number of frames needed to show a two-hour-long movie would be 172800 frames

Finding the product of two or more numbers in mathematics is done by multiplying the numbers. It is one of the fundamental operations in mathematics that we perform on a daily basis.

As given in the problem If each frame of a motion picture film is 35 mm high, and 24 frames go by in a second,

The number of seconds in 2 hours = 2 ×3600

                                                        =7200 seconds

The number of frames goes by in one second =  24 frames

the number of frames in a 2-hour movie = 24×7200

                                                                =172800 frames

Thus, the number of frames needed to show a two-hour-long movie would be 172800 frames

Learn more about multiplication here,

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Answer:

33,458.71 turns

Explanation:

Given: L = 37 cm = 0.37 m, B= 0.50 T, I = 4.4 A, n= number of turn per meter

μ₀ = Permeability of free space = 4 π × 10 ⁻⁷

Solution:

We have B = μ₀ × n × I

⇒ n = B/ (μ₀ × I)

n = 0.50 T / ( 4 π × 10 ⁻⁷ × 4.4 A)

n = 90,428.94 turn/m

No. of turn through 0.37 m long solenoid = 90,428.94 turn/m × 0.37

= 33,458.71 turns

Given Information:  

Diameter of solenoid = d = 1.8 cm = 0.018 m

Length of solenoid = L = 37 cm = 0.37 m

Current = I = 4.4 A  

Magnetic field = B = 0.50 T  

Required Information:  

Number of turns = N = ?  

Answer:  

Number of turns ≈ 33,498 or 33,458

Step-by-step explanation:  

The magnetic field at the center of the solenoid is given by

B = μ₀NI/√ (L²+4r²)

N = B√ (L²+4r²)/μ₀I

Where L is the length and r is the radius of the solenoid, N is the number of turns and B is the magnetic field.

r = d/2 = 0.018/2 = 0.009 m

N = 0.50√ (0.37)²+(4*0.009²)/4πx10⁻⁷*4.4

N ≈ 33,498 Turns

Please note that we can also use a more simplified approximate model for this problem since the length of the solenoid is much greater than the radius of the solenoid

L = 0.37 >> r = 0.009

The approximate model is given by

B = μ₀NI/L

N = BL/μ₀I

N = 0.50*0.37/4πx10⁻⁷*4.4

N ≈ 33,458 Turns

As you can notice the results with the approximate model are very close to the exact model.

The expression 6m(2m + 18) simplifies to 12m² + 108m. Therefore, the equivalent expression using the variable 'm' is 12m² + 108m.

The given expression is 6m(2m + 18). First, we distribute 6m across the terms in the parentheses, resulting in 12m² + 108m. So, an expression that uses the variable 'm' and makes the equation true when the given expression is simplified would be 12m² + 108m. This answer is fully simplified, and the variable 'm' is assumed to be an integer.

The expression 6m(2m + 18) simplifies to 12m² + 108m. Therefore, the equivalent expression using the variable 'm' is 12m² + 108m.

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Answer:

Explanation:

We shall apply conservation of momentum  here because it is a case of inelastic collision

u₁ , u₂ initial velocity of knife and target . v₁ , v₂ be their final velocity.

m₁ u₁ + m₂u₂ = m₁v₁ + m₂v₂

u₂ will be negative as it is coming from opposite direction.

36 x 22.5 - 300 x 2.45 = 22.5 x v₁ + 0

810 - 735 = 22.5 x v₁

v₁ = 3.33 m /s

Answer:

 V = k Q/l  ln [(l +√(l² + x²)) / x]

Explanation:

The electrical potential for a continuous distribution of charges is

      V = k ∫ dq / r

Let's apply this expression to our case, define a linear charge density for the bar

           λ = dq / dy

          dq = λ dy

The distance from a point on the bar to the x-axis is

            r = √ (x² + y²)

Let's replace

          V = K ∫ λ dy /√ (x² + y²)

We integrate

         V = k λ ln (y + √ (x² + y²))

Let's evaluate between y = 0 and y = l

         V = k λ [ ln (l +√(x² + l²) - ln x]

We substitute the linear density

           V = k Q/l  ln [(l +√(l² + x²)) / x]

Final answer:

To calculate the electric potential at a point on the x-axis due to a uniformly charged rod, we use the concept of charge density and integrate the contributions from each infinitesimal charge element along the rod with respect to the position.

Explanation:

The student is asking about the electric potential due to a uniformly charged rod on the x-axis. To find an expression for the electric potential at a point on the x-axis, we can imagine dividing the rod into infinitesimally small segments of charge dq. Given that the charge Q is evenly distributed along the rod of length L, the linear charge density λ is Q/L. The potential dV due to a small element dq at a distance x from the rod is given by Coulomb's law as dV = k * dq / r, where k is Coulomb's constant and r is the distance from the charge element to the point on the x-axis.

