A Pitot-static tube is used to measure the velocity of helium in a pipe. The temperature and pressure are 44oF and 24 psia. A water manometer connected to the Pitot-static tube indicates a reading of 3.0 in. (a) Determine the helium velocity. (b) Is it reasonable to consider the flow as incompressible?

The speed of sound in air increases with temperature. By taking the difference between the given temperature and 0°C, then multiplying by the rate of speed increase per degree Celsius, you can find the speed of sound. This comes out to approximately 365.8 m/s at 58.0°C.

The speed of sound in air varies depending on the temperature of the air. In general, the speed of sound increases by approximately 0.6 m/s for each degree Celsius increase in temperature. Therefore, we need to find the difference between the given temperature (58.0°C) and 0°C, which is 58, and then multiply that by 0.6 to find the increase in speed due to temperature.

Step 1: Find the temperature difference = 58.0°C - 0°C = 58°C

Step 2: Multiply the temperature difference by the rate of speed increase, which is 0.6 m/s/°C. We get: (58°C) x (0.6 m/s/°C) = 34.8 m/s.

Step 3: To find the speed of sound at the higher temperature, add this increase to the speed of sound at 0°C, which is 331 m/s.

Step 4: So, the speed of sound in air at 58.0°C is 331 m/s + 34.8 m/s = 365.8 m/s (approximately).

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Q: How much heat must be absorbed by 125 g of ethanol to change its temperature from 21.5 oC to 34.8 oC?  The specific heat of ethanol is 2.44 J/(gC).

Answer:

4056.5 J

Explanation:

The formula for the specific heat capacity of ethanol is given as

Q = cm(t₂-t₁)..................... Equation 1

Where q = quantity of heat, c = specific heat capacity of ethanol, m = mass of ethanol, t₁ = initial temperature of ethanol, t₂ = final temperature of ethanol.

Given: m = 125 g, t₁ = 25.5 °C, t₂ = 34.8 °C

Constant; c = 2.44 J/g.°C

Substitute into equation 1

Q = 125(2.44)(34.8-21.5)

Q = 125(2.44)(13.3)

Q = 4056.5 J.

Hence the amount of heat absorbed = 4056.5 J

Answer:

unit (v) = [ -0.199 i - 0.8955 j + 0.39801 k ]

Explanation:

Given:

                            v =  (-23.2, -104.4, 46.4) m/s

Above expression describes spacecraft's velocity vector v.

Find:

Find unit vector in the direction of spacecraft velocity v.

Solution:

Step 1: Compute magnitude of velocity vector.

                            mag (v) = sqrt ( 23.2^2 + 104.4^2 + 46.4^2)

                            mag (v) = 116.58 m/s

Step 2: Compute unit vector unit (v)

                            unit (v) = vec (v) / mag (v)

                            unit (v) = [ -23.2 i -104.4 j + 46.4 k ] / 116.58

                            unit (v) = [ -0.199 i - 0.8955 j + 0.39801 k ]

The unit vector in the direction of the spacecraft’s velocity is found by dividing the velocity vector by its magnitude. Accordingly, if one knows the component values of the spacecraft's velocity, one can calculate the components for the unit vector by dividing each component by the magnitude of the velocity.

The unit vector in the direction of the spacecraft's velocity can be determined from its velocity vector. The unit vector is a vector of length 1 that points in the same direction as the given vector. It can be found by dividing the velocity vector of the spacecraft by its magnitude. If V represents the velocity vector, then the unit vector U is calculated as U = V / |V| where |V| is the magnitude of the vector V. Thus, if you know the component values of the spacecraft's velocity, you can calculate the respective components for the unit vector by dividing each component of the velocity vector by the velocity's magnitude.

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Answer:

Please see below as the answer is self - explanatory

Explanation:

a) 10³ = kilo (kilogram = 10³ grams, kilometer= 10³ meters)

Symbol : k.

b) 10⁻² = centi (it is the 100th part of a unit, like centimeter, or centigram) Symbol:  c.

c) .1 = deci (it is the tenth part (deci comes from the word that means "ten"in latin) of a unit: decimeter, decigram)

Symbol: d.

d) 10⁻3 = mili (it is a 1000th part of a unit: milimeter, miligram), the name comes from the word used in latin to mean "one thousand".

Symbol: m

e) 1,000,000 = 10⁶ = mega (megawatt)

Symbol: M

f= 0.000001 = 10⁻6 = micro (micrometer, microsecond),

Symbol: μ.

Answer:

0.072 seconds

Explanation:

t = Time taken

u = Initial velocity = 100 km/h

v = Final velocity

s = Displacement = 1 m

a = Acceleration

[tex]v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{0^2-(\dfrac{100}{3.6})^2}{2\times 1}\\\Rightarrow a=-385.802469136\ m/s^2[/tex]

[tex]v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{0-\dfrac{100}{3.6}}{-385.802469136}\\\Rightarrow t=0.072\ s[/tex]

The time taken is 0.072 seconds

Answer:

Explanation:

The experiment confirm the inverse square law which relates the force between two charged particles to the product of the charges and the distance between the charges.

