A Rankine oval is formed by combining a source-sink pair, each having a strength of 36 ft2/s and separated by a distance of 15 ft along the x axis, with a uniform velocity of 12 ft/s (in the positive x direction). Determine the length of the oval.

To solve this problem we will apply the concept of wavelength, which warns that this is equivalent to the relationship between the speed of the air (in this case in through the air) and the frequency of that wave. The air is in standard conditions so we have the relation,

Frequency [tex]= f = 562Hz[/tex]

Speed of sound in air [tex]= v = 331m/s[/tex]

The definition of wavelength is,

[tex]\lambda = \frac{v}{f}[/tex]

Here,

v = Velocity

f = Frequency

Replacing,

[tex]\lambda = \frac{331m/s}{562Hz}[/tex]

[tex]\lambda = 0.589m[/tex]

Therefore the wavelength of that tone in air at standard conditions is 0.589m

The wavelength of the tone in air 0.59 Hz

The trombone can produce pitches wavelength ranging from 85 Hz to 660 Hz

The trombone produces a tone of 562 Hz

The tone of air is at standard conditions, hence the velocity of the sound in air is 331 m/s

velocity=  frequency/wavelength

331= 562/wavelength

wavelength= 331/562

= 0.59 HZ

Hence the wavelength of the tone is 0.59 Hz

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Answer with Explanation:

Mass of pendulum bob, m=0.453 kg

Speed, [tex]v_1=[/tex]2.58 m/s

a.r=75.1 cm=[tex]75.1\times 10^{-2}m[/tex]=0.751 m

[tex] 1cm=10^{-2} m[/tex]

Tension in the pendulum cable is given  by

Tension=Centripetal force+force due to gravity

[tex]T=\frac{mv^2}{r}+mg[/tex]

Where [tex]g=9.8 m/s^2[/tex]

Substitute the values

[tex]T=\frac{0.453(2.58)^2}{75.1\times 10^{-2}}+0.453\times 9.8[/tex]

[tex]T=8.45 N[/tex]

b.When the pendulum reaches its highest point,then

Final velocity, [tex]v_2=0[/tex]

According to law of conservation of energy

[tex]mgh_1+\frac{1}{2}mv^2_1=mgh_2+\frac{1}{2}mv^2_2[/tex]

[tex]gh_1+\frac{1}{2}v^2_1=gh_2+\frac{1}{2}v^2_2[/tex]

[tex]h_1=0[/tex]

Substitute the values

[tex]9.8\times 0+\frac{1}{2}(2.58)^2=9.8\times h_2+\frac{1}{2}(0)^2[/tex]

[tex]3.3282=9.8h_2[/tex]

[tex]h_2=\frac{3.3282}{9.8}=0.34 m[/tex]

The angle mad  by cable with the vertical=[tex]cos\theta=\frac{0.751-0.34}{0.751}=0.55[/tex]

[tex]\theta=cos^{-1}(0.55)=56.6^{\circ}[/tex]

c.When the pendulum reaches at highest point then

Acceleration, a=0

Therefore, the tension  in the pendulum cable

[tex]T=mgcos\theta[/tex]

Substitute the values

[tex]T=0.453\times 9.8cos56.6[/tex]

[tex]T=2.4 N[/tex]

Answer:

GRadient= 0.192 mb / km ,   the correct answer is a

Explanation:

The pressure gradient would be considered linear so we can use a proportional rule to find the gradient

          Gradient = Dp / Dd

          Gradient = (1045 -997) / 250

          Gradient = 48/250

          GRadient= 0.192 mb / km

The correct answer is a

Explanation:

Beloware attachments containing the complete question and solution.

Answer:

2.8m-3.4m

Explanation:

Radio waves are electromagnetic waves and they all travel with a speed of

[tex]3*10^8m/s[/tex] in air.

The range of frequencies in the FM band is from 88MHz to 108MHz.

Generally, the relationship between velocity, frequency and wavelength for electromagnetic waves is given by equation (1);

[tex]v=\lambda f..............(1)[/tex]

From equation (1), we can write,

[tex]\lambda=\frac{v}{f}...............(2)[/tex]

For the upper frequency of 108MHz, the wavelength is given by;

[tex]\lambda_1=\frac{3*10^8}{108*10^6}\\\lambda_1=0.02778*10^2\\\lambda_1=2.78m[/tex]

Similarly, for the lower frequency of 88MHz, the wavelength is given by;

[tex]\lambda_2=\frac{3*10^8}{88*10^6}\\\lambda_2=0.0341*10^2\\\lambda_2=3.41m[/tex]

The range of wavelength therefore  is [tex]\lambda_1-\lambda_2=2.8m-3.4m[/tex] approximately.