To find the total potential V, we integrate this expression from one end of the rod to the other. The distance r varies along the rod, so we integrate with respect to y, the position along the rod. Setting up the integral, we have V = k * ∫_{-L/2}^{L/2} (dq / √((L/2 - y)^2 + x^2)) where the limits of integration account for the rod's position along the y-axis with one end at the origin and λ = Q/L. After substituting dq with λdy, performing the integration, and simplifying, we obtain the electric potential V(x) as a function of position along the x-axis.

Answer:

F = 100.0 N

Explanation:

       [tex]F_{c} = m *\frac{v^{2} }{r} = 2.0 kg *\frac{(5.0m/s)^{2} }{0.5m} =100.0 N[/tex]

100.0N

As the ball moves round the circle, a centripetal acceleration which is directed towards the center of the circle will keep the ball from falling off. This acceleration produces a force called centripetal force (F).

Since this is the only force acting on the ball, then the net force acting on the ball to keep it moving round the circle is the centripetal force.

F = m a         [according to Newton's second law of motion]    --------------(i)

Where;

m = mass of the ball.

a = centripetal acceleration. = [tex]\frac{v^2}{r}[/tex]

v = speed of the ball.

r = radius of the circle.

Substitute a = [tex]\frac{v^2}{r}[/tex] into equation (i) as follows;

F = m x [tex]\frac{v^2}{r}[/tex]      --------------------(ii)

From the question;

m = 2.0kg

v = 5.0m/s

r = diameter / 2     [diameter = 1.0m]

r = 1.0 / 2

r = 0.5m

Substitute these values into equation (ii) as follows;

F = 2.0 x [tex]\frac{5.0^{2} }{0.5}[/tex]

F = 2.0 x [tex]\frac{25.0}{0.5}[/tex]

F = 2.0 x 50.0

F = 100.0N

Therefore, the magnitude of the net force on the ball is 100.0N

Answer:

The magnitude of F₁ is 3.7 times of F₂

Explanation:

Given that,

Time = 10 sec

Speed = 3.0 km/h

Speed of second tugboat = 11 km/h

We need to calculate the speed

[tex]v_{1}=\dfrac{3.0\times10^{3}}{3600}[/tex]

[tex]v_{1}=0.833\ m/s[/tex]

The force F₁is constant acceleration is also a constant.

[tex]F_{1}=ma_{1}[/tex]

We need to calculate the acceleration

Using formula of acceleration

[tex]a_{1}=\dfrac{v}{t}[/tex]

[tex]a_{1}=\dfrac{0.833}{10}[/tex]

[tex]a_{1}=0.083\ m/s^2[/tex]

Similarly,

[tex]F_{2}=ma_{2}[/tex]

For total force,

[tex]F_{3}=F_{2}+F_{1}[/tex]

[tex]ma_{3}=ma_{2}+ma_{1}[/tex]

The speed of second tugboat is

[tex]v=\dfrac{11\times10^{3}}{3600}[/tex]

[tex]v=3.05\ m/s[/tex]

We need to calculate total acceleration

[tex]a_{3}=\dfrac{v}{t}[/tex]

[tex]a_{3}=\dfrac{3.05}{10}[/tex]

[tex]a_{3}=0.305\ m/s^2[/tex]

We need to calculate the acceleration a₂

[tex]0.305=a_{2}+0.083[/tex]

[tex]a_{2}=0.305-0.083[/tex]

[tex]a_{2}=0.222\ m/s^2[/tex]

We need to calculate the factor of F₁ and F₂

Dividing force F₁ by F₂

[tex]\dfrac{F_{1}}{F_{2}}=\dfrac{m\times0.83}{m\times0.22}[/tex]

[tex]\dfrac{F_{1}}{F_{2}}=3.7[/tex]

[tex]F_{1}=3.7F_{2}[/tex]

Hence, The magnitude of F₁ is 3.7 times of F₂

Answer:

U = 12 J.

Explanation:

The potential energy in a spring is given by the following formula

[tex]U = \frac{1}{2}kx^2[/tex]

where k is the spring force constant and x is the displacement from the equilibrium.

If U = 3 J when x = 0.05 m, then k is

[tex]3 = \frac{1}{2}k(0.05)^2\\k = 2400~N/m[/tex]

Using this constant, we can calculate the potential energy at x = 0.10 m:

[tex]U = \frac{1}{2}(2400)(0.1)^2 = 12 ~J[/tex]

The potential energy when the mass is at [tex]\( x = 0.10 \)[/tex] m is 12 J.