From the general equation, we notice the force is inversely related to the distance between the charges, when the distance is halved, the force increase by a factor of 4, hence a decrease in distance leads to a corresponding increase in the force value.

Also the electric field intensity a charge exert on another charge within the region of its field is independent on the distance because distance has no effect on electric field strength.

Answer and Explanation

• Heterotopic protons are those that when substituted by the same substituent, are structurally different. They are not similar, diastereotopic or enantiotopic.

• Diastreotopic protons refers to two protons in a molecule which, if replaced by the same substituent, would generate compounds that are diastereomers. Diastereotopic groups are often, but not always, identical groups attached to the same atom in a molecule containing at least one chiral center.

For example, the two hydrogen atoms of the C3 carbon in (S)-2-bromobutane are diastereotopic (shown in the attached image). Replacement of one hydrogen atom with a bromine atom will produce (2S,3R)-2,3-dibromobutane. Replacement of the other hydrogen atom with a bromine atom will produce the diastereomer (2S,3S)-2,3-dibromobutane.

• Homotopic protons in a compound are equivalent protons. Two protons A and B are homotopic if the molecule remains the same (including stereochemically) when the protons are interchanged with some other atom (substituent) while the remaining parts of the molecule stay fixed. Homotopic atoms are always identical, in any environment.

For example, ethane, the two H atoms on C1 and C2 carbons on the same side (as shown in the attached image) are homotopic as they exhibit the phenomenon described above.

• Enantiotopic protons are two protons in a molecule which, if one or the other were replaced (by the same substituent), would generate a chiral compound. The two possible compounds resulting from that replacement would be enantiomers.

For example, in the attached image to this answer, the two hydrogen atoms attached to the second carbon in butane are enantiotopic. Replacement of one hydrogen atom with a bromine atom will produce (R)-2-bromobutane. Replacement of the other hydrogen atom with a bromine atom will produce the enantiomer (S)-2-bromobutane.

Hope this helps!!!

Answer:

If the radius of the disk is much greater than the point where the electric field is calculated, then the field of the disk matches the field of the infinite sheet.

Explanation:

First, we have to calculate the electric field of the disk.

We should choose an infinitesimal area, 'da', on the disk and calculate the E-field of this small portion, 'dE'. Then we will integrate dE over the entire disk using cylindrical coordinates.

According to the cylindrical coordinates: da = rdrdθ

The small portion is chosen at a distance r from the axis. Let's find the dE at a point on the axis and a distance z from the center of the disk.

[tex]dE = \frac{1}{4\pi\epsilon_0}\frac{dQ}{r^2 + z^2}[/tex]

Here dQ can be found by the following relation: The charge density of the disk is equal to the total charge divided by the total area of the disk. The small portion of the disk will have the same charge density, therefore:

[tex]\frac{Q}{\pi R^2} = \frac{dQ}{da}\\dQ = \frac{Qda}{\pi R^2}[/tex]

Furthermore, we need to separate the vertical and horizontal components of dE, because it is a vector and cannot be integrated without separating the components. By symmetry, the horizontal components of dE will cancel out each other, leaving only the vertical components in the z-direction.

[tex]dE_z = dE\sin(\alpha) = dE \frac{z}{\sqrt{z^2+r^2}}\\dE_z = \frac{1}{4\pi\epsilon_0}\frac{Qzda}{\pi R^2(z^2+r^2)^{3/2}} = \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2}\frac{rdrd\theta}{(z^2+r^2)^{3/2}}[/tex]

We have to use a double integral over the radius and the angle to find the total electric field due to a uniformly charged disk:

[tex]E = \int \int dE = \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2}\int\limits^{2\pi}_0 {\int\limits^R_0 {\frac{1}{(z^2+r^2)^{3/2}}} \, rdr} \, d\theta\\E = \frac{1}{2\epsilon_0}\frac{Q}{\pi R^2}[1 - \frac{1}{\sqrt{(R^2/z^2) + 1}}][/tex]

If the radius of the disk is much greater than the point z, R >> z, than the term in the denominator becomes very large, and the fraction becomes zero. In that case electric field becomes

[tex]E = \frac{1}{2\epsilon_0}\frac{Q}{\pi R^2}[/tex]

This is equal to the electric field of an infinite sheet.

As a result, the condition for the field of a disk to be equal to that of a infinite sheet is R >> z.