Please note the following:

[tex]108MHz=108*10^6Hz\\88MHz=88*10^6Hz[/tex]

Answer:

[tex]8.64283\times 10^{-14}\ N[/tex]

Explanation:

q = Charge of proton = [tex]1.6\times 10^{-19}\ C[/tex]

v = Velocity of proton = [tex]0.267\times c[/tex]

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

B = Magnetic field = 0.00687 T

[tex]\theta[/tex] = Angle = [tex]101^{\circ}[/tex]

Magnetic force is given by

[tex]F=qvBsin\theta\\\Rightarrow F=1.6\times 10^{-19}\times (0.267\times 3\times 10^8)\times 0.00687\times sin101\\\Rightarrow F=8.64283\times 10^{-14}\ N[/tex]

The magnetic force acting on the proton is [tex]8.64283\times 10^{-14}\ N[/tex]

The given question is incomplete. The complete question is as follows.

For your senior project, you would like to build a cyclotron that will accelerate protons to 10% of the speed of light. The largest vacuum chamber you can find is 60 cm in diameter.

What magnetic field strength will you need?

Explanation:

Formula for the strength of magnetic field is as follows.

      B = [tex]\frac{mv}{qr}[/tex]

Here,    m = mass of proton = [tex]1.67 \times 10^{-27}[/tex] kg

      v = velocity = 10% of [tex]3 \times 10^{8}[/tex] = [tex]3 \times 10^{7}[/tex] m/s

      q = charge of proton = [tex]1.6 \times 10^{-19} C[/tex]

      r = radius = [tex]\frac{60}{2}[/tex] = 30 cm = 0.30 m   (as 1 m = 100 cm)

Therefore, magnetic field will be calculated as follows.

            B = [tex]\frac{mv}{qr}[/tex]

               = [tex]\frac{1.67 \times 10^{-27} \times 3 \times 10^{7}}{1.6 \times 10^{-19} C \times 0.30 m}[/tex]

               = [tex]\frac{5.01 \times 10^{-20}}{0.48 \times 10^{-19}}[/tex]

               = 1.0437 T

Thus, we can conclude that magnetic field strength is 1.0437 T.

The student queries about the construction of a cyclotron to accelerate protons to a specific velocity. The radius and rotational period of protons within the cyclotron are calculated using the magnetic field strength and the desired velocity.

The student is interested in building a cyclotron that can accelerate protons to 10% of the speed of light, with specific constraints on the vacuum chamber dimensions. In physics, particularly in the field of particle accelerators, a cyclotron is a type of particle accelerator that uses a combination of an electric field and a constant magnetic field to increase the kinetic energy of charged particles. The radius of the cyclotron, which determines the maximum orbit size for the particles being accelerated, is a critical design parameter and can be calculated based on the desired kinetic energy of the particles and the strength of the magnetic field.

In the context of a cyclotron, the student might need to calculate the rotational period and maximum radius of proton orbits within given specifications such as the strength of the magnetic field and desired velocity. Understanding the principles behind cyclotrons and particle acceleration is essential for this project, which falls under the umbrella of advanced physics topics.

Answer:

the value of vA that will allow the car to coast in neutral so as to just reach the top of the 150 m high hill at B with vB = 0 is 31.3 m/s

Explanation:

given information

car's mass, m = 1200 kg

[tex]h_{A}[/tex] = 100 m

[tex]v_{A}[/tex] = [tex]v_{A}[/tex]

[tex]h_{B}[/tex] = 150 m

[tex]v_{B}[/tex] = 0

according to conservative energy

the distance from point A to B, h = 150 m - 100 m = 50 m

the initial speed [tex]v_{A}[/tex]

final speed  [tex]v_{B}[/tex] = 0

thus,

[tex]v_{B}[/tex]² = [tex]v_{A}[/tex]² - 2 g h

0 = [tex]v_{A}[/tex]² - 2 g h

[tex]v_{A}[/tex]² = 2 g h

[tex]v_{A}[/tex] = √2 g h

    = √2 (9.8) (50)

    = 31.3 m/s

Answer and Explanation:

Using Gauss's law,

If r>a

then charge enclosed in all the three cases is same as Q.