The potential energy (U) of a mass-spring system is given by the equation:

[tex]\[ U = \frac{1}{2} k x^2 \][/tex]

Given that the potential energy is 3.0 J when \( x = 0.050 \) m, we can use this information to find the spring constant \( k \). Plugging the values into the equation, we get:

[tex]\[ 3.0 \, \text{J} = \frac{1}{2} k (0.050 \, \text{m})^2 \] \[ k = \frac{2 \times 3.0 \, \text{J}}{(0.050 \, \text{m})^2} \] \[ k = \frac{6.0 \, \text{J}}{0.0025 \, \text{m}^2} \] \[ k = 2400 \, \text{N/m} \][/tex]

Now that we have the spring constant, we can find the potential energy when [tex]\( x = 0.10 \)[/tex]m using the same formula:

[tex]\[ U = \frac{1}{2} k x^2 \] \[ U = \frac{1}{2} (2400 \, \text{N/m}) (0.10 \, \text{m})^2 \] \[ U = \frac{1}{2} (2400 \, \text{N/m}) (0.010 \, \text{m}^2) \] \[ U = 1200 \, \text{N/m} \times 0.010 \, \text{m}^2 \] \[ U = 12 \, \text{J} \][/tex]

Therefore, the potential energy when the mass is at[tex]\( x = 0.10 \)[/tex]m is 12 J."

Given Information:

frequency = 1240 Hz

width = a = 1.13 m

speed of sound = c = 344 m/s

Required Information:

angle = θ = ?

Answer:

θ = 14.18 rad

Explanation:

We can find out the angle relative to the centerline perpendicular to the doorway by using the following relation

sin(θ) = λ/a

Where λ is the wavelength of the sound wave and a is width

λ = c/f

Where c is the speed of the sound and f is the frequency

λ = 344/1240

λ = 0.277

sin(θ) = λ/a

θ =sin⁻¹(λ/a)

θ =sin⁻¹(0.277/1.13)

θ =sin⁻¹(0.277/1.13)

θ = 14.18 rad

Answer:

[tex]6.68\times 10^{15}\ kg[/tex]

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

R = Radius of orbit = 45 km

T = Time period = 1.04 days

Mass of Eros would be given by the following equation

[tex]M=\dfrac{4\pi^2R^3}{GT^2}\\\Rightarrow M=\dfrac{4\pi^2\times (45\times 10^3)^3}{6.67\times 10^{-11}\times (1.04\times 24\times 3600)^2}\\\Rightarrow M=6.68\times 10^{15}\ kg[/tex]

The mass of Eros is [tex]6.68\times 10^{15}\ kg[/tex]

Answer:

The conditions are not given in the question. Here is the complete question.

High-altitude mountain climbers do not eat snow, but always melt it first with a stove. To see why, calculate the energy absorbed from a climber's body under the following conditions. The specific heat of ice is 2100 J/kg⋅C∘, the latent heat of fusion is 333 kJ/kg, the specific heat of water is 4186 J/kg⋅C∘

a) eat 1.0kg of -15 degrees Celsius snow which your body warms to body temperature of 37 degrees Celsius ?

b) melt 1.0kg of -15 degrees Celsius snow using a stove and drink the resulting 1.0kg of water at 2 degrees Celsius, which your body has to warm to 37 degrees Celsius ?

Explanation:

Let's calculate the heat required to convert ice from -15°C to 0°C and then the heat required to convert it into the water.

a)

Heat required to convert -15°C ice to 0°C.

  [tex]ms_{ice}[/tex]ΔT = (1.0)(2.100×10³)(15) = 3.150×[tex]10^{4}[/tex]J

Heat required to convert 1.0 kg ice to water.

    [tex]mL_{ice} = 1[/tex]×[tex]3.33[/tex]×[tex]10^{5}[/tex] = 3.33×[tex]10^{5}[/tex]J

Heat required to convert 1.0 kg water at 0°C to 37°C.

   [tex]ms_{water}[/tex]ΔT = 1×4.186×[tex]10^{3}[/tex]×37 = 1.548×[tex]10^{5}[/tex] J

Total heat required = 3.150×[tex]10^{4}[/tex] + 3.33×[tex]10^{5}[/tex] + 1.548×[tex]10^{5}[/tex]

                              = 5.19×[tex]10^{5}[/tex] J

b)

Heat required to warm 1.0 kg water at 2°C to water at 37°C.

[tex]ms_{water}[/tex]ΔT = 1×4.186×[tex]10^{3}[/tex]×35

                = 1.465×[tex]10^{5}[/tex]J

At the start of meiosis II, the cell exhibits characteristics of a haploid cell preparing for mitosis. It has one set of homologous chromosomes, each with two chromatids, equivalent to a haploid cell in the G₂ phase of interphase.

