Equipotential surfaces are to be drawn 100 V apart near a very large uniformly charged metal plate carrying a surface charge density σ = 0.75 μC/m2. How far apart (in space) are the equipotential surfaces?

Answer:

(a) The energy  extracted from the hot reservoir (Qh) is 3316.67J

(b) The energy deposited in the cold reservoir (Qc) is 2321.67J

Explanation:

Part (a) The energy  extracted from the hot reservoir (Qh)

e = W/Qh

where;

e is the maximum efficiency of the system = 30% = 0.3

W is the the work done on the system = 995 J

Qh is the heat absorbed from the hot reservoir

Qh = W/e

Qh = 995/0.3

Qh = 3316.67J

Part (b) The energy deposited in the cold reservoir (Qc)

e = W/Qh

W = Qh - Qc

where;

Qc is the heat deposited in the cold reservoir

e = (Qh - Qc)/Qh

Qh - Qc = e*Qh

Qc = Qh - e*Qh

Qc = 3316.67J - 0.3*3316.67J

Qc = 3316.67J - 995J

Qc = 2321.67J

Answer:

Explanation:

Given

Gauge Pressure of a tank is

[tex]P_{gauge}=55\ kPa[/tex]

at that place atmospheric Pressure is  [tex]h=72.1\ cm\ of\ Hg[/tex]

Density of mercury [tex]\rho _{Hg}=13600\ kg/m^3[/tex]

Atmospheric Pressure in kPa is given by

[tex]P_{atm}=\rho _{Hg}\times g\times h[/tex]  

[tex]P_{atm}=13600\times 9.8\times 0.721[/tex]

[tex]P_{atm}=96.09\ kPa[/tex]

and Absolute Pressure is summation of gauge pressure and atmospheric Pressure

[tex]P_{abs}=P_{gauge}+P_{atm}[/tex]

[tex]P_{abs}=55+96.09=151.09\ kPa[/tex]                        

Answer:

The correct option is 1.  720 m

Explanation:

Projectile Motion

When an object is launched in free air (no friction) with an initial speed vo at an angle [tex]\theta[/tex], it describes a curve which has two components: one in the horizontal direction and the other in the vertical direction. The data provided gives us the initial conditions of the survival package's launch.

[tex]\displaystyle V_o=250\ m/s[/tex]

[tex]\displaystyle \theta =-37^o[/tex]

The initial velocity has these components in the x and y coordinates respectively:

[tex]\displaystyle V_{ox}=250\ cos(-37^o)=199.7\ m/s[/tex]

[tex]\displaystyle V_{oy}=250\ sin(-37^o)=-150.5\ m/s[/tex]

And we know the plane has an altitude of 600 m, so the package will reach ground level when:

[tex]\displaystyle y=-600\ m[/tex]

The vertical distance traveled is given by:

[tex]\displaystyle y=V_{oy}\ t-\frac{g\ t^2}{2}=-600[/tex]

We'll set up an equation to find the time when the package lands

[tex]\displaystyle -150.5t-4.9\ t^2=-600[/tex]

[tex]\displaystyle -4.9\ t^2-150.5\ t+600=0[/tex]

Solving for t, we find only one positive solution:

[tex]\displaystyle t=3.6\ sec[/tex]

The horizontal distance is:

[tex]\displaystyle x=V_{ox}.t=199.7\times3.6=720\ m[/tex]

The correct option is 1.  720 m

The horizontal distance of the point of impact from the plane at the moment of the package's release is approximately 1760 m.

The time taken for the package to reach the ground can be found using the equation y = v0y * t + (1/2) * g * t2, where y is the initial altitude, v0y is the vertical component of the initial velocity, t is the time taken, and g is the acceleration due to gravity. Solving for t gives us a value of approximately 5 seconds. The horizontal distance traveled by the package can be found using the equation x = v0x * t, where x is the horizontal distance, v0x is the horizontal component of the initial velocity, and t is the time taken. Plugging in the values gives us x = 250 m/s * cos(37º) * 5 s, which simplifies to approximately 1760 m. So the horizontal distance of the point of impact is approximately 1760 m.