Answer:

Explanation:

If u is the initial velocity at an angle \theta with horizontal then

Horizontal range of Frog can be given by

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

where u=initial velocity

a=acceleration

t=time

Here initial horizontal velocity

[tex]u_x=u\cos \theta [/tex]

and there is no acceleration in the horizontal motion

Therefore

[tex]R=u\cos \theta \times t+0[/tex]

Considering vertical motion

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

here Initial vertical velocity [tex]u_y=u\sin \theta [/tex]

acceleration [tex]a=g[/tex]

for complete motion Y=0 i.e.displacement is zero

[tex]0=u\sin \theta \times t-\frac{1}{2}gt^2[/tex]

[tex]t=\frac{2u\sin \theta }{g}[/tex]

Therefore Range is

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

Range will be maximum when [tex]\theta =45[/tex]

[tex]R=\frac{u^2}{g}----1[/tex]

and Maximum height [tex]h_{max}=\frac{u^2\sin ^2 \theta }{2g}[/tex]

for [tex]\theta =45[/tex]

[tex]h_{max}=\frac{u^2}{4g}----2[/tex]

Divide 1 and 2

[tex]\frac{R_{max}}{h_{max}}=\frac{\frac{u^2}{g}}{\frac{u^2}{4g}}[/tex]

[tex]\frac{R_{max}}{h_{max}}=4[/tex]

Answer:

a) [tex]t=5\ s[/tex]

b) [tex]s=61.5\ m[/tex]

Explanation:

Given:

acceleration of the car, [tex]a=-4.92\ m.s^{-1}[/tex]

initial velocity of the car, [tex]u=24.6\ m.s^{-1}[/tex]

final velocity of the car, [tex]v=0\ m.s^{-1}[/tex]

a)

Using eq. of motion:

[tex]v=u+a.t[/tex]

[tex]0=24.6-4.92\times t[/tex]

[tex]t=5\ s[/tex]

b)

Distance travelled before stopping:

[tex]s=u.t+\frac{1}{2} a.t^2[/tex]

[tex]s=24.6\times 5-0.5\times 4.92\times 5^2[/tex]

[tex]s=61.5\ m[/tex]

c)

The car takes deceleration in 5 seconds to stop and travels a distance of 61.5 meters.

a) To find the time it takes for the car to stop, we can use the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the deceleration, and t is the time elapsed. Rearranging the equation to solve for t, we have t = (v - u) / a. Substituting the given values, we get t = (0 - 24.6) / -4.92 = 5 seconds.

b) To find the distance traveled during this time, we can use the equation s = ut + (1/2)at^2, where s is the distance, u is the initial velocity, a is the deceleration, and t is the time elapsed. Substituting the given values, we have s = 24.6(5) + (1/2)(-4.92)(5)^2 = 61.5 meters.

c) The graph of x vs t would be a straight line with a negative slope, representing the car's distance decreasing over time. The graph of v vs t would also be a straight line with a negative slope, representing the car's velocity decreasing over time.

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Answer:

6.8 m/s2

Explanation:

Let g = 9.8 m/s2. The total weight of both the rope and the mouse-robot is

W = Mg + mg = 1*9.8 + 2*9.8 = 29.4 N

For the rope to fails, the robot must act a force on the rope with an additional magnitude of 43 - 29.4 = 13.6 N. This force is generated by the robot itself when it's pulling itself up at an acceleration of

a = F/m = 13.6 / 2 = 6.8 m/s2

So the minimum magnitude of the acceleration would be 6.8 m/s2 for the rope to fail

The robot must have a minimum acceleration of 16.6 m/s^2 for the rope to fail.

To determine the minimum magnitude of acceleration the robot should have for the rope to fail, we need to consider the forces acting on the rope. The weight of the robot is 2.0 kg multiplied by the acceleration, which we need to find. The tension in the rope is equal to the weight of the rope plus the weight of the robot. Since the rope breaks if the tension exceeds 43 N, we can set up the equation:

Tension = Weight of rope + Weight of robot
43 N = (1.0 kg)(9.8 m/s^2) + (2.0 kg)(acceleration)

Solving for the acceleration, we get:

acceleration = (43 N - 9.8 N) / 2.0 kg = 16.6 m/s^2

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Answer:

(a)  104 N

(b) 52 N

Explanation:

Given Data

Angle of inclination of the ramp: 20°

F makes an angle of 30° with the ramp

The component of F parallel to the ramp is Fx = 90 N.  

The component of F perpendicular to the ramp is Fy.

(a)  

Let the +x-direction be up the incline and the +y-direction by the perpendicular to the surface of the incline.  

Resolve F into its x-component from Pythagorean theorem:  

Fx=Fcos30°

Solve for F:  

F= Fx/cos30°  

Substitute for Fx from given data:  

Fx=90 N/cos30°

   =104 N

(b) Resolve r into its y-component from Pythagorean theorem:

     Fy = Fsin 30°

   Substitute for F from part (a):

     Fy = (104 N) (sin 30°)  

          = 52 N  

To find the force FS necessary for the component Fx parallel to the ramp to be 90.0 N, we can use the equation Fx = FS * cos(30°) = 90.0 N. To find the component Fy perpendicular to the ramp, we can use the equation Fy = FS * sin(30°).

To find the force FS necessary for the component Fx parallel to the ramp to be 90.0 N, we can use the equation Fx = FS * cos(30°) = 90.0 N. Rearranging the equation, FS = 90.0 N / cos(30°). Therefore, FS ≈ 103.9 N.