At the beginning of meiosis II, a cell appears much like a haploid cell preparing to undergo mitosis. After completion of meiosis I, the cell does not duplicate its chromosomes, so at the onset of meiosis II, each dividing cell has only one set of homologous chromosomes, each with two chromatids. This results in half the number of sister chromatids to separate out as a diploid cell undergoing mitosis. In terms of chromosomal content, cells at the start of meiosis II are similar to those of haploid cells in the G₂ phase of interphase, where they are preparing for mitosis.

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Answer:

The angle through which each bicycle wheel rotates is 35.03 rad.

Explanation:

Given;

the radius of the circular path, R = 9.30 m

the angular displacement of the bicycle, θ = 1.130 rad

the radius of the bicycle wheel, r = 0.3

S = θR

where;

s is the distance of the circular path

S = 1.13 x 9.3 = 10.509 m

The angle (in radians) through which each bicycle wheel of radius 0.300 m rotates is given as;

θr = 10.509 m

θ = 10.509 / 0.3

θ = 35.03 rad.

Therefore, the angle through which each bicycle wheel of radius 0.300 m rotates is 35.03 rad.

Answer:

Explanation:

Resonant frequency is 240

4π² x 240² = 1 / LC

230400π² = 1 / LC

Let the required frequency = n

inductive reactance = 2 πn L

capacitative reactance = 1 /  2 π n C

inductive reactance / capacitative reactance

= 4π² x n ² x LC = 5.68

4π² x n ² = 1 / LC x 5.68

= 230400π²  x 5.68

4n ²= 230400 x 5.68

n ²= 57600 x 5.68

n ² = 327168

n = 572 approx

The generator's frequency in this RLC circuit, where the ratio of inductive to capacitive reactance is 5.68 and the resonant frequency is 240 Hz, is approximately 571 Hz.

To solve this problem, we need to use the relationship between inductive and capacitive reactance in an RLC circuit. Given that the ratio of the inductive reactance (XL) to the capacitive reactance (XC) is 5.68,

Reactance is frequency-dependent, with the following formulas for inductive and capacitive reactance:

Given the resonant frequency as 240 Hz, we start by calculating the resonance reactances:

Using the non-resonant frequency, we use the reactance ratio:

Rearranging gives us:

Since 240 Hz is the resonant frequency, substituting f:

Simplifying yields:

Therefore, the generator frequency is approximately 571 Hz.

Answer:

Explanation:....

The range of the projectile is 4675 m. The maximum height reached by the projectile is 2124 m. The speed of the projectile at impact is 234 m/s.

(a) Range is defined as the maximum horizontal distance covered by a projectile during its motion. It is given by the formula:

[tex]R = \frac{u^2 sin2\theta}{g}[/tex],

where u = initial speed of the projectile = 230 m/s

θ = angle of projection = 60°

g = acceleration due to gravity = 9.8 m/s²

Using these in formulas, we get:

[tex]R = \frac{(230 \hspace{0.8mm} m/s)^2 \hspace{0.8mm} sin2(60)}{9.8 \hspace{0.8mm} m/s^2}[/tex]

or, R = 4674.77 m ≈ 4675 m

(b) The maximum height reached by the projectile is given by the formula:

[tex]H = \frac{u^2 \hspace{0.8mm} sin^2\theta}{2g}[/tex]

using the numeric values, we get:

[tex]H = \frac{(230 \hspace{0.8mm}m/s)^2 sin^2(60)}{2 \times (9.8 \hspace{0.8mm} m/s^2)}[/tex]

or, H = 2024.23 m

However, since the projectile was fired 100 m above the ground, hence, the maximum height would be:

h = 100 m + H = 100 m + 2024.23 m = 2124.23 m ≈ 2124 m

(c) The speed of the projectile comprises of two parts, a horizontal velocity and a vertical velocity.

The horizontal velocity is given as [tex]v_x[/tex]. It is constant throught the motion and its magnitude can be determined by the formula:

[tex]v_x = v cos\theta = 230 \hspace{0.8 mm} m/s \times cos(60)[/tex]

or, [tex]v_x = 115 \hspace{0.8mm} m/s[/tex]

The vertical velocity is given as [tex]v_y[/tex]. It is not constant throught the motion. At the highest point, the magnitude of vertical velocity is zero. Using the kinetic equation,

[tex]v_y^2[/tex] = u² + 2gs, where s = h = 2124 m, we get:

[tex]v_y^2[/tex] = 0 + (2 × 9.8 m/s² × 2124 m)

or, [tex]v_y^2[/tex] = 41630.4 m²/s²

or, [tex]v_y[/tex] = 204.03 m/s ≈ 204 m/s

Now, the magnitude of speed will be equal to the resultant of both the horzontal and vertical velocities.