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Answer:

The fraction of collision is [tex]4.00\times10^{-15}[/tex]

Explanation:

Given that,

Temperature = 47°C

Activation energy = 88.20 KJ/mol

From Arrhenius equation,

[tex]k=Ae^{-\dfrac{E_{a}}{RT}}[/tex]

Here, [tex]e^{-\dfrac{E_{a}}{RT}}[/tex]=fraction of collision

We need to calculate the fraction of collisions

Using formula of fraction of collisions

[tex]f=e^{-\dfrac{E_{a}}{RT}}[/tex]

Where f = fraction of collision

E = activation energy

R = gas constant

T = temperature

Put the value into the formula

[tex]f=e^{-\dfrac{88.20}{8.314\times10^{-3}\times(47+273)}}[/tex]

[tex]f=4.00\times10^{-15}[/tex]

Hence, The fraction of collision is [tex]4.00\times10^{-15}[/tex]

Answer:

Vab =17.5kV

A = 16.9 cm2

C   = 4.27pF

Explanation:

a) Find the voltage difference:

Vab = Ed

E Electric field

d distance between plates

Vab potential difference

d = 3.5mm

 = 3.5 * 10^(-3) m    

Q = 75.0nC

    = 75 * 10^(-9)  

E = 5.00 * 10^6 V/m

Vab = (5.00 * 10^6) * (3.5 * 10^(-3))

        = 17.5 * 10^3 V

        =17.5kV

b. What is the area of the plate?

 The relation between the electric field and area is given as:

E = Q/(ϵ0 * A)

A = Q/(ϵ0 *E)

Where ϵ0 is the electric constant and equals 8.854 × 10^ (-12) C2/N•m2    

A = 75 * 10^ (-9) / (8.854 × 10^ (-12) (5.00 * 10^6)

 = 1.69 X 10^ (-3) m2

  = 16.9 cm2

c. Find the capacitance

   The equation relating capacitance, area of plate and plate distance is given by:

C = ϵ0 A/d

plug in the values of d, ϵ0 and A above to get the capacitance:

C = (8.854 × 10^ (-12) * 1.69 X 10^ (-3) / 3.5 * 10^ (-3)  

 = 4.27 * 10^ (-12) F

 = 4.27pF

Answer:

The ratio of [tex]E_{app}[/tex] and [tex]E_{act}[/tex] is 0.9754

Explanation:

Given that,

Distance z = 4.50 d

First equation is

[tex]E_{act}=\dfrac{qd}{2\pi\epsilon_{0}\times z^3}[/tex]

[tex]E_{act}=\dfrac{Pz}{2\pi\epsilon_{0}\times (z^2-\dfrac{d^2}{4})^2}[/tex]

Second equation is

[tex]E_{app}=\dfrac{P}{2\pi\epsilon_{0}\times z^3}[/tex]

We need to calculate the ratio of [tex]E_{act}[/tex] and [tex]E_{app}[/tex]

Using formula

[tex]\dfrac{E_{app}}{E_{act}}=\dfrac{\dfrac{P}{2\pi\epsilon_{0}\times z^3}}{\dfrac{Pz}{2\pi\epsilon_{0}\times (z^2-\dfrac{d^2}{4})^2}}[/tex]

[tex]\dfrac{E_{app}}{E_{act}}=\dfrac{(z^2-\dfrac{d^2}{4})^2}{z^3(z)}[/tex]

Put the value into the formula

[tex]\dfrac{E_{app}}{E_{act}}=\dfrac{((4.50d)^2-\dfrac{d^2}{4})^2}{(4.50d)^3\times4.50d}[/tex]

[tex]\dfrac{E_{app}}{E_{act}}=0.9754[/tex]

Hence, The ratio of [tex]E_{app}[/tex] and [tex]E_{act}[/tex] is 0.9754

The student determine the actual speed of the object when it reaches the ground as 12.52 m/s.

Given data:

The distance from the Earth's surface is, = 10 m.

Time taken to reach the ground is, t = 1.43 s.

The speed of object is, v = 14.3 m/s.

Experimental value of time interval is, t' = 3.2 s.

Use kinematic equation of motion to compute true value for acceleration of the ball as it reaches the ground:

[tex]h=ut+\dfrac{1}{2}a't'^{2} \\\\10=0 \times t+\dfrac{1}{2} \times a' \times 3.2^{2} \\\\a'=\dfrac{20}{3.2^{2}}\\\\a'= 1.95 \;\rm m/s^{2}[/tex]

Now, use the principle of conservation of total energy of system:

Potential energy - work done by air resistance = Kinetic energy

[tex]mgh-(ma) \times h=\dfrac{1}{2}mv^{2} \\\\gh-(a) \times h=\dfrac{1}{2}v^{2} \\\\v=\sqrt{2h(g-a)}[/tex]

Here, v is the actual speed of object while reaching the ground.

Solving as,

[tex]v=\sqrt{2 \times 10(9.8-1.95)}\\\\v=12.52 \;\rm m/s[/tex]

Thus, we can conclude that  the student determine the actual speed of the object when it reaches the ground as 12.52 m/s.

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The actual speed of the object when it reaches the ground is approximately [tex]31.392 \, m/s.[/tex]

The actual speed of the object when it reaches the ground can be determined using the kinematic equation that relates initial velocity, acceleration, and time to the final velocity. Since the object is released from rest, the initial velocity [tex]\( u \)[/tex] is 0 m/s, and the acceleration [tex]\( a \)[/tex] is due to gravity, which is approximately [tex]\( 9.81 \, m/s^2 \)[/tex] near the Earth's surface.