So Electric field for all three is same.

So {1,1,1}.

(b) r<a,

Charge enclosed in case of shell is zero since all charge is present on the surface. So E = 0.

Charge enclosed by incase of point charge is Q.

Charge enclosed in case of sphere is Qr3/a3 which is less than Q.

So ranking {2,3,1}

Answer:

(a) 0.3 C

(b) 0.9 J

Explanation:

(a)

Given:

Current drawn (I) = 0.50 mA = 0.50 × 10⁻³ A

Terminal voltage (V) = 3.0 V

Time of operation (t) = 10 min = 10 × 60 = 600 s

Charge flowing through the toy car 'Q' is given as:

[tex]Q=It[/tex]

Plug in the given values and solve for 'Q'. This gives,

[tex]Q=(0.50\times 10^{-3}\ A)(600\ s)\\\\Q=0.3\ C[/tex]

Therefore, 0.3 C charge flows the toy car.

(b)

Energy lost by the battery is equal to the product of power consumed by the battery and time of operation.

Power consumed by the battery is given as:

[tex]P=VI[/tex]

Plug in the given values and solve for 'P'. This gives,

[tex]P=(3.0\ V)(0.50\times 10^{-3}\ A)\\\\P=1.5\times 10^{-3}\ W[/tex]

Therefore, the energy lost by the battery is given as:

[tex]E=P\times t\\\\E=1.5\times 10^{-3}\ W\times 600\ s\\\\E=0.9\ J[/tex]

Therefore, the energy lost by the battery is 0.9 J

Answer:

Negative

Explanation:

First law of thermodynamic also known as the  law of conservation of energy states that the total energy of an isolated system is constant; energy can be transformed from one form to another, but can be neither created nor destroyed.

The first law relates  relates changes in internal energy to heat added to a system and the work done by a system by the conservation of energy.

The first law is mathematically given as ΔU  = [tex]U_{F}[/tex] - [tex]U_{0}[/tex] = Q - W

Where Q  = Quantity of heat

            W = Work done

From the first law The internal energy has the symbol U. Q is positive if heat is added to the system, and negative if heat is removed; W is positive if work is done by the system, and negative if work is done on the system.

Analyzing the pistol when it raises in isothermal and when it falls in  isobaric state.The following can be said:

In the Isothermal compression of a gas there is work done on the system to decrease the volume and increase the pressure. For work to be done on the system it is a negative work done then.

In the Isobaric State An isobaric process occurs at constant pressure. Since the pressure is constant, the force exerted is constant and the work done is given as PΔV.If a gas is to expand at a constant pressure, heat should be transferred into the system at a certain rate.Isobaric is a fuction of heat which is Isothermal Provided the pressure is kept constant.

In Isobaric definition above it can be seen that " Heat should be transferred into the system ata certain rate. For heat to be transferred into the system work is deinitely been done on the system thereby favouring the negative work done.

Answer: 4.08kg/J

Explanation: Please find the attached file for the solution

Answer:

Entropy = 4.08 kj/k

Explanation:

From energy balance in first law of thermodynamics, we have;

Δv(i)+ ΔU(h2o) = 0

Thus;

[MCp(T2 - T1)]iron + [MCp(T2 - T1)]water = 0

Where Cp is specific heat capacity

For iron, Cp = 0.45 Kj/kg°C and for water, Cp = 4.18 Kj/kg°C

From question, Mass of iron =25kg while mass of water = 100kg

And Initial temperature of iron (T1) = 350°C while initial temperature of water(T1) = 18°C

Thus,

[25 x 0.45(T2 - 350)] + [100 x 4.18(T2 - 18)] = 0

11.25T2 - 3937.5 + 418T2 - 7524 = 0

So,

429.25T2 = 11461.5

T2 = 26.7 °C

Now for entropy, we have convert the temperature from degree celsius to kelvins.

Thus, for iron T1 = 350 + 273 = 623K and for water, T1 = 18 + 273 = 291 K. Also, T2 = 26.7 + 273 = 299.7K.