To find the component Fy perpendicular to the ramp, we can use the equation Fy = FS * sin(30°). Substituting the value of FS, Fy ≈ 103.9 N * sin(30°). Therefore, Fy ≈ 51.9 N.

Answer:

V = 6.23 cm³

Explanation:

given,

Length of the cylinder, h = 5 cm

Diameter of the cylinder, d = 1.26 cm

                                          r = 0.63 cm

Volume of the cylinder = ?

We know,

  [tex]V = \pi r^2 h[/tex]

  [tex]V = \pi \times 0.63^2\times 5[/tex]

        V = 6.23 cm³

Volume of the cylinder is equal to V = 6.23 cm³

The volume of a metal cylinder with a length of 5.0 cm and a diameter of 1.26 cm (-approximately 4.9 cm^3- when calculated using the formula: Volume = π * (d/2)^2 * h.

To calculate the volume of a cylinder, you use the formula: Volume = π * (d/2)^2 * h. Here, d represents the diameter of the cylinder, h represents the height, and π (pi) is a constant approximately equal to 3.14159. Plugging the values into the formula, we get:
Volume = π * (1.26 cm/2)^2 * 5.0 cm
After performing the above computations, the volume of the metal cylinder is approximately 4.9 cm^3 when rounded to two significant figures.

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Answer:

197.263157895 m/s

169.491525424 m/s

Explanation:

x Denotes position

t Denotes time

Average velocity is given by

[tex]v_a=\dfrac{x_2-x_1}{t_2}\\\Rightarrow v_a=\dfrac{1000-63}{4.75}\\\Rightarrow v_a=197.263157895\ m/s[/tex]

The average velocity is 197.263157895 m/s

[tex]v_a=\dfrac{x_2-x_1}{t_2}\\\Rightarrow v_a=\dfrac{1000-0}{5.9}\\\Rightarrow v_a=169.491525424\ m/s[/tex]

The average velocity is 169.491525424 m/s

The magnitude of the average velocity of the rocket during the 4.75-second part of the flight is 197.3 m/s, while for the first 5.90 seconds of the flight, the average velocity is 169.5 m/s.

To find the magnitude of the average velocity of the rocket, we use the formula average velocity = displacement / time.

(a) The displacement during the 4.75-second part of the flight is 1.00 km - 63 m = 937 m (we converted km to m to keep units consistent). Hence, the average velocity in this part of the flight is 937 m / 4.75 s = 197.3 m/s.

(b) For the first 5.90 seconds of flight, the displacement is 1.00 km = 1000 m (the height above the ground) while time is 5.90 s. Therefore, for this duration, the average velocity is 1000 m / 5.90 s = 169.5 m/s.

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Answer:

[tex]m=1864.68\ g[/tex]

Explanation:

Given:

volume of air in the room, [tex]V=410\ m^3[/tex]

temperature of the room, [tex]T=25+273=298\ K[/tex]

Saturation water vapor pressure at any temperature T K is given as:

[tex]p_{sw}=\frac{e^{(77.3450 + 0.0057\times T - \frac{ 7235}{T} )}}{T^{8.2}}[/tex]

putting T=298 K we have

[tex]p_{sw}=3130\ Pa[/tex]

The no. of moles of water molecules that this volume of air can hold is:

Using Ideal gas law,

[tex]P.V=n.R.T[/tex]

[tex]n=\frac{P_{sw}.V}{R.T}[/tex]

[tex]n=\frac{3130\times 410}{8.314\times 298}[/tex]

[tex]n=518\ moles[/tex] is the maximum capacity of the given volume of air to hold the moisture.

Currently we have 80% of n, so the mass of 20% of n:

[tex]m=(20\%\ of\ n)\times M}[/tex]

where;

M= molecular mass of water

[tex]m=0.2\times 518\times 18[/tex]

[tex]m=1864.68\ g[/tex] is the mass of water that can vaporize further.

Answer:

x = 2 x 10¹⁶ Km

Explanation:

distance between two star,d = 10⁹ Km

separation between them, θ = 10⁻⁵ radians

focal length of the eyepiece = 1.5 cm = 0.015 m

focal length of the objective = 3 m

observer distance from star, x = ?

we know,

[tex]tan \theta = \dfrac{d}{x}[/tex]

for small angle

[tex]\theta = \dfrac{d}{x}[/tex].......(1)

angular magnification of telescope

[tex]M = \dfrac{f_{objective}}{f_{eyepiece}}=\dfrac{3}{0.015} = 200[/tex]

Angular magnification of the telescope is also calculated by

[tex]M = \dfrac{observed\ angle}{original\ angle}[/tex]

[tex]M = \dfrac{\theta_0}{\theta}[/tex]

now,

[tex]\dfrac{\theta_0}{\theta}=200[/tex]

[tex]\dfrac{\theta_0}{200}=\theta[/tex]

from equation (1)

[tex]\dfrac{\theta_0}{200}=\dfrac{d}{x}[/tex]

[tex]x=\dfrac{200d}{\theta_0}[/tex]

[tex]x=\dfrac{200\times 10^9}{10^{-5}}[/tex]

x = 2 x 10¹⁶ Km

Distance between the observer and the star is x = 2 x 10¹⁶ Km

To solve this problem we will apply the linear motion kinematic equations.