The kinematic equation that relates these quantities to the final velocity [tex]\( v \)[/tex] is:

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

Given that [tex]\( u = 0 \, m/s \)[/tex] and [tex]\( a = 9.81 \, m/s^2 \)[/tex], and the time [tex]\( t \)[/tex] to reach the ground is [tex]\( 3.2 \, s \)[/tex], we can substitute these values into the equation to find the actual final velocity:

[tex]\[ v = 0 + (9.81 \, m/s^2)(3.2 \, s) \] \[ v = (9.81)(3.2) \, m/s \] \[ v = 31.392 \, m/s \][/tex]

Therefore, the actual speed of the object when it reaches the ground is approximately [tex]31.392 \, m/s.[/tex]

Answer:

False

Explanation:

Sun mass is dominating in Solar system as compared to other planets, asteroids and comets. Sun itself accounting for the 99.9% of the mass of the solar system. Hence the gravitational force exerted by the Sun dominates the other objects in the solar system. So we can conclude that solar system has non-uniform composition. The given statement is false

Answer:

a) [tex] \theta_2 = 0.05 * \frac{3.5}{0.7} = 0.25[/tex]

b) [tex] \theta_2 = 0.05 * \frac{140}{700} = 0.01[/tex]

Explanation:

We are comparing two wavelengths with the radius and diameter constant, and if we want to compare it, we need to use the following formula:

[tex]\frac{\theta_1}{\theta_2}= \frac{\lambda_1}{\lambda_2}[/tex]

Where [tex] \theta[/tex] represent the angular resolution and [tex]\lambda[/tex] the wavelength.

So if we have a fixed resolution and wavelength 1 and we want to find the resolution for a new condition we can solve for [tex] \theta_2[/tex] and we got

[tex] \theta_2 = \theta_1 \frac{\lambda_2}{\lambda_1}[/tex]

Part a

For this case the subindex 1 is for the color red and we know that:

[tex] \lambda_1 = 700 nm *\frac{1 \mu m}{1000 nm} = 0.7 \mu m[/tex]

And the angular resolution for the color red is specified as [tex] \theta_1 = 0.05[/tex]

And for the infrared case we know that [tex] \lambda_2 = 3.5 \mu m[/tex], so if we replace we got:

[tex] \theta_2 = 0.05 * \frac{3.5}{0.7} = 0.25[/tex]

Part b

For this case the subindex 1 is for the color red and we know that:

[tex] \lambda_1 = 700 nm[/tex]

And the angular resolution for the color red is specified as [tex] \theta_1 = 0.05[/tex]

And for the ultraviolet case we know that [tex] \lambda_2 = 140 nm[/tex], so if we replace we got:

[tex] \theta_2 = 0.05 * \frac{140}{700} = 0.01[/tex]

Answer:

(a) 5.6 m/s, 7.2 m/s, and 8.8 m/s, respectively.

(b) 0.8 m/s^2

(c) 4.8 m/s

(d) 6 s

(e) 5.2 m

Explanation:

(a) The average velocity is equal to the total displacement divided by total time.

For the first 2s. interval:

[tex]V_{\rm avg} = \frac{\Delta x }{\Delta t} = \frac{25.6 - 14.4}{2} = 5.6~{\rm m/s}[/tex]

For the second 2s. interval:

[tex]V_{\rm avg} = \frac{40 - 25.6}{2} = 7.2~{\rm m/s}[/tex]

For the third 2s. interval:

[tex]V_{\rm avg} = \frac{57.6 - 40}{2} = 8.8~{\rm m/s}[/tex]

(b) Every 2 s. the velocity increases 1.6 m/s. Therefore, for each second the velocity increases 0.8 m/s. So, the acceleration is 0.8 m/s2.

(c) The sled starts from rest with an acceleration of 0.8 m/s2.

[tex]v^2 = v_0^2 + 2ax\\v^2 = 0 + 2(0.8)(14.4)\\v = 4.8~{\rm m/s}[/tex]

(d) The following kinematics equation will yield the time:

[tex]\Delta x = v_0 t + \frac{1}{2}at^2\\14.4 = 0 + \frac{1}{2}(0.8)t^2\\t = 6~{\rm s}[/tex]

(e) The same kinematics equation will yield the displacement:

[tex]\Delta x = v_0t + \frac{1}{2}at^2\\\Delta x = (4.8)(1) + \frac{1}{2}(0.8)1^2\\\Delta x = 5.2~{\rm m}[/tex]

Answer:

If the frequency of an electromagnetic wave increases, the number of waves passing by you increase.