[tex]v = \sqrt{v_x^2 + v_y^2}[/tex]

or, [tex]v = \sqrt{204^2 + 115^2} \hspace{0.8 mm} m/s[/tex]

or, v = 234.21 m/s ≈ 234 m/s

Answer:

0.2289

Explanation:

Power required to climb= Fv where F is force and v is soeed. We know that F= mg hence Power, P= mgv and substituting 700 kg for m, 9.81 for g and 2.5 m/s for v then

P= 700*9.81*2.5=17167.5 W= 17.1675 kW

To express it as a fraction of 75 kw then 17.1675/75=0.2289 or 22.89%

The minimum width b that can be used for the beam is 4.32 in

To determine the minimum width (b) of the solid rectangular wood beam, we employ principles of structural engineering.

The maximum bending moment (M) for a simply supported beam with a uniformly distributed load (\(w_0\)) occurs at the center and is given by [tex]\(M = \frac{w_0 L^2}{8}\).[/tex]

The section modulus (S) for a rectangular cross-section is [tex]\(S = \frac{b \times h^2}{6}\).[/tex]

The bending stress (\(σ_b\)) is given by[tex]\(σ_b = \frac{M}{S}\)[/tex].

Setting \(σ_b\) equal to the allowable bending stress [tex](\(1.95 \, \text{ksi}\))[/tex]and using the specified aspect ratio (h/b = 1.75), we can solve for b.

This ensures that the wood beam meets structural safety criteria.

The calculation yields a minimum width b of 4.32 in, ensuring that the beam is structurally sound, withstanding the specified uniformly distributed load and adhering to the aspect ratio constraint in compliance with the allowable bending stress.

Answer:

Area = 25km² in a century.

Explanation:

Given

Spread Rate = 5m/century

Length of rift = 5000km

Convert to metres

Length of rift = 5000 * 1000m

Length of rift = 5,000,000m

In one century, the additional area to Atlantis = (Rift Length) * (Spread rate) * 1 century

Atea = 5,000,000 m * 5m/century * 1 century

Area = 25,000,000m² in a century

----- Convert to km²

Area = 25,000,000km² * 1m²/1,000,000km²

Area = 25km² in a century.

Hence, the total area of new ocean floor created in the Atlantic each century is 25km²

Final answer:

The total area of new ocean floor created along the mid-Atlantic Rift each century is 5,000,000 m², or 25 km² of new oceanic crust formed every century.

Explanation:

The question asks how much new ocean floor is created in the Atlantic Ocean each century along the mid-Atlantic Rift, which is known to be 5000 km long, as Europe and North America drift apart by about 5 m per century.

To calculate the total area of the new ocean floor created, we consider the length of the rift (5000 km) and the rate of separation (5 m per century).

First, we need to convert kilometers to meters for uniform units:

5000 km * 1000 m/km = 5,000,000 m.

Now we can calculate the area:

Area = Length * Width

Area = 5,000,000 m * 5 m

Area = 25,000,000

So, every century, an area of 25,000,000 m2 of new ocean floor is created along the mid-Atlantic Rift. This is equivalent to 25 km2 of new oceanic crust, since 1 km2 = 1,000,000 m2.

The magnitude of the frictional force between the person and the wall is

[tex]\[ f = \frac{mv^2}{r} \][/tex]

To find the magnitude of the frictional force between the person and the wall, we can use the concept of centripetal force.

First, let's define the variables:

m = mass of the person

v = tangential velocity of the person

R = radius of the circular path (distance from the person to the axis of rotation, in this case, the distance from the person to the wall)

The centripetal force [tex](\( F_c \))[/tex] required to keep the person moving in a circular path is given by:

[tex]\[ F_c = \frac{{mv^2}}{R} \][/tex]

Since the person is held against the wall, the frictional force [tex](\( F_f \))[/tex] provides the centripetal force. Therefore,

[tex]\[ F_f = F_c \][/tex]

[tex]\[ F_f = \frac{{mv^2}}{R} \][/tex]

Therefore, the magnitude of the frictional force between the person and the wall is [tex]\( \frac{{mv^2}}{R} \)[/tex].

Answer:

Upper scale reads 4.4kg while lower scale reads 7.6 kg

Explanation:

The buoyant force of the water on the aluminum piece would equal to the weight of the water displaced by the aluminum.