The equation that describes the position as a function of the initial velocity, acceleration and time is given by the relation

[tex]s = v_0 t +\frac{1}{2} at^2[/tex]

Here,

[tex]v_0 =[/tex] Initial velocity

t = Time

a = Acceleration, at this case due to gravity

There is not initial velocity then we have that the equation to the given time is

[tex]s = \frac{1}{2} (9.8)(6.8)^2[/tex]

[tex]s = 226.8m[/tex]

If the ball is held at a height of 1m before it is dropped, we have that the Building height is

[tex]h = 226.8-1[/tex]

[tex]h = 225.8m[/tex]

Answer: Option (b) is the correct answer.

Explanation:

Since, there is a negative charge present on the ball and a positive charge present on the rod. So, when the negatively charged metal ball will come in contact with the rod then positive charges from rod get conducted towards the metal ball.

Hence, the rod gets neutralized. But towards the metal ball there is a continuous supply of negative charges. Therefore, after the neutralization of positive charge from the rod there will be flow of negative charges from the metal ball towards the rod.

Thus, we can conclude that negative charge spread evenly on both ends.

Option (b) is the correct answer.

b. There is negative charge spread evenly on both ends.

The following information should be considered:

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Answer:

Explanation:

Given

Wavelength of first timpani [tex]\lambda _1=2.2\ m[/tex]

Frequency corresponding to this Wavelength

[tex]f_1=\frac{v}{\lambda _1}[/tex]

where [tex]v=velocity\ of\ sound (343 m/s)[/tex]

[tex]f_1=\frac{343}{2.2}[/tex]

[tex]f_1=155.9\ Hz[/tex]

Wavelength of Second timpani [tex]\lambda _2=2.1\ m[/tex]

Frequency corresponding to this Wavelength

[tex]f_2=\frac{v}{\lambda _2}[/tex]

[tex]f_2=\frac{343}{2.1}=163.33\ Hz[/tex]

Beat frequency [tex]=f_2-f_1[/tex]

Beat frequency [tex]=163.33-155.9=7.43[/tex]

so approximately 7 beats per second

Final answer:

The beat frequency heard when two timpani are played together, with wavelengths of 2.20 m and 2.10 m and the speed of sound being 343 m/s, is 7.42 Hz.

Explanation:

When two timpani (tunable drums) are played at the same time with wavelengths of 2.20 m and 2.10 m, respectively, and the speed of sound is 343 m/s, we can calculate the frequencies of these sounds and subsequently determine the beat frequency heard.

The frequency of a wave is given by the formula f = v / λ (where f is the frequency, v is the speed of sound, and λ is the wavelength).

For the first timpani, the frequency is 343 m/s / 2.20 m = 155.91 Hz; for the second, it is 343 m/s / 2.10 m = 163.33 Hz.

The beat frequency, which is the absolute difference between these two frequencies, will be -

= 163.33 Hz - 155.91 Hz

= 7.42 Hz.

Answer:

a. 11.5kv/m

b.102nC/m^2

c.3.363pF

d. 77.3pC

Explanation:

Data given

[tex]area=7.60cm^{2}\\ distance,d=2mm\\voltage,v=23v[/tex]

to calculate the electric field, we use the equation below

V=Ed

where v=voltage, d= distance and E=electric field.

Hence we have

[tex]E=v/d\\E=\frac{23}{2*10^{-3}} \\E=11.5*10^{3} v/m\\E=11.5Kv/m[/tex]

b.the expression for the charge density is expressed as

σ=ξE

where ξ is the permitivity of air with a value of 8.85*10^-12C^2/N.m^2

If we insert the values we have

[tex]8.85*10^{-12} *11500\\1.02*10^{-7}C/m^{2} \\102nC/m^{2}[/tex]

c.

from the expression for the capacitance

[tex]C=eA/d[/tex]

if we substitute values we arrive at

[tex]C=\frac{8.85*10^{-12}*7.6*10^{-4}}{2*10^{-3} } \\C=\frac{6.726*10^{-15} }{2*10^{-3} } \\C=3.363*10^{-12}F\\C=3.363pF[/tex]

d. To calculate the charge on each plate, we use the formula below

[tex]Q=CV\\Q=23*3.363*10^{-12}\\ Q=7.73*10^{-12}\\ Q=77.3pC[/tex]

For the air-filled capacitor:

(a) The electric field is 11.5 kV/m

(b) The surface charge density is 1.02×10⁻⁷C/m²

(c) The capacitance is 3.36pF

(d) Charge on the plate: 76pC

Given that an air-filled capacitor consists of two parallel plates such that the:

Area of the plates, A = 7.6 cm² = 7.6×10⁻⁴ m²

distance between the plates, d = 2mm = 2×10⁻³m

voltage applied, V = 23V

(a) Electric field is given by:

E = V\d

E = 23/2×10⁻³

E = 11.5 kV/m

(b) The electric field for a parallel plate capacitor in terms of the surface charge density is given by:

E = σ/ε₀

where σ is the surface charge density:

σ = ε₀E

σ = 8.85×10⁻¹²×11.5×10³

σ = 1.02×10⁻⁷C/m²

(c) the capacitance is given by:

C = ε₀A/d

C = (8.85×10⁻¹²)×(7.6×10⁻⁴)/2×10⁻³

C = 3.36×10⁻¹²F

C = 3.36 pF

(d) The charge (Q) on the plate is given by:

Q = σA

Q = 1.02×10⁻⁷× 7.6×10⁻⁴

Q = 76 pC

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Final answer:

To raise the boiling point of 2.85 kg of water by 2°C, one needs to add approximately 320.41 grams of sodium chloride (NaCl), calculated based on the boiling point elevation formula and considering NaCl's dissociation into ions.

Explanation:

To calculate how much NaCl is needed to increase the boiling point of 2.85 kg of water by 2°C, we use the boiling point elevation formula: ΔT = i*Kb*m, where ΔT is the change in boiling point, i is the van 't Hoff factor (which is 2 for NaCl because it dissociates into Na+ and Cl- ions), Kb is the ebullioscopic constant of water (0.52 °C kg/mole), and m is the molality of the solution. First, we solve for m knowing that we want to increase the boiling point by 2°C. With Kb = 0.52 °C kg/mole and i=2, we have 2°C = (2)*(0.52 °C kg/mol)*m. From here, m = 1.923 mol/kg.

Next, to find the mass of NaCl needed, we convert molality to moles of solute needed using the mass of the solvent (water) in kg, then multiply by the molar mass of NaCl. Since molality = moles of solute / kg of solvent, moles of NaCl = molality * kg of solvent = 1.923 mol/kg * 2.85 kg = 5.48 moles. The mass of NaCl required = moles * molar mass = 5.48 mol * 58.44 g/mol = 320.41 g.

Therefore, to increase the temperature of the boiling water by 2°C, we need to add approximately 320.41 grams of NaCl.

Answer:

(c) more than 500

Explanation:

Until 2019, more than 3000 planetary systems have been discovered that contain more than 4000 exoplanets, since some of these systems contain multiple planets. Most known extrasolar planets are gas giants equal to or more massive than the planet Jupiter, with orbits very close to its star.

Final answer:

The number of confirmed exoplanets is more than 5000. Through missions like Kepler, the catalog of exoplanets includes systems with a variety of planet arrangements and types, indicating an incredibly diverse universe with potentially billions of Earth-size planets. The correct answer is option is d) more than 5000.

Explanation:

The number of confirmed exoplanets is (d) more than 5000. As of July 2015, NASA's Kepler mission had detected a total of 4,696 possible exoplanets, and 1,030 of those candidates had been confirmed as planets. Advancing to 2018, astronomers had data on nearly 3,000 exoplanet systems.

By 2022, this knowledge expanded to include data on over 800 systems, many with multiple planets, and an understanding that one quarter of stars may have exoplanet systems. This implies the existence of at least 50 billion planets in our Galaxy alone. Recent studies have shown that planets like Earth are the most common type of planet, leading to an estimated 100 billion Earth-size planets around Sun-like stars in the Galaxy.

It's important to note that detections have been made possible using the Doppler and transit techniques, and while most of the exoplanets found are more massive than Earth, the shortage of small rocky planets detected is an observational bias, as they are more difficult to detect.

The ensemble of exoplanets discovered is incredibly diverse, suggesting a variety of planet formations and arrangements in these systems. Some systems have been observed to have rocky planets closer to their stars than in our solar system, and others have large gas giants, known as 'hot Jupiters', very close to their stars.

Answer:

Q= 835J

Explanation:

Specific heat capacity of Helium=5.193J/gk

Q=msΔT

ΔT=T2-T1

Q=change in energy

m=mass of substance

S= Specific heat capacity

Q= Change in energy

T2= 38.3 degrees Celsius =(38.3+273)k= 311.3k

T1=24.2 degree Celsius =24.2+273k

T1=297.2k

ΔT=T2-T1=(311.3-297.2)k=14.1K

Q=(11.4g)*(5.193J/gk)*(14.1k)

Q=834.72282J

Approximately, Q= 835J

Answer:

f = 735 Hz

Explanation:

given,

Person distance from speakers

r₁ = 4.1 m      r₂ = 4.8 m

Path difference

d = r₂ - r₁ = 4.8 - 4.1 = 0.7 m

For destructive interference

[tex]d = \dfrac{n\lambda}{2}[/tex]

where, n = 1, 3,5..

we know, λ = v/f

[tex]d = \dfrac{n v}{2f}[/tex]

v is the speed of the sound = 343 m/s

f is the frequency

[tex]f = \dfrac{n v}{2d}[/tex]

for n = 1

[tex]f = \dfrac{343}{2\times 0.7}[/tex]

     f = 245 Hz

for n = 3

[tex]f = \dfrac{3\times 343}{2\times 0.7}[/tex]

     f = 735 Hz

Hence,the second lowest frequency of the destructive interference is 735 Hz.