Explanation:

The total number  of vibration or oscillation per unit time is called frequency of a waver. It is given by :

[tex]f=\dfrac{n}{t}[/tex]

n is the number of waves passing

t is the time taken

It is clear that the frequency is directly proportional to the number of waves. Hence, if the frequency of an electromagnetic wave increases, the number of waves passing by you increase.

The amount of current that flows through this given portion of a cell membrane, calculated using Ohm's law and the properties of the membrane, is 1.60 µA. If the side dimensions of the membrane are halved, the current will decrease by a factor of 4.

The relevant concept needed to answer these questions is Ohm's Law, defined as Voltage = Current x Resistance. In this context, Resistance = Resistivity x (Thickness/Area) and the area is a square.

First, calculate the resistance: R = ρ x (Thickness/ Area)
Remove the micrometers units of the area and convert it into meters to match the ρ units. So, you get an area of 1.3 x 10^-6 m x 1.3 x 10^-6 m = 1.69 x 10^-12 m^2. Then, R = 1.30 x 10^7 Ω*m x (7.50 x 10^-9 m / 1.69 x 10^-12 m^2) = 57.404 Ω.

By plugging the calculated resistance and given voltage into Ohm's Law, we can find the current: I = V/R = 92.2 x 10^-3 V / 57.4 Ω = 1.60 μA

If the side dimensions are halved, the area of the membrane becomes one-fourth of the original, thus the resistance increases by a factor of 4. According to Ohm's Law, as resistance increases, the current decreases, meaning that if the resistance is multiplied by 4, the current will decrease by a factor of 4.

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a) Therefore, the amount of current that flows through this portion of the membrane is approximately [tex]\({1.60 \times 10^{-6} \, \text{A}} \)[/tex]. b) The correct answer is decrease by a factor of 5208. The current decreases by a factor of approximately 5208.

Part (a): Determine the amount of current that flows through this portion of the membrane

To find the current ( I ) flowing through the membrane portion, we use Ohm's law and the given potential difference ( V ) across the membrane.

1. Calculate the resistance ( R ) of the membrane:

The resistivity [tex](\( \rho \))[/tex] is given as [tex]\( 1.30 \times 10^7 \) ohms\·m.[/tex]

First, calculate the cross-sectional area ( A ) of the membrane portion:

[tex]\[ A = 1.3 \, \mu \text{m} \times 1.3 \, \mu \text{m} = (1.3 \times 10^{-6} \, \text{m})^2 = 1.69 \times 10^{-12} \, \text{m}^2 \][/tex]

Then, calculate the resistance ( R ):

[tex]\[ R = \frac{\rho \cdot L}{A} \][/tex]

[tex]\[ R = \frac{1.30 \times 10^7 \, \text{ohm} \cdot \text{m} \cdot 7.50 \times 10^{-9} \, \text{m}}{1.69 \times 10^{-12} \, \text{m}^2} \][/tex]

[tex]\[ R = \frac{9.75 \times 10^{-2}}{1.69 \times 10^{-12}} \approx 5.77 \times 10^7 \, \text{ohms} \][/tex]

2. Calculate the current ( I ):

Ohm's law states [tex]\( I = \frac{V}{R} \).[/tex]

Given potential difference [tex]\( V = 92.2 \, \text{mV} = 92.2 \times 10^{-3} \, \text{V} \):[/tex]

[tex]\[ I = \frac{92.2 \times 10^{-3} \, \text{V}}{5.77 \times 10^7 \, \text{ohms}} \approx 1.60 \times 10^{-6} \, \text{A} \][/tex]

Part (b): By what factor does the current change if the side dimensions of the membrane portion is halved?

If the side dimensions of the membrane portion are halved, the cross-sectional area ( A ) of the membrane will decrease by a factor of ( 4 ) (since both length and width are halved).

1. New cross-sectional area ( A' ):

[tex]\[ A' = \left( \frac{1.3 \, \mu \text{m}}{2} \right) \times \left( \frac{1.3 \, \mu \text{m}}{2} \right) = \left( \frac{1.3}{2} \times 10^{-6} \, \text{m} \right)^2 = 0.325 \times 10^{-12} \, \text{m}^2 \][/tex]

2. New resistance ( R' ):

Using the same resistivity [tex]\( \rho \)[/tex] and thickness ( L ):

[tex]\[ R' = \frac{\rho \cdot L}{A'} = \frac{1.30 \times 10^7 \cdot 7.50 \times 10^{-9}}{0.325 \times 10^{-12}} \approx 3.00 \times 10^8 \, \text{ohms} \][/tex]

3. New current ( I' ):

[tex]\[ I' = \frac{V}{R'} = \frac{92.2 \times 10^{-3}}{3.00 \times 10^8} \approx 3.07 \times 10^{-10} \, \text{A} \][/tex]