The mass of the water displaced by aluminum is

[tex]m_w = V_w \rho_w = \frac{m_a}{\rho_a}\rho_w = \frac{7}{2700}1000 = 2.6 kg[/tex]

As 7kg aluminum piece is supported by 2.6 kg buoyant force, the scale that the aluminum is hung on would read

Mass of aluminum - mass of water displaced

7 - 2.6 = 4.4 kg

This buoyant force would also created a reaction on the lower scale of 2.6 kg (according to Newton's 3rd law). So the lower scale would read:

Mass of water + mass of bleak + mass of buoyant reaction force

3 + 2 + 2.6 = 7.6 kg

Answer:

Explanation:

W (Ce) = 2.1 eV

W (Al) = 4.1 eV

W (Be) = 5 eV

W (K) = 2.3 eV

W (Pt) = 6.4 eV

W (Mg) = 3.7 eV

Part A:

Work function is directly proportional to the cut off frequency.

let f denotes the cut off frequency.

So, f (Be) > f (Be) > f (Al) > f (Mg) > f (K) > f (Ce)

Part B:

Maximum wavelength for the emission is inversely proportional to the cut off frequency

So, λ (Ce) > λ (K) > λ (Mg) > λ (Al) > λ (Be) > λ (Pt)

Part C:

E = 3.10 eV

Let K is the maximum kinetic energy

K = E - W

K (Ce) = 3.1 - 2.1 = 1 eV

K (Al) 3.1 - 4.1 = not possible

K (Be) = 3.1 - 5 = not possible

K (K)  = 3.1 - 2.3 = 0.8 eV

K (Pt) = .1 - 6.4 = not possible

K (Mg) = 3.1 - 3.7 = not possible

So, K (Ce) > K (K) > K (Al) = K (Be) = K (Pt) = K (Mg)

Metals can be ranked based on their work functions, which determine cutoff frequency, the maximum wavelength of light needed to free electrons, and maximum kinetic energy of emitted electrons under specific light. Lower work functions result in higher cutoff frequencies and longer required wavelengths, and more energetic emitted electrons when under 400 nm light.

The ranking of the metals, in terms of cutoff frequency, maximum wavelength of light, and maximum kinetic energy, is based on the work function of each metal, which is the minimum energy required to free an electron from the metal's surface. The cutoff frequency is inversely related to the work function, so the metal with the lowest work function (cesium, 2.1 eV) will have the highest cutoff frequency, and the metal with the highest work function (platinum, 6.4 eV) will have the lowest cutoff frequency.

Conversely, the maximum wavelength of light needed to free electrons is directly related to the work function, so cesium will require the longest wavelength and platinum will require the shortest wavelength. When illuminated with 400 nm (3.10 eV) light, the kinetic energy of the emitted electrons will be the difference between the energy of the light and the work function, so cesium (with the smallest work function) will have the most energetic electrons, and metals with a work function larger than 3.10 eV (like aluminum, beryllium, and platinum) will not emit electrons at all.

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Answer:

 ΔV = 2k λ ln (r₁ / r₂)

Explanation:

The electric potential is

         ΔV = - ∫ E. ds

The expression for the electric field of a charge line is given

           E = 2k λ / r

Let's replace and integrate

         ΔV = - 2k λ ∫ dr / r

         ΔV = -2 k λ ln r

Let's evaluate

       ΔV = - 2k λ ln r₂- ln r₁

       ΔV = -2k λ ln (r₂ / r₁)

       ΔV = 2k λ ln (r₁ / r₂)

Explanation:

Given:

m = 1.673 × 10^-27 kg

Q = q = 1.602 × 10^-19 C

r = 0.75 nm

= 0.75 × 10^-9 m

A.

Energy, U = (kQq)/r

Ut = 1/2 mv^2 + 1/2 mv^2

1.673 × 10^-27 × v^2 = (8.99 × 10^9 × (1.602 × 10^-19)^2)/0.75 × 10^-9

v = 1.356 × 10^4 m/s

B.

F = (kQq)/r^2

F = m × a

1.673 × 10^-27 × a = ((8.99 × 10^9 × (1.602 × 10-19)^2)/(0.075 × 10^-9)^2

a = 2.45 × 10^17 m/s^2.

Answer:

φ = -7.16 × 10⁴ Nm²/C

Explanation:

q = -3.80 × 10⁻⁶ C

ε₀ = 8.85 10⁻¹² C²/Nm²

By Gauss's law

φ = q/ε₀

as the cube has 6 faces so for one side the flux will

φ =1/6 ( q/ε₀)

φ = 1/6 ( -3.80 × 10⁻⁶ C / 8.85 10⁻¹² C²/Nm²)

φ = -71,563.088 Nm²/C

φ = -7.16 × 10⁴ Nm²/C

-7.00 x 10⁴Nm²/C

Let the single point charge of -3.80µC be Q

i.e

Q = -3.80µC = -3.80 x 10⁻⁶C

From Gauss's law, the total electric flux,φ, through an enclosed surface is equal to the quotient of the enclosed charge Q, and the permittivity of free space, ε₀. i.e;