Answer:

C. Charge on the capacitor

Explanation:

Read further: Capacitors consist of two parallel conductive

plates (usually a metal) which are prevented

from touching each other (separated) by an

insulating material called the “dielectric”. When

a voltage is applied to these plates an

electrical current flows charging up one plate

with a positive charge with respect to the

supply voltage and the other plate with an

equal and opposite negative charge.

Then, a capacitor has the ability of being able

to store an electrical charge Q (units in

Coulombs ) of electrons. When a capacitor is

fully charged there is a potential difference,

p.d. between its plates, and the larger the area

of the plates and/or the smaller the distance

between them (known as separation) the

greater will be the charge that the capacitor

can hold and the greater will be its

Capacitance.

The capacitors ability to store this electrical

charge ( Q ) between its plates is proportional

to the applied voltage, V for a capacitor of

known capacitance in Farads.

The correct option is c which is charge on the capacitance. When a parallel-plate capacitor is disconnected from the battery, and the separation between its plates is doubled, the charge on the capacitor remains constant. Changes in plate separation affect capacitance and voltage, but not the existing charge on the plates.

The question asks which of the following remains constant when the battery is disconnected from a parallel-plate capacitor and the separation between the plates is doubled: voltage across the capacitor, capacitance of the capacitor, or the charge on the capacitor. The key to answering this question lies in understanding how capacitors work and the relationship between charge (Q), capacitance (C), and voltage (V).

Capacitance is given by C = εA/d, where A is the area of the plates, d is the separation between the plates, and ε is the permittivity of the material between the plates. When the battery is disconnected, the external voltage source is removed, but the charge on the plates does not have a path to dissipate. Therefore, the charge on the capacitor remains constant, even when the plate separation is changed. Doubling the separation would affect the capacitance and the voltage across the capacitor but not the charge.

Answer:

[tex]6\times 10^{10}\ Hz[/tex]

Explanation:

d = Slit gap = 5 mm

Slit distance = 16 cm

c = Speed of light = [tex]3\times 10^8\ m/s[/tex]

[tex]\lambda[/tex] = Wavelength

We have the relation

[tex]dsin\theta=\lambda[/tex]

Here, [tex]\theta=90[/tex]

So

[tex]d=\lambda\\\Rightarrow \lambda=5\ mm[/tex]

Frequency is given by

[tex]f=\dfrac{c}{\lambda}\\\Rightarrow f=\dfrac{3\times 10^8}{5\times 10^{-3}}\\\Rightarrow f=6\times 10^{10}\ Hz[/tex]

The frequency is [tex]6\times 10^{10}\ Hz[/tex]

Answer: °C = 4.44°C

Explanation: The centigrade scale and Fahrenheit scale are related by the formulae below

9 *°C = 5(°F - 32)

Where °C = measurement of temperature in centigrade scale.

°F = measurement of temperature in Fahrenheit scale =40°F

By substituting the parameters, we have that

9 * °C = 5 (40 - 32)

9* °C = 5(8)

9 * °C = 40

°C = 40/9

°C = 4.44°C

Temperature decreases by approximately 3.5°F for every 1000 feet increase in altitude. Given the altitude difference of 3130 feet from Boulder to Bear Peak and Boulder's temperature of 40°F, we can calculate that the temperature at Bear Peak is expected to be approximately 29°F.

The question asks for the temperature at a higher altitude given the temperature at a lower one. This involves understanding how temperature changes with altitude. It is stated that, on average, temperature decreases by about 3.5°F for every 1000 feet you climb in altitude, a phenomenon known as the lapse rate.

Given that, if the starting temperature in Boulder (altitude: 5330’) is 40F, and we need to find the temperature on Bear Peak (altitude: 8460’), we need to calculate the difference in altitude and the corresponding temperature decrease.

The altitude difference is 8460 - 5330 = 3130 feet. From this height difference, we can determine the corresponding temperature change by multiplying 3130 feet by 3.5°F per 1000 feet, yielding a temperature decrease of about 11°F.

To find the temperature at the higher altitude (Bear Peak), we then subtract this temperature change from the starting temperature. Thus,  40°F - 11°F gives 29°F as the expected temperature on Bear Peak.

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Final answer:

The charge enclosed by the cubical box with a total electric flux of 1840 N·m2/C is calculated using Gauss's law and is found to be 16.29 nC.