The entropy changes will be;

For iron ;

Δs(i) = MCp(In(T2/T1)) = 25 x 0.45(In(299.7/623)) = -8.23 Kj/k

Now, for water;

Δs(water) = MCp(In(T2/T1)) = 100 x 4.18(In(299.7/291)) = 12.31 kj/k

Thus, total entropy will be the sum of that of iron and water.

Δs(total) = 12.31 kj/k - 8.23 Kj/k = 4.08 kj/k

Answer:

(a). The resultant of these forces is 1216.55 N.

(b).  The direction of the resultant forces is 80.53°.

Explanation:

Given that,

First force = 1200 N

Second force = 200 N

(a). We need to calculate the resultant of these forces

Using cosine law

[tex]F=\sqrt{F_{1}^2+F_{2}^2+2F_{1}F_{2}\cos\theta}[/tex]

Put the value into the formula

[tex]F=\sqrt{1200^2+200^2+2\times1200\times200\cos90}[/tex]

[tex]F=\sqrt{1200^2+200^2}[/tex]

[tex]F= 1216.55\ N[/tex]

The resultant of these forces is 1216.55 N.

(b). We need to calculate the direction of the resultant forces

Using formula of direction

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

Put the value into the formula

[tex]\alpha=\tan^{-1}(\dfrac{1200}{200})[/tex]

[tex]\alpha=80.53^{\circ}[/tex]

Hence, (a). The resultant of these forces is 1216.55 N.

(b).  The direction of the resultant forces is 80.53°.

Answer:

a) [tex]F_r=1216.55\ N[/tex]

b) [tex]\theta=80.54^{\circ}[/tex]

Explanation:

Given:

a)

Now the resultant of these forces:

Since the forces are mutually perpendicular,

[tex]F_r=\sqrt{F_x^2+F_y^2}[/tex]

[tex]F_r=\sqrt{200^2+1200^2}[/tex]

[tex]F_r=1216.55\ N[/tex]

b)

The direction of this force from the positive x-direction:

[tex]\tan\theta=\frac{F_y}{F_x}[/tex]

[tex]\tan\theta=\frac{1200}{200}[/tex]

[tex]\theta=80.54^{\circ}[/tex]

Answer:

You could use Malus's Law. Malus's Law tells us that if you have a polarized wave (of intensity I 0 0 ) passing through a polarizer the emerging intensity ( Y OR ) will be proportional to the square cosine of the angle between the polarization direction of the incoming wave and the axis of the polarizer.

Explanation:

OR: I = I 00 ⋅ cos two ( e )

Answer:

12 volts.

Explanation:

Equal to the emf of battery. internal resistance won't count because the internal resistance is only apparent when a current passes through the battery.

The voltmeter reading is the terminal voltage, which is slightly less than the EMF due to the internal resistance of the battery and the small current drawn by the voltmeter. The exact value in volts is not provided due to the unknown internal resistance.

Answer:

Better Equilibrium Maintenance for better accuracy...

Explanation:

In the Galileo's experiment, there is no utilization of two equal masses at a time. However, as we can see in a Atwood Machine, there are two equal masses involved that make the whole system to be in a state of equilibrium and ultimately the better measurements of acceleration due to gravity.

The magnifying power of an astronomical telescope will be:

"0.095".

According to the question,

Radius of curvature, R = 5.5 mm

Focal length of eyepiece, [tex]F_e[/tex] = 2.9 cm

We know that,

→ Focal length of mirror,

F₀ = [tex]\frac{Radius \ of \ curvature}{2}[/tex]

By substituting the values,

    = [tex]\frac{5.5}{2}[/tex]

    = 2.75 mm or,

    = 0.278 cm

hence,

The telescope's magnification be:

= [tex]\frac{F_0}{F_e}[/tex]

= [tex]\frac{0.275}{2.9}[/tex]

= 0.095

Thus the above approach is correct.

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Final answer:

The magnifying power of the astronomical telescope using the given values is approximately 0.19.

Explanation:

In order to find the magnifying power of an astronomical telescope, we need to use the formula:

Magnifying Power = Angular Magnification = (focal length of objective) / (focal length of eyepiece)

Given that the radius of curvature of the reflecting mirror is 5.5 mm (which is equal to 0.55 cm) and the focal length of the eyepiece is 2.9 cm, we can substitute these values into the formula to find the magnifying power.

