Determine the inductive reactance for a 50 mH inductor that is across a 15 volt, 400 Hz source.

Answer:

The Reynolds number is 5600.

Explanation:

Given that,

Density = 1400 kg/m³

Viscosity = 0.5 Pa's

Length = 2 m

Speed = 1 m/s

We need to calculate the Reynolds number

Using formula of Reynolds number

[tex]R_{e}=\dfrac{\rho V\times L}{\mu}[/tex]

Where, [tex]\rho[/tex] = density of fluid

v = speed of syrup

l = length of a person

[tex]\mu[/tex]=Viscosity

Put the all value into the formula

[tex]R_{e}=\dfrac{1400\times1\times2}{0.5}[/tex]

[tex]R_{e}=5600[/tex]

Hence, The Reynolds number is 5600.

Answer:

work =p×v =3.27×10^5×1.5×10^-3 =490.5 joule

efficiency =w/q in

:. qin= w/efficiency =490.5/0.225=2180 joule

qout =q in - work =1689.5 joule

q out is work given as heat

The engine must give up 1689.5 J of heat to the low-temperature reservoir after calculating the total work done by the gas and accounting for the engine's efficiency.

To find the amount of work the engine gives up as heat, we first calculate the total work done by the gas using the formula W = PΔV, where W is work, P is pressure, and ΔV is the change in volume. Given a constant pressure of 3.27 x 105 Pa and a change in volume of 1.50 x 10-3 m3, the work done is:

W = PΔV = 3.27 x 105 Pa x 1.50 x 10-3 m3 = 490.5 J.

The efficiency of the engine is the ratio of the useful work output to the total work input, given by  ext_eta = useful work / total work. The equation that relates efficiency, work done (W), and heat given up (Q) is  ext_eta = W / (Q + W). We rearrange the equation to solve for Q:

Q = W /  ext_eta - W

Substituting the known values:

Q = 490.5 J / 0.225 - 490.5 J = 2180 J - 490.5 J = 1689.5 J.

Therefore, the engine must give up 1689.5 J of heat to the low-temperature reservoir.

Answer:

Answer to the question is: 1837.65 millimeters of mercury are equal to 245 kPa.

Explanation:

1 kPa are equal to 7.50062 millimeters of mercury.

Final answer:

To convert the pressure from 245 kPa to mmHg, first convert kPa to atm, then multiply by the conversion factor from atm to mmHg. The pressure is 1837.68 mmHg.

Explanation:

To convert the pressure in a container from kilopascals (kPa) to millimeters of mercury (mmHg), we use the conversion factor that 1 atmosphere (atm) is equivalent to 760 mmHg. First, we convert the given pressure in kilopascals to atmospheres:

1 atm = 101.325 kPa

So, to convert 245 kPa to atm, we divide 245 kPa by 101.325 kPa/atm:

245 kPa / 101.325 kPa/atm = 2.418 atm

Next, we convert atmospheres to millimeters of mercury (mmHg) using the conversion factor:

2.418 atm x760 mmHg/atm = 1837.68 mmHg

Therefore, the pressure in the container is 1837.68 mmHg.

Answer:

d

Explanation:

HOPE THIS HELPS!!

Answer:

Well..

Explanation:

That's impossible. I know because I once weighed 11.1 kN, and I was temporarily immobile. It's probably the same for a car, and therefore it can not be "traveling" anywhere at all.. unless you put the car on an airplane or a boat or something.

Answer:

rate of change in temperature of copper is more than the rate of change in temperature of aluminium.

so here copper will reach to our body temperature first

Explanation:

As we know that rate of energy absorb by the two sphere is same

so here we will have

[tex]\frac{dQ}{dt} = ms\frac{\Delta T}{\Delta t}[/tex]

now for copper sphere we will have

[tex]\frac{dQ}{dt} = 50(0.4)\frac{\Delta T}{\Delta t}[/tex]

[tex]\frac{\Delta T}{\Delta t} = \frac{1}{20}\frac{dQ}{dt}[/tex]

now for Aluminium sphere we will have

[tex]\frac{dQ}{dt} = 25(0.9)\frac{\Delta T}{\Delta t}[/tex]

[tex]\frac{\Delta T}{\Delta t} = \frac{1}{22.5}\frac{dQ}{dt}[/tex]

So rate of change in temperature of copper is more than the rate of change in temperature of aluminium.

so here copper will reach to our body temperature first

Using the principle of continuity for incompressible fluids, if the diameter of a blood vessel is reduced by 59.0% due to plaque, the speed of the blood just as it enters this section will be approximately 2.44 times its initial speed.