Explanation:

The question deals with the concept of electric flux and its relation to the enclosed charge using Gauss's law, which is a fundamental principle in electromagnetism. According to Gauss's law, the total electric flux through a closed surface is equal to the charge enclosed by the surface divided by the permittivity of free space (ε0).

The formula for Gauss's law in integral form is Φ = Q / ε0, where Φ is the electric flux, Q is the charge enclosed, and ε0 is the electric constant (approximately 8.854 x 10-12 C2/N·m2). Given that the total electric flux from a cubical box is 1840 N·m2/C, and using the value of ε0, the enclosed charge (Q) can be calculated.

To find the charge, we rearrange the equation as Q = Φ·ε0 and substitute the given values to get Q = 1840 N·m2/C × 8.854 x 10-12 C2/N·m2, resulting in Q = 1.629 x 10-8 C or 16.29 nC (nanocoulombs).

The charge enclosed by the cubical box,is approximately 1.63 × 10⁻⁸ C .

To determine the charge enclosed by a cubical box, we can use Gauss's Law. According to Gauss's Law, the electric flux (Φ) through a closed surface is given by:

Φ = Q / ε₀

where:

Rearranging this formula to solve for Q, we get:

Q = Φ × ε₀

Substitute the given values:

Q = 1840 N·m²/C × 8.854 × 10⁻¹² C²/N·m²

Q ≈ 1.63 × 10⁻⁸ C

Therefore, the charge enclosed by the cubical box is approximately 1.63 × 10⁻⁸ C.

Answer:

8.07 m/s, 81.7º NE.

Explanation:

        [tex]vocx = voc* cos 40 = 1.53 m/s * 0.766 = 1.17 m/s\\vocy = voc* sin 40 = 1.53 m/s * 0.643 = 0.98 m/s[/tex]

        [tex]vshy = vsw + vwy = 7.00 m/s + 0.98 m/s = 7.98 m/s[/tex]

        vshx = 1.17 m/s

        [tex]v = \sqrt{vshx^{2} + vshy^{2} } =\sqrt{(7.98m/s)^{2} +(1.17m/s)^{2} } =8.07 m/s[/tex]

        [tex]tg \theta = \frac{vshy}{vshx} = \frac{7.98}{1.17} = 6.82 \\ \theta = tg^{-1} (6.82)\\ \theta= 81.7\deg[/tex]

Answer:

 In a circuit with parallelresistors, the smaller resistance dominates; in a circuit with resistors inseries, the larger one dominates. Make some specific examples of resistorsin parallel and in series to explain this rule of thumb.

Explanation:

Lets start with Resistors in series:

Suppose a 12v battery is connected in a circuit with 3 resistors in series.

Voltage = 12 V

R1          = 1 ohm

R2         = 6 ohm

R3         = 13 ohm

Total resistance is simply sum of all resistors connected in series as following:

Rs   = R1 + R2 + R3

Rs   = 1 + 6 + 13

Rs   = 20 ohm

Here Rs = Total Resistance in series circuit

Now total current can be found by using ohm's law V = IR

I = V/R

Here R = R total

I = 12/20 = 0.6 Ampere

In series circuit current remains same.

so I1 = I2 = I3 = I

Now we can find Power dissipation across every resistor to show the dominant Resistor as:

P1 = I² R1 = (0.6 A)² ( 1 ohm) = 0.36 watt.          (a)

P2 = I² R2 = (0.6 A)² ( 6 ohm) = 2.16 watt.        (b)

P3 = I² R3 = (0.6 A)² ( 13 ohm) = 4.68 watt.      (c)

From equation a, b and c  we can conclude that more power is dissipated across R3 which is the Larger than both other resistors R1 and R2.

Now lets take the case of Parallel Circuit with same Values of Voltage and resistors But now in parallel.

In parallel circuit Total resistance is calculated as following:

1/Rp = 1/R1 + 1/R2 + 1/R3

1/Rp = 1/1 + 1/6 + 1/13 = 1 + 0.1667 + 0.0769

1/Rp = 1.2436

Rp   = 0.8041 ohm

Now total current is as following:

I = V/R

I = 12/0.8041 = 14.92 A

Now we calculate individual currents :

In parallel circuit Voltage remains same.

I1 = V/R1

I1 = 12/1 = 12 A

Similarly,

I2 = V/R2

I2 = 12/ 6 = 2 A

and

I3 = V/R3

I3 = 12/13 = 0.92 A

Now Power dissipation can be calculated as :

P1 = I1² * R1

P1 = 12² * 1 = 144 * 1

P1 = 144 Watt            (1)

Similarly,

P2 = I2² * R2

P2 = 2² * 6 = 4 * 6

P2 = 24 Watt            (2)

And

P3 = I3² * R3

P3 = (0.92)² * 13 = 0.8464 * 13

P3 = 11 Watt.              (3)

Now from Equation 1, 2 and 3 we can conclude that More power is dissipated across Lower resistor .

Hence in a circuit with parallel resistors, the smaller resistance dominates; in a circuit with resistors in series, the larger one dominates.