4. Calculate the factor by which the current changes:

[tex]\[ \frac{I'}{I} = \frac{3.07 \times 10^{-10}}{1.60 \times 10^{-6}} \approx 1.92 \times 10^{-4} \][/tex]

Since the current decreases, we consider the reciprocal:

[tex]\[ \frac{I}{I'} \approx \frac{1}{1.92 \times 10^{-4}} \approx 5208 \][/tex]

Answer:

Explanation:

Given

mass of first child [tex]m_1=20\ kg[/tex]

mass of second child [tex]m_2=30\ kg[/tex]

Distance between two children is [tex]d=3\ m[/tex]

Suppose light weight child is placed at a distance of x m from Pivot point

therefore

Torque due to heavy child [tex]T_1=m_2g\times (3-x)[/tex]

Torque due to small child [tex]T_2=m_1g\times x[/tex]

Net Torque about Pivot must be zero

Therefore [tex]T_1=T_2[/tex]

[tex]30\times g\times (3-x)=20\times g\times x[/tex]

[tex]9-3x=2x[/tex]

[tex]9=5x[/tex]

[tex]x=\frac{9}{5}[/tex]

[tex]x=1.8\ m[/tex]

Answer:

The electron began in the quantum level of 4

Explanation:

Using the formula of wave number:

Wave Number = 1/λ = Rh(1/n1² - 1/n2²)

where,

Rh = Rhydberg's Constant = 1.09677 x 10^7 /m

λ = wavelength of light emitted = 486.4 nm = 486.4 x 10^-9 m

n1 = final shell = 2

n2 = initial shell = ?

Therefore,

1/486.4 x 10^-9 m = (1.09677 x 10^7 /m)(1/2² - 1/n2²)

1/4 - 1/(486.4 x 10^-9 m)(1.09677 x 10^7 /m) = 1/n2²

1/n2² = 0.06082

n2² = 16.44

n2 = 4

The principal quantum level in which the electron began is 5.

Given the following data:

Rydberg constant = [tex]1.09 \times 10^7\; m^{-1}[/tex]

To determine the principal quantum level in which the electron began, we would use the Rydberg equation:

Mathematically, the Rydberg equation is given by the formula:

[tex]\frac{1}{\lambda} = R(\frac{1}{n_f^2} -\frac{1}{n_i^2})[/tex]

Where:

Substituting the parameters into the formula, we have;

[tex]\frac{1}{486.4 \times 10^{-9}} = 1.09 \times 10^7(\frac{1}{2^2} -\frac{1}{n_i^2})\\\\\frac{1}{486.4 \times 10^{-9} \times 1.09 \times 10^7}=\frac{1}{4} -\frac{1}{n_i^2}\\\\\frac{1}{5.3018} =\frac{1}{4} -\frac{1}{n_i^2}\\\\\frac{1}{n_i^2}=\frac{1}{4}-\frac{1}{5.3018}\\\\\frac{1}{n_i^2}=0.25-0.1886\\\\\frac{1}{n_i^2}=0.0614\\\\n_i^2=\frac{1}{0.0614} \\\\n_i=\sqrt{16.29} \\\\n_i=4.0[/tex]

Initial transition = 4.0

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Answer:

Explanation:

A block of mass  M is placed on a semicircular track and released from rest at point  P , which is at vertical height H₁ above the track’s lowest point.

Its initial potential energy = mgH₁

Kinetic energy = 0

Total energy = mgH₁

When  block slides to a vertical height  H₂ on the other side of the track

Its final potential energy = mgH₂

Kinetic energy = 0

Total  final energy = mgH₂

As negative work is done by frictional force while block moves ,

final energy < initial energy

mgH₂ < mgH₁

H₂ < H₁

H₂  will be less than H₁ .

The angular displacement of the solid disk after 5 seconds is 0 rad.

To determine the angular displacement of the solid disk after 5 seconds, we can use the formula:

Δθ = ΔErot / (I * ω)

where Δθ is the angular displacement, ΔErot is the change in rotational kinetic energy, I is the moment of inertia of the disk, and ω is the angular velocity of the disk.

The moment of inertia of a solid disk rotating around an axis through its center perpendicular to its plane is given by:

I = (1/2) * m * r2

where m is the mass of the disk and r is the radius.

Given that ΔErot = 20 J/s, m = 4 kg, r = 2 m, and the disk starts from rest, we can calculate the angular displacement:

Δθ = ΔErot / (I * ω) = 20 / [(1/2) * 4 * 22 * ω]

Since the disk starts from rest, the initial angular velocity ω is 0. Therefore, the angular displacement after 5 seconds is:

Δθ = 20 / [(1/2) * 4 * 22 * 0] = 0 rad

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Answer:

Explanation:

False

When two waves overlap or superimpose over each other then they either undergo Constructive or destructive interference.

waves are the disturbance created by a force and add up to gives constructive interference when they are in the same line i.e. in the same phase.