The subject of the question falls under the topic of fluid flow in physics, specifically concerning the principle of continuity for incompressible fluids. This principle, often applied in fluid dynamics, suggests that in an area of steadily flowing fluid, the mass passing through one cross-section in a unit of time equals the mass passing through other sections.

Given this principle, if the cross-sectional area of the blood vessel decreases due to plaque buildup, the speed of the blood flow must increase accordingly to maintain a steady flow rate. If the diameter of the vessel decreases by 59.0%, the cross-sectional area A, which is proportional to the square of the diameter (A ~ D²), will be reduced to 0.41 of its original value (because (1 - 59/100)² = 0.41). Therefore, the speed V would be 1/0.41, or approximately 2.44 times the original speed V0.

So, if the blood vessel's diameter is reduced by 59.0%, then just as the blood enters the blocked portion of the vessel, its speed V will be 2.44 times the initial speed V0.

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Answer:

2.07 m/s

Explanation:

m = 1.07 x 10^5 kg, u1 = 3.41 m/s, u2 = 1.4 m/s

Let the speed of three coupled car after collision is v

Use conservation of momentum

m x u1 + 2 m x u2 = 3 m x v

u1 + 2 u2 = 3 v

3.41 + 2 x 1.4 = 3 v

v = 2.07 m/s

Answer:

Work done by the gravitational force is 37 mJ.

Explanation:

It is given that,

Mass of the raindrop, [tex]m=3.26\times 10^{-5}\ kg[/tex]

It falls from a height of, h = 115 m

It falls vertically at constant speed under the influence of gravity and air resistance. We need to find the  work done on the raindrop by the gravitational force. It is given by :

[tex]W=mgh[/tex]

[tex]W=3.26\times 10^{-5}\ kg\times 9.8\ m/s^2\times 115\ m[/tex]

W = 0.0367 J

or

W = 0.037 J = 37 mJ

So, the work done on the raindrop by the gravitational force is 37 mJ. Hence, this is the required solution.

Final answer:

The work done on a raindrop of mass 3.26 times [tex]10^{-5}[/tex] kg by the gravitational force.

Explanation:

To calculate the work done on a raindrop by the gravitational force as it falls, we use the formula for work: W = mgh, where W is the work done, m is the mass of the object, g is the acceleration due to gravity (9.8 m/s2), and h is the height the object falls through.

In this case, the mass m of the raindrop is 3.26 times 10-5 kg, and the height h is 115 m. So:

W = (3.26 times 10-5 kg)(9.8 [tex]m/s^2[/tex])(115 m) = 0.0368134 J

Therefore, the work done on the raindrop by the gravitational force as it falls 115 m is approximately 0.0368134 joules.

The average current passing through a device is given by:

I = Q/Δt

I is the average current

Q is the amount of charge that has passed through the device

Δt is the amount of elapsed time

Given values:

Q = 4.00C

Δt = 4.00hr = 14400s

Plug in the values and solve for I:

I = 4.00/14400

I = 0.000277777778A

I = 0.278mA

Final answer:

The current produced by the solar cells of a pocket calculator through which 4.00 C of charge passes in 4.00 hours is 0.278 milliamperes.

Explanation:

The current produced by the solar cells of a pocket calculator when 4.00 C of charge passes through it in 4.00 hours can be calculated using the formula for electric current I = Q / t, where I is the current in amperes, Q is the charge in coulombs, and t is the time in seconds.

To find the current in milliamperes (mA), first convert the time to seconds:

4.00 hours × 3600 seconds/hour = 14400 seconds.

Next, use the formula to calculate current:

I = 4.00 C / 14400 s = 0.00027778 A,

which is equivalent to 0.278 mA

Answer:

4.4 cm

Explanation:

B = 1.98 mT = 1.98 x 10^-3 T, v = 1.53 x 10^7 m/s, m = 9.1 x 10^-31 kg

q = 1.6 x 10^-19 C

(a) The force due to the magnetic field is balanced by the centrpetal force

mv^2 / r = q v B

r = m v / q B

r = (9,1 x 10^-31 x 1.53 x 10^7) / (1.98 x 10^-3 x 1.6 x 10^-19)

r = 0.044 m = 4.4 cm

Explanation:

It is given that,

Mass of the shot, m = 7.27 kg

Time taken to accelerate, t = 1.3 s

It is shot from rest to 16 m/s and it raises to a height of 0.9 m. We need to find the power output of the shot-putter. It is given by :

[tex]P=\dfrac{energy}{time}[/tex]

Energy = kinetic energy + potential energy

[tex]E=\dfrac{1}{2}\times 7.27\ kg\times (16\ m/s)^2+7.27\ kg\times 9.8\ m/s^2\times 0.9\ m[/tex]

E = 994.68 J

Power, [tex]P=\dfrac{994.68\ J}{1.3\ s}[/tex]

P = 765.13 Watts

We know that, 1 horse power = 745.7 watts

Or P = 1.02 horse power

Hence, this is the required solution.