φ = Q / ε₀     ----------------------(i)

Where;

ε₀ = known constant = 8.85 x 10⁻¹²F/m

Now, If the total flux through the enclosed cube is given by equation (i), then the flux, φ₁, through one of the six sides of the cube is found by dividing the equation by 6 and is given as follows;

φ₁ = Q / 6ε₀     --------------------(ii)

Substitute the values of  Q and ε₀ into equation (ii) as follows;

φ₁ = -3.80 x 10⁻⁶ / (6 x 8.85 x 10⁻¹²)

φ₁ = -0.07 x 10⁶

φ₁ = -7.00 x 10⁴

Therefore, the electric flux through one side of the cube is -7.00 x 10⁴ Nm²/C

Note:

The result (flux) above is negative to show that the electric field lines from the point charge points radially inwards in all directions.

After 9 seconds, an object dropped on the moon will have fallen 65.61 meters, be traveling at 14.58 meters per second, with an acceleration of 1.62 meters per second squared. The second derivative of the distance function represents the moon's gravitational acceleration.

An object is dropped on the moon from a height of 200 meters, and the distance function of free-fall due to lunar gravity is given by s(t)=0.81t^2. We are asked to determine various properties of the object's motion after 9 seconds.

a) Distance Fallen

To find how far the object has fallen, we plug t=9 into the distance function to get s(9)=0.81(9)^2=65.61 meters. This is the distance the object has fallen, not the distance from its starting height.

b) Velocity

The object's velocity can be determined by finding the derivative of the distance function, which gives us v(t) = 2(0.81)t = 1.62t. Substituting t=9, we find v(9) = 1.62(9) = 14.58 meters per second.

c) Acceleration

The acceleration of the object is constant and equal to twice the coefficient of t^2 in the distance function, which is 1.62 meters per second squared on the moon.

d) Meaning of the Second Derivative

The second derivative of the free-fall distance function represents the acceleration due to gravity on the moon. It's constant at 1.62 meters per second squared, indicating uniform acceleration during free-fall.

Answer:

b. Both stars will have the same shift.

Explanation:

It's a very simple problem to solve. Star 1 is approaching toward Earth with a speed v, so let's assume that the change in Doppler Shift is +F and Star 2 is moving away so the change in Doppler shift is -F. But it's time to notice the speed of both stars and that is same but only directions are different. speed is the main factor here. The magnitude of both shifts is F as we can see and + and - are showing there direction of motion. So, because of same amount of speed, both stars will have same shift magnitude. (Just the directions are different)

Answer:

A) v/ sq. rt. of 2

Explanation:

The speed of the wave in the string is defined as:

[tex]v=\sqrt{\frac{T}{\mu}}[/tex]

Where T is the tension on the string and [tex]\mu[/tex] is the linear density, that is, the mass per unit length:

[tex]\mu=\frac{m}{L}[/tex]

Where m is the mass of the string and L its length. We have [tex]m'=2m[/tex], [tex]T'=T[/tex] and [tex]L'=L[/tex]:

[tex]v'=\sqrt\frac{T'}{m'/L'}\\v'=\sqrt\frac{T}{2m/L}\\v'=\frac{1}{\sqrt2}\sqrt\frac{T}{m/L}\\v'=\frac{v}{\sqrt2}[/tex]

Explanation:

Below is an attachment containing the solution.

Answer:

160 m

Explanation:

The intensity, I, of the sound is inversely proportional to the square of the distance, r, from the source.

[tex]I\propto \dfrac{1}{r^2}[/tex]

Hence,

[tex]I_1r_1^2 = I_2r_2^2[/tex]

[tex]r_2 = r_1\sqrt{\dfrac{I_1}{I_2}}[/tex]

From the question, [tex]I_2[/tex] is half of [tex]I_1[/tex]

[tex]r_2 = r_1\sqrt{\dfrac{I_1}{0.5I_1}}[/tex]

[tex]r_2 = r_1\sqrt{2}[/tex]

[tex]r_2 = 113\text{ m}\sqrt{2} = 160 \text{ m}[/tex]

Answer:

Distance ,d= 159.81m

Explanation:

The intensity, I, of the sound is inversely proportional to the square of the distance, d,from the source.

Using the equation d/do=sqrt2

d= dosqrt2

Where d=113m

d= 113sqrt2

d= 159.81m

Answer:

Check Explanation.

Explanation:

Momentum before collision = (2)(2) + (2)(0) = 4 kgm/s

a) Scenario A

After collision, Mass A sticks to Mass B and they move off with a velocity of 1 m/s

Momentum after collision = (sum of the masses) × (common velocity) = (2+2) × (1) = 4 kgm/s

Which is equal to the momentum before collision, hence, momentum is conserved.