When these disturbances are in the opposite phase then they superimpose to give destructive interference where the amplitude of the resulting wave will be much smaller as compared to original waves.

Relation Between Velocity And Wavelength

Wavelength is the measure of the length of a complete wave cycle. The velocity of a wave is the distance travelled by a point on the wave. In general, for any wave the relation between Velocity and Wavelength is proportionate. It is expressed through the wave velocity formula.

Velocity And Wavelength

For any given wave, the product of wavelength and frequency gives the velocity. It is mathematically given by wave velocity formula written as-

V=f×λ

Where,

V is the velocity of the wave measure using m/s.

f is the frequency of the wave measured using Hz.

λ is the wavelength of the wave measured using m.

Velocity and Wavelength Relationtion

Amplitude, Frequency, wavelength, and velocity are the characteristic of a wave. For a constant frequency, the wavelength is directly proportional to velocity.

Given by:

V∝λ

Example:

For a constant frequency, If the wavelength is doubled. The velocity of the wave will also double.

For a constant frequency, If the wavelength is made four times. The velocity of the wave will also be increased by four times.

Hope you understood the relation between wavelength and velocity of a wave. You may also want to check out these topics given below!

Relation between phase difference and path difference

Relation Between Frequency And Velocity

Relation Between Escape Velocity And Orbital Velocity

Relation Between Group Velocity And Phase Velocity

The relationship between wavelength, wave frequency, and wave velocity is described by the equation v = fλ. Wavelength and frequency are inversely proportional given a constant wave velocity – high frequency correlates with short wavelength and vice versa.

In Physics, there's a mathematical relationship between wavelength, wave frequency, and wave velocity for any type of wave motion. This relationship is often stated as v = fλ, where v is wave velocity, f is the frequency of the wave, and λ is the wavelength. The wavelength is the distance between identical parts of the wave, while the velocity is the speed at which the disturbance moves, and the frequency is the rate of oscillation of the wave.

When you look at this formula, it becomes clear that if the wave velocity (v) is constant, a wave with a longer wavelength (λ) will have lower frequency (f). On the other side, higher frequency means shorter wavelength. This is because frequency and wavelength are inversely proportional in the given formula.

For instance, the speed of light in vacuum is a constant value (approximately 3.00×108 m/s). So, if a certain light wave has a larger wavelength, its frequency will be lower to ensure this speed remains consistent.

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Answer:

11.48 degree N of W

Explanation:

We are given that

Wind velocity=[tex]v_w=-43[/tex]km/h

Because wind is blowing towards south

Air speed=[tex]v_a=-212km/h[/tex]

Because the captain want to move with air speed in west direction.

x component of relative velocity=-212 km/h

y-Component of relative velocity=-43km/h

Direction=[tex]\theta=tan^{-1}(\frac{y}{x})[/tex]

[tex]\theta=tan^{-1}(\frac{-43}{-212}=11.48^{\circ}[/tex]N of W

Hence, the direction in which the pilot should set her course to travel due west=11.48 degree N of W

The Poisson's ratio definition is given as the change in lateral deformation over longitudinal deformation. Mathematically it could be expressed like this,

[tex]\upsilon = \frac{\text{Lateral Strain}}{\text{Longitudinal Strain}}[/tex]

[tex]\upsilon = \frac{(\delta a/a)}{(\delta l/l)}[/tex]

Replacing with our values we would have to,

[tex]\upsilon = \frac{(2.89933-2.9/2.9)}{(9.70710-9.7/97)}[/tex]

[tex]\upsilon = 0.1977[/tex]

Therefore Poisson's ratio is 0.1977.

The strength of an electric field that will balance the weight is 1.023 × 10⁷ N/C.

An electric field is a physical field that surrounds electrically charged particles and acts as an attractor or repellent to all other charged particles in the vicinity. Additionally, it refers to a system of charged particles' physical field.

Electric charges and time-varying electric currents are the building blocks of electric fields.

The strength of an electric field that will balance the plastic sphere is = weight of the object/charge on the object

= (  9.6 ×10⁻³×9.8)/(9.2×10⁻⁹) N/C

= 1.023 × 10⁷ N/C

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Explanation:

A.

Given:

V = 13 m/s

t = 4 s

Constant acceleration, a= (V-Vi)/t

= 13/4

= 3.25 m/s^2

F = mass * acceleration

= 2.7 * 3.25

= 8.775 N.

Using equations of motion,

distance,S = (13 * 4) - (1/2)(3.25)(4^2)

= 26 m

Workdone, W = force * distance

= 8.775 * 26

= 228.15 J

B.