Magnifying Power = (radius of curvature of mirror) / (focal length of eyepiece)

Magnifying Power = 0.55 cm / 2.9 cm

Magnifying Power = 0.19

Therefore, the magnifying power of the astronomical telescope is approximately 0.19.

Answer:

The length of the lake is 340 meters.

Explanation:

It is given that, a hiker determines the length of a lake by listening for the echo of her shout reflected by a cliff at the far end of the lake. She hears the echo 2 s after shouting. We need to find the length of the lake.

The distance covered by the person in 2 s is :

[tex]d=vt[/tex]

v is the speed of sound

[tex]d=340\ m/s\times 2\ s[/tex]

[tex]d=680\ m[/tex]

The length of the lake is given by :

[tex]l=\dfrac{d}{2}[/tex]

[tex]l=\dfrac{680\ m}{2}[/tex]

l = 340 meters

So, the length of the lake is 340 meters. Hence, this is the required solution.

Answer:

(a) 0.336 A m²

(b) 0 Nm

Explanation:

(a) Magnetic dipole moment, μ, is given by

[tex]\mu = IA[/tex]

I is the current in the loop and A is the area of the loop.

The loop is a triangle. To find its area, observe that the dimensions form a Pythagorean triple, making it a right-angled triangle with base and height of 30 cm and 40 cm.

[tex]A = \frac{1}{2}\times(0.3\text{ m})\times(0.4\text{ m})=0.06\text{ m}^2[/tex]

[tex]\mu = (5.6\text{ A})(0.06\text{ m}^2) = 0.336\text{ A}\,\text{m}^2[/tex]

(b) Torque is given by

[tex]\tau = \mu B\sin\theta[/tex]

where B is the magnetic field and [tex]\theta[/tex] is the angle between the loop and the magnetic field. Since the field is parallel, [tex]\theta[/tex] is 0.

[tex]\tau = \mu B\sin0 = 0\text{ Nm}[/tex]

Answer:

The statements which are true are:

1) A person's power output limits the amount of work that he or she can do in a given time span.

2) The SI unit of power is the watt.

4) Power can be considered the rate at which work is done.

Explanation:

Power can be defined as the the rate at which the work is done. Mathematically,

[tex]\large{P = \dfrac{W}{t}}[/tex]

where '[tex]W[/tex]' is the amount of work done in time '[tex]t[/tex]'.

Also one can perform some work in the expense of the energy that he/she consumes. So alternatively power can also be defined as the rate at which the energy is expended.

The SI unit of work done is Joule. So the unit of power in SI system is

[tex]Js^{-1}[/tex] or Watt.

Answer:

180 N

Explanation:

We know that acceleration is the rate of change of speed per unit time hence

[tex]a=\frac {v_f-v_i}{t}[/tex] where v and t are velocity and time respectively, f and i represent final and initial.

Also, from Newton's law of motion, F=ma and replacing a with the above then

[tex]F=m\frac {v_f-v_i}{t}[/tex]

Substituting 1.5 Kg for mass, m -14 m/s for i and 10 m/s for for v then

[tex]F=1.5\times \frac {10--14}{0.2}=180 N[/tex]

Therefore, the force is 180 N

The average force exerted by a 1.5 kg ball on the floor, falling with a speed of 14 m/s and rebounding with a speed of 10 m/s over 0.20 seconds, is 180 N.

The subject of this question is Physics and from the concept of impulse and momentum. The change in momentum equals the product of force and the time over which the force is applied. So, we can calculate the force using this formula: Force = Change in momentum / Time.

In this case, the ball's momentum changes by the difference in velocity multiplied by the mass of the ball, which is (14 m/s + 10 m/s) * 1.5 kg. The reason the velocities are added is because the direction of velocity changes, making the speed of ball before and after striking the floor of equal magnitude but opposite in direction. So, the change in momentum becomes 36 kg*m/s. Given that the time is 0.20 seconds, the force would be 36 kg*m/s divided by 0.20 s, or 180 N.

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Answer:

q = 1.96 10⁴ C

Explanation:

The elective force is given by

         [tex]F_{e}[/tex] = q E

Where E is the electric field and q the charge.