Answer:

[tex]T_C=256.2^{\circ}C[/tex]

Explanation:

Given that,

Efficiency of heat engine, [tex]\eta=40\%=0.4[/tex]

Temperature of hot source, [tex]T_H=427^{\circ}C[/tex]

We need to find the temperature of cold sink i.e. [tex]T_C[/tex]. The efficiency of heat engine is given by :

[tex]\eta=1-\dfrac{T_C}{T_H}[/tex]

[tex]T_C=(1-\eta)T_H[/tex]

[tex]T_C=(1-0.4)\times 427[/tex]

[tex]T_C=256.2^{\circ}C[/tex]

So, the temperature of the cold sink is 256.2°C. Hence, this is the required solution.

Answer:

0.94 m/s^2 downwards

Explanation:

m = 70 kg, m g = 70 x 9.8 = 686 N

R = 620 N

Let the acceleration be a, as the apparent weight decreases so the elevator is moving downwards with an acceleration a.

mg - R = ma

686 - 620 = 70 x a

a = 0.94 m/s^2

Answer:

[tex]C = 2.48 \times 10^{-4} Farad[/tex]

Explanation:

As per the equation of voltage on capacitor we know that

[tex]V = V_{max}(1 - e^{-\frac{t}{\tau}})[/tex]

now we know that voltage reached to its 80% of maximum value in 4 second time

so we will have

[tex]0.80 V_{max} = V_{max}(1 - e^{-\frac{4}{\tau}})[/tex]

[tex]0.20 = e^{-\frac{4}{\tau}}[/tex]

[tex]-\frac{4}{\tau} = ln(0.20)[/tex]

[tex]-\frac{4}{\tau} = -1.61[/tex]

[tex]\tau = 2.48[/tex]

as we know that

[tex]\tau = RC[/tex]

[tex](10 k ohm)(C) = 2.48[/tex]

[tex]C = 2.48 \times 10^{-4} Farad[/tex]

Answer:

Part a)

[tex]V = 7.2 \times 10^{11} Volts[/tex]

Part b)

this is a large potential which can not be possible because at this high potential the air will break down and the charge on the sphere will decrease.

Part C)

here we can assume the sphere is placed at vacuum so that there is no break down of air.

Explanation:

Part a)

As we know that the potential near the surface of metal sphere is given by the equation

[tex]V = \frac{kQ}{R}[/tex]

here we have

Q = 8 C

R = 10.0 cm

now we have

[tex]V = \frac{(9\times 10^9)(8 C)}{0.10}[/tex]

[tex]V = 7.2 \times 10^{11} Volts[/tex]

Part b)

this is a large potential which can not be possible because at this high potential the air will break down and the charge on the sphere will decrease.

Part C)

here we can assume the sphere is placed at vacuum so that there is no break down of air.

Final answer:

The voltage near a 10.0 cm diameter metal sphere with 8.00 C of excess charge is calculated to be 1.438 x 10^12 V, which is unreasonable due to the high value leading to inevitable discharge. The assumption of an 8.00 C charge on such a small sphere is responsible for this unrealistic result.

Explanation:

Calculating the Voltage near a Charged Sphere

To find the voltage near a 10.0 cm diameter metal sphere with an excess positive charge of 8.00 C, we use the formula V = kQ/r, where V is the voltage, k is Coulomb's constant (8.99 x 10^9 N m^2/C^2), Q is the charge, and r is the radius of the sphere. For a diameter of 10.0 cm, the radius (r) is 0.05 m. Thus, V = (8.99 x 10^9 N m^2/C^2 * 8.00 C) / 0.05 m = 1.438 x 10^12 V.

Unreasonable Voltage

This voltage is extremely high and unreasonable because a metal sphere of that size could not sustain such a high voltage without discharging. The consequence of such a high voltage would include electric breakdown of the air around the sphere, leading to sparks or lightning-like discharges.

Erroneous Assumptions

The assumption responsible for this unreasonable result is the magnitude of charge being considered. An 8.00 C charge on a small metal sphere is significantly larger than what could realistically accumulate on the surface, given the limits of charge density and material breakdown thresholds.

Answer:

(a) 0.33 second

(b) 6 cm/s

Explanation:

Frequency, f = 3 waves per second

wavelength, λ = 2 cm = 0.02 m

(a) The period of wave is defined as the time taken by the wave to complete one oscillation. It is the reciprocal of frequency.