Instantaneous power, P = Force *Velocity

= 8.775 * 13

= 114. 075 W

C.

t = 2 s,

Constant acceleration, a= (V-Vi)/t

= 13/2

= 6.5 m/s^2

Force = mass * acceleration

= 2.7 * 6.5

= 17.55 N

Instantaneous power, P = Force *Velocity

= 17.55 * 13

= 228.15 W.

= 114. 075 W.

Answer:

The force of gravitational attraction increases by 9 as the two point masses increase by 3.

Explanation:

Gravitational force of attraction, F is the force that pulls two point masses, m and M which are separated by a distance, d.

Mathematically,

Fg = GMm/r^2

Initially,

M1 = M1

M2 = M2

The remaining parameters are unchanged.

Fg1 = G * M1 * m1/(d/2)^2

Then,

M1 = 3M1

M2 = 3M2

Fg2 = G * 3M1 * 3M2/(d/2)^2

Making the constants G/(d/2)^2 the subject of formula and then comparing both equations,

= Fg1 = (M1 * M2); Fg2 = (9 * M1 * M2)

= Fg2 = 9 * Fg1

The force of gravitational attraction increases by 9 as the two point masses increase by 3.

Answer:

F_x = 750.7 N

Explanation:

Given:

- Length of the pipe between flanges L = 75 cm

- Weight Flow rate is flow(W) = 230 N/c

- P_1 = 165 KPa

- P_2 = 134 KPa

- P_atm = 101 KPa

- Diameter of pipe D = 0.05 m

Find:

The total force that the flanges must withstand F_x.

Solution:

- Use equation of conservation of momentum.

            (P_1 - P_a)*A + (P_2 - P_a)*A - F_x = flow(m)*( V_2 - V_1)

- From conservation of mass:

                                       A*V_1 = A*V_2

                       V_1 = V_2 ( but opposite in directions)

- Hence,

             (P_1 - P_a)*A + (P_2 - P_a)*A - F_x = - 2*flow(m)*V_1

                                     flow(m) = flow(W) / g

                                     p*A*V_1 = flow(W) / g

                                    V_1 =  flow(W) / g*p*A    

Hence,      

             (P_1 - P_a)*A + (P_2 - P_a)*A - F_x = - 2*flow(W)^2 / g^2*p*A

Hence, compute:

   64*10^3 *pi*0.05^2 /4 + 33*10^3 *pi*0.05^2 /4 - F_x = - 2*(230/9.81)^2 / 997*pi*0.05^2 /4

                            125.6 + 64.7625 - F_x = -560.33

                                        F_x = 750.7 N

Answer:

Explanation:

Given

Speed while running towards east is [tex]v_1=3\ m/s[/tex]

Distance traveled in east direction [tex]x_1=120\ m[/tex]

For Another interval you  run with velocity

[tex]v_2=5\ m/s[/tex]

[tex]x_2=240\ m[/tex]

Total displacement[tex]=x_1+x_2[/tex]

[tex]=120+120=240\ m[/tex]

Time for first interval

[tex]t_1=\frac{x_1}{v_1}=\frac{120}{3}[/tex]

[tex]t_1=\frac{120}{3}=40\ s[/tex]

Time for second interval

[tex]t_2=\frac{x_2}{v_2}=\frac{120}{5}=24\ s[/tex]

total time [tex]t=t_1+t_2[/tex]

[tex]t=40+24=64\ s[/tex]

average velocity [tex]v_{avg}=\frac{x_1+x_2}{t}[/tex]

[tex]v_{avg}=\frac{240}{64}=3.75\ m/s[/tex]

Therefore average velocity is less than [tex]4 m/s[/tex]  

Answer:

On moon time period will become 2.45 times of the time period on earth

Explanation:

Time period of simple pendulum is equal to [tex]T=2\pi \sqrt{\frac{l}{g}}[/tex] ....eqn 1 here l is length of the pendulum and g is acceleration due to gravity on earth

As when we go to moon, acceleration due to gravity on moon is [tex]\frac{1}{6}[/tex] times os acceleration due to gravity on earth

So time period of pendulum on moon is equal to

[tex]T_{moon}=2\pi \sqrt{\frac{l}{\frac{g}{6}}}=2\pi \sqrt{\frac{6l}{g}}[/tex] --------eqn 2

Dividing eqn 2 by eqn 1

[tex]\frac{T_{moon}}{T}=\sqrt{\frac{6l}{g}\times \frac{g}{l}}[/tex]

[tex]T_{moon}=\sqrt{6}T=2.45T[/tex]

So on moon time period will become 2.45 times of the time period on earth

Final answer:

The period of a pendulum on the Moon would be longer because the Moon's gravity is weaker. To achieve the same one-second period as on Earth, a pendulum needs to be much shorter due to the Moon's 1/6th gravitational acceleration. Consequently, a pendulum's frequency would decrease if taken from Earth to the Moon.