Let's use Newton's law of equilibrium for the case of the suspended drop

               F_{e} –W = 0

               F_{e} = W

               q E = m g

               q = m g / E

Let's calculate

              q = 2.0 10⁵ 9.8 / 100

              q = 1.96 10⁴ C

To maintain suspension in the Earth’s electric field, an oil drop with a mass of 2.0 × 10^-15 kg requires a charge of 1.96 × 10^-16 Coulombs, calculated by equating the electric force with the gravitational force.

To determine the charge needed for an oil drop of mass 2.0 × 10-15 kg to remain suspended by the Earth’s electric field of 100 N/C, we can apply the equilibrium condition between the electric force and the gravitational force. The electric force (Felectric) is equal to the charge (q) multiplied by the electric field (E), so Felectric = q × E. The gravitational force (Fgravity) is the mass (m) multiplied by the acceleration due to gravity (g), which is approximately 9.8 m/s2.

For the oil drop to remain suspended, these two forces must be equal: q × E = m × g, which gives us q = (m × g) / E. Using the provided values, the charge q is calculated as follows:

q = (2.0 × 10-15 kg × 9.8 m/s2) / 100 N/C

q = (2.0 × 10-15 × 9.8) / 100

q = 1.96 × 10-16 C

Therefore, the oil drop must have a charge of 1.96 × 10-16 Coulombs to remain suspended in the Earth’s electric field.

Answer:

C) equal to zero

Explanation:

Electric potential is calculated by multiplying constant and charge, then dividing it by distance. The location that we want to measure is equidistant from two particles, mean that the distance from both particles is the same(r2=r1). The charges of the particle have equal strength of magnitude but the opposite sign(q2=-q1). The resultant will be:V = kq/r

ΔV= V1 + V2= kq1/r1 + kq2/r2

ΔV= V1 + V2= kq1/r1 + k(-q1)/(r)1

ΔV= kq1/r1 - kq1/r1

ΔV=0

The electric potential equal to zero

Answer:

T = 2 T₀

Explanation:

To answer this question let's write the expression for electrical conductivity

    σ = n e2 τ / m*

The relationship with resistivity is

       ρ = 1 /σ

Whereby the resistance

        R = ρ L / A = 1 /σ  L / A

We see that there is no explicit relationship between time and resistance, there is only a dependence on the life time (τ) that depends on the properties of the material, not on its diameter or length.

As also the average velocity or electron velocity of electrons is constant, the time to cross 2 mm in length is twice as long as the time to cross a mm in length

 T = 2 T₀

Answer:

[tex]\mu_k=\frac{a}{g}[/tex]

Explanation:

The force of kinetic friction on the block is defined as:

[tex]F_k=\mu_kN[/tex]

Where [tex]\mu_k[/tex] is the coefficient of kinetic friction between the block and the surface and N is the normal force, which is always perpendicular to the surface that the object contacts. So, according to the free body diagram of the block, we have:

[tex]N=mg\\F_k=F=ma[/tex]

Replacing this in the first equation and solving for [tex]\mu_k[/tex]:

[tex]ma=\mu_k(mg)\\\mu_k=\frac{a}{g}[/tex]

Answer:

[tex]Q_{2}=1200cm^{3}/s[/tex]

Explanation:

Given data

Q₁=200cm³/s

We know that:

[tex]F=n\frac{vA}{l}\\[/tex]

can be written as:

ΔP=F/A=n×v/L

And

Q=ΔP/R

As

n₂=6.0n₁

So

Q=ΔP/R

[tex]Q=\frac{nv}{lR}\\ \frac{Q_{2}}{n_{2}}= \frac{Q_{1}}{n_{1}}\\ Q_{2}=\frac{Q_{1}}{n_{1}}*(n_{2})\\Q_{2}=\frac{200}{n_{1}}*6.0n_{1}\\ Q_{2}=1200cm^{3}/s[/tex]

Answer:

Incomplete question

This is the complete question

The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 100 V potential difference is suddenly applied to the initially uncharged plates through a 1000 Ω resistor in series with the deflection plates. How long does it take for the potential difference between the deflection plates to reach 95 V?

Explanation:

Given that,

The dimension of 10cm by 2cm

0.1m by 0.02m

Then, the area is Lenght × breadth

Area=0.1×0.02=0.002m²

The distance between the plate is d=1mm=0.001m

Then,

The capacitance of a capacitor is given as

C=εoA/d

Where

εo is constant and has a value of

εo= 8.854 × 10−12 C²/Nm²

C= 8.854E-12×0.002/0.001

C=17.7×10^-12

C=17.7 pF

Value of resistor resistance is 1000ohms

Voltage applied is V = 100V

This Is a series resistor and capacitor (RC ) circuit

In an RC circuit, voltage is given as

Charging system

V=Vo[1 - exp(-t/RC)]

At, t=0, V=100V

Therefore, Vo=100V

We want to know the time, the voltage will deflect 95V.