T = 1 / f = 1 / 3 = 0.33 second

(b) the relation between wave velocity, frequency and wavelength is given by

v = f x λ

v = 3 x 0.02 = 0.06 m /s

v = 6 cm /s

Final answer:

The period of the waves generated by a grasshopper in water is 0.333 seconds, and the wave velocity is 0.06 m/s.

Explanation:

Calculating the Period and Wave Velocity

When dealing with waves generated by a grasshopper floating in water, two key properties to determine are the period of the waves and the wave velocity.

(a) The Period of the Waves

The period (T) of a wave is the amount of time it takes for one complete wave cycle to pass. It can be calculated as the inverse of the frequency (f), which is the rate at which waves are generated. The formula to find the period is:

T = 1/f

In this case, the grasshopper generates waves at a frequency of 3 waves per second (3 Hz). Therefore, the period is:

T = 1/3 Hz = 0.333 seconds

(b) The Wave Velocity

The velocity (v) of a wave is determined by multiplying the frequency (f) of the wave by its wavelength (λ). The formula for wave velocity is:

v = f × λ

Here, the wavelength given is 2 cm, which we need to convert to meters (since the SI unit for velocity is m/s). Thus:

λ = 2 cm = 0.02 m

The velocity of the waves generated by the grasshopper is then:

v = 3 Hz × 0.02 m = 0.06 m/s

The speed of an electron with a kinetic energy of 1.25 eV is approximately 1.57 x 10⁶ m/s, and the speed of a proton with the same kinetic energy is approximately 5.29 x 10⁵ m/s.

To calculate the speed of an electron and a proton with a kinetic energy of 1.25 electron volts (eV), we can use the kinetic energy formula and relate it to the speed of the particles. The kinetic energy (KE) of a particle is given by:

KE = (1/2) * m * v²

Where:

KE = kinetic energy

m = mass of the particle

v = speed of the particle

We are given the kinetic energy in electron volts (eV), but we need to convert it to joules (J) since the mass is given in kilograms (kg). The conversion factor is 1 eV = 1.60219 x 10⁻¹⁹ J.

So, the kinetic energy in joules is:

KE = 1.25 eV * 1.60219 x 10⁻¹⁹ J/eV = 2.0027375 x 10⁻¹⁹ J

Now, we can rearrange the kinetic energy formula to solve for the speed (v):

v = √((2 * KE) / m)

For an electron:

Mass of electron (mₑ) = 9.11 x 10⁻³¹ kg

v(electron) = √((2 * 2.0027375 x 10⁻¹⁹ J) / (9.11 x 10⁻³¹ kg))

Calculating this gives us the speed of the electron.

For a proton:

Mass of proton (m_p) = 1.67 x 10⁻²⁷ kg

v(proton) = √((2 * 2.0027375 x 10⁻¹⁹ J) / (1.67 x 10⁻²⁷ kg))

Calculating this gives us the speed of the proton.

Now, let's calculate these speeds.

After performing the calculations, the speed of the electron is approximately 1.57 x 10⁶ m/s, and the speed of the proton is approximately 5.29 x 10⁵ m/s.

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Answer:

The distance between the person and the building is 68.48 meters.

Explanation:

It is given that,

Angle of elevation, θ = 30.6 degrees

Height of building, MP = 42 m

Height of person, AB = 1.5 m

We need to find the distance between person and building. It is given by BP.

Since, MN + NP = 42

So, MN = 40.5 m

Using trigonometric equation as :

[tex]tan\theta=\dfrac{MN}{AN}[/tex]

[tex]tan(30.6)=\dfrac{40.5}{AN}[/tex]

AN = 68.48 meters.

So, the distance between the person and the building is 68.48 meters. Hence, this is the required solution.

To determine the distance from a building, we use trigonometry and the formula adjacent = opposite / tangent(angle), taking into account the height of the building minus your height. The distance is calculated to be approximately 68.88 meters.

To find out how far you are from the building, we need to calculate the horizontal distance from the building's base to the point where you are standing. To do this, we can use trigonometry, specifically the tangent function which relates the angle of elevation to the opposite side and the adjacent side of a right-angle triangle. We need to consider the height of the building minus your height to find the correct opposite side.

Since the building is 42.0 meters tall, and you are 1.50 meters tall, the effective height we are looking at is 42.0 m - 1.50 m = 40.5 m. The angle of elevation you are looking at is 30.6 degrees. By using the formula tangent (angle) = opposite / adjacent, we can rearrange this to find the adjacent side (the distance from you to the building): adjacent = opposite / tangent (angle).

Therefore, the distance from you to the building is approximately adjacent = 40.5 m / tan(30.6°). Plugging in the values, we get:

Distance = 40.5 m / tan(30.6°) ≈ 40.5 m / 0.588 ≈ 68.88 m.

So, you are approximately 68.88 meters away from the building.

Answer:

Explanation:