Explanation:

The period of a simple pendulum is affected by the acceleration due to gravity, which is less on the Moon than on Earth. Hence, if you took a pendulum clock, like a grandfather clock, to the Moon, its pendulum would swing more slowly because of the Moon's weaker gravity. To maintain a steady tick-tock of one second per period on the Moon, the pendulum would need to be much shorter. A grandfather clock pendulum designed to have a two-second period on Earth with a length of 50 cm would need to be only 8.2 cm long on the Moon to achieve the same period, since the Moon's gravity is 1/6th that of Earth. Therefore, if a pendulum from Earth was taken to the Moon, its frequency would decrease because the acceleration due to gravity on the Moon is less than that on Earth.

Answer:

9. The force is a force of attraction and it is 2.95N

10. The magnitude of acceleration 35.12m/s^2 and the direction of this acceleration is away from the other balloon.

Explanation:

Parameters given:

Q1 = 3.4 * 10^-6C

Q2 = - 5.1 * 10^-6C

Distance between the two balloons = 23cm = 0.23m

9. Force acting between the two balloons is a force of attraction because they are unlike charges. Hence, the force between them is:

F = kQ1Q2/r^2

F = (9 *10^9 * 3.4 * 10^-6 * -5.1 * 10^-6)/(2.3 * 10^-1)^2

F = (1.56 * 10^-1)/(5.29 * 10^-2)

F = - 2.95N

10. Assuming that Balloon A has a mass, m, of 0.084kg, then:

F = ma

Where a = acceleration

a = F/m

a = -2.95/0.084

a = - 35.12m/s^2

The acceleration has a magnitude of 35.12m/s^2 and its direction is away from balloon B.

The negative sign shows that the balloon A is slowing down as it moves towards balloon B. Hence, it's velocity is reducing slowly.

Answer:

W = 7591.56 J

Explanation:

given,

distance of the box, d = 37 m

Force for pulling the box, F = 217 N

angle of inclination with horizontal,θ = 19°

We know,

Work done is equal to product of force and the displacement.

W = F.d cos θ

W = 217 x 37 x cos 19°

W = 7591.56 J

Hence, the work done to pull the box is equal to W = 7591.56 J

Final answer:

The work done to slide the box is 7586.09 Joules.

Explanation:

To calculate the work done to slide a box along the ground, we can use the formula:

Work = Force x Distance x cos(theta)

Where:

Force = 217 N (the force applied to pull the box)

Distance = 37 meters (the distance the box is being slid)

theta = 19° (the angle between the applied force and the horizontal)

Plugging in these values into the formula, we get:

Work = 217 N x 37 m x cos(19°)

Calculating this using a calculator, we find that the work done to slide the box is approximately 7586.09 Joules.

This process occurs because there is a contact between the carpet and the person's feet. Basically that contact generates the transfer of some electrons to the carpet on dry winter days.

In this way a person is charged when dragging bare feet on the carpet on a dry winter day.

Therefore, the net positive charge occurs on the surface of the carpet.

A person becomes charged when shuffling across a carpet due to the transfer of electrons from the feet to the carpet, leaving a net positive charge. The lack of humidity on a dry winter day allows the static charge to build up, leading to noticeable static shocks when touching a metal object. Humidity helps in dissipating the charge, making shocks less common on humid days.

When a person shuffles across a carpet with bare feet, they can become "charged" through a process known as charging by friction. This occurs when electrons are transferred from one surface to another due to the contact and relative motion between them. In this case, electrons move from the person's feet to the carpet, leaving the feet with a net positive charge.

Materials have different affinities for electrons, and when they come into close contact, the one with the higher affinity will take on electrons from the other. Since a dry winter day has low humidity, there is less moisture in the air to carry away the excess electrons. Therefore, the static charge you accumulate is less likely to be neutralized by the surrounding air, making static shocks more frequent and noticeable when you touch a metal object like a doorknob.

The reason for the shock is the rapid movement of electrons as they try to redistribute themselves to reach a state of electrical neutrality. When you touch a metal object, the excess electrons on your body rapidly transfer to the metal, causing the shock. On a humid day, the air's moisture helps electrons move away from your body more easily, preventing the build-up of a significant static charge.

Final answer:

The problem requires calculating the work done to pump water out of a full swimming pool using given dimensions, the density of water, and gravity.

Explanation:

The question involves finding the amount of work required to pump the water out of a swimming pool when it is full. The dimensions of the pool are given, along with the density of water and the acceleration due to gravity. Using the density of water (1000 kg/m3), the volume of the pool can be calculated to determine the total mass of the water. The work done in pumping the water is found by multiplying the mass by the gravitational constant (9.8 m/s2) and the vertical distance the water needs to be moved (2.2 m, which is the uniform depth of the pool). This distance can be different depending on the location of the pump, but for this problem, we assume the water is being pumped from the very bottom.