Then applying our parameters

V=Vo[1 - exp(-t/RC)]

95=100[1-exp(-t/1000×17.7×10^-12)]

95/100=1-exp(-t/17.7×10^-9)

0.95=1-exp(-t/17.7×10^-9)

0.95 - 1 = -exp(-t/17.7×10^-9)

-0.05=-exp(-t/17.7×10^-9)

Divide both side by -1

0.05=exp(-t/17.7×10^-9)

Take In of both sides

In(0.05)=-t/17.7×10^-9

-2.996=-t/17.7×10^-9

-2.996×17.7×10^-9=-t

-t=-53.02×10^-9

Divide both side by -1

t= 53.02×10^-9s

t=53.02 ns

The time to deflect 95V is 53.02nanoseconds

Answer:a

Explanation:

Given

First stone is thrown [tex]15^{\circ}[/tex] above the horizontal with some speed let say u

Second stone is thrown at [tex]15^{\circ}[/tex] below the horizontal with speed u

For a height h of building

For first stone (motion in vertical direction)

using

[tex]v^2-u^2=2ah [/tex]

where v=final velocity

u=initial velocity

a=acceleration

h=displacement

[tex]h=u\sin 15(t_1)-\frac{1}{2}gt_1^2---1[/tex]

For second stone

[tex]h=(-u\sin 15)(t_2)-\frac{1}{2}gt_2^2----2[/tex]

Equating 1 and 2

[tex]u\sin 15(t_1+t_2)-\frac{1}{2}g(t_1-t_2)(t_1+t_2)=0[/tex]

[tex](t_1+t_2)[u\sin 15-4.9(t_1-t_2)]=0[/tex]

as [tex]t_1+t_2[/tex] cannot be zero

so [tex]t_1-t_2=1.05\ s[/tex]

[tex]t_1=t_2+1.056[/tex]

therefore time taken by first stone(thrown upward) will be more.

Answer:

a. The stone thrown upward spends more time in the air.

Explanation:

Given:

projection of first stone, [tex]\theta_1=15^{\circ}[/tex] above the horizontal

initial velocity of projectiles, [tex]u_1=u_2=20\ m.s^{-1}[/tex]

projection of second stone,[tex]\theta_2=15^{\circ}[/tex] below the horizontal

The stone thrown upward will spend more time in the air because it travels more distance than the one thrown downwards.

The stone thrown upwards faces deceleration due to the gravity because it goes opposite to the gravity initially, then reaches a velocity zero for a moment and then falls freely from a greater height.

While the second stone posses an initial velocity downward in the direction of the gravity and which further increases its velocity and it travels a short distance.

Final answer:

The coefficient of elasticity is a measure of the elasticity of a collision, defined as the ratio of speeds after and before the collision. In this case, the coefficient of elasticity is 0.3.

Explanation:

The coefficient of elasticity, also known as the coefficient of restitution (c), is a measure of the elasticity of a collision between a ball and an object. It is defined as the ratio of the speeds after and before the collision.

In this case, the ball strikes a surface with a speed of 10 m/s and rebounds with a speed of 3 m/s. To find the coefficient of elasticity, we can use the formula c = (v_final / v_initial), where v_final is the final velocity and v_initial is the initial velocity.

Therefore, the coefficient of elasticity in this case would be c = (3 m/s / 10 m/s) = 0.3.

Answer:

0.05T

Explanation:

Data given,

area, A=20cm*33cm=0.2m*0.33m=0.066m^2

Number of turns, N=110 turns,

Emf= 72v,

angular speed, W= 200rad/s

magnetic field strength, B= ??

from the expression showing the relationship between induced emf and magnetic field is shown below

[tex]E=NBAW[/tex]

Where N is the number of turns,

E=is the emf,

Bis the magnetic field strength

if we substitute values, we arrive at

[tex]E=NABW\\72=110*0.066*B*200\\B=\frac{72}{1452}\\ B=0.05T[/tex]

Explanation:

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