Scenario B

They bounce off of each other and move off in the same direction, mass A moves with a speed of 0.5 m/s and mass B moves with a speed of 1.5 m/s

Momentum after collision = (2)(0.5) + (2)(1.5) = 1 + 3 = 4.0 kgm/s

This is equal to the momentum before collision too, hence, momentum is conserved.

Scenario C

Mass A comes to rest after collision and mass B moves off with a speed of 2 m/s

Momentum after collision = (2)(0) + (2)(2) = 0 + 4 = 4.0 kgm/s

This is equal to the momentum before collision, hence, momentum is conserved.

b) Kinetic energy is normally conserved in a perfectly elastic collision, if the two bodies do not stick together after collision and kinetic energy isn't still conserved, then the collision is termed partially inelastic.

Kinetic energy before collision = (1/2)(2.00)(2.00²) + (1/2)(2)(0²) = 4.00 J.

Scenario A

After collision, Mass A sticks to Mass B and they move off with a velocity of 1 m/s

Kinetic energy after collision = (1/2)(2+2)(1²) = 2.0 J

Kinetic energy lost = (kinetic energy before collision) - (kinetic energy after collision) = 4 - 2 = 2.00 J

Kinetic energy after collision isn't equal to kinetic energy before collision. This collision is evidently totally inelastic.

Scenario B

They bounce off of each other and move off in the same direction, mass A moves with a speed of 0.5 m/s and mass B moves with a speed of 1.5 m/s

Kinetic energy after collision = (1/2)(2)(0.5²) + (1/2)(2)(1.5²) = 0.25 + 3.75 = 4.0 J

Kinetic energy lost = 4 - 4 = 0 J

Kinetic energy after collision is equal to kinetic energy before collision. Hence, this collision is evidently elastic.

Scenario C

Mass A comes to rest after collision and mass B moves off with a speed of 2 m/s

Kinetic energy after collision = (1/2)(2)(0²) + (1/2)(2)(2²) = 4.0 J

Kinetic energy lost = 4 - 4 = 0 J

Kinetic energy after collision is equal to kinetic energy before collision. Hence, this collision is evidently elastic.

c) An impossible outcome of such a collision is that A stocks to B and they both move off together at 1.414 m/s.

In this scenario,

Kinetic energy after collision = (1/2)(2+2)(1.414²) = 4.0 J

This kinetic energy after collision is equal to the kinetic energy before collision and this satisfies the conservation of kinetic energy.

But the collision isn't possible because, the momentum after collision isn't equal to the momentum before collision.

Momentum after collision = (2+2)(1.414) = 5.656 kgm/s

which is not equal to the 4.0 kgm/s obtained before collision.

This is an impossible result because in all types of collision or explosion, the second law explains that first of all, the momentum is always conserved. And this evidently violates the rule. Hence, it is not possible.

Answer:

The magnitude of the electric field inside the membrane is 8.8×10⁶V/m

Explanation:

The electric field due to electric potential at a distance Δs is given by

E=ΔV/Δs

We have to find the magnitude electric field in the membrane

Ecell= -ΔV/Δs

[tex]E_{cell}=-\frac{V_{in}-V_{out}}{s} \\E_{cell}=\frac{V_{out}-V_{in}}{s}\\E_{cell}=\frac{0.070V}{8*10^{-9}m } \\E_{cell}=8.8*10^{6}V/m[/tex]

The magnitude of the electric field inside the membrane is 8.8×10⁶V/m

The magnitude of the electric field inside the membrane, if it were empty, would be approximately [tex]\( 8.75 \times 10^6 \) V/m.[/tex]

To find the magnitude of the electric field inside the membrane, we can use the formula for electric field strength ( E ) due to a uniform field between two parallel plates, which is given by:

[tex]\[ E = \frac{V}{d} \][/tex]

where ( V ) is the potential difference across the plates (in volts) and ( d ) is the separation between the plates (in meters). In this case, the potential difference ( V ) is given as 70 mV, which we need to convert to volts:

[tex]\[ V = 70 \text{ mV} = 70 \times 10^{-3} \text{ V} \][/tex]

The thickness of the membrane ( d ) is given as 8 nm, which we need to convert to meters:

[tex]\[ d = 8 \text{ nm} = 8 \times 10^{-9} \text{ m} \][/tex]

Now we can plug these values into the formula for the electric field:

[tex]\[ E = \frac{70 \times 10^{-3} \text{ V}}{8 \times 10^{-9} \text{ m}} \][/tex]

[tex]\[ E = \frac{70}{8} \times 10^{(3 - (-9))} \text{ V/m} \][/tex]

[tex]\[ E = 8.75 \times 10^{6} \text{ V/m} \][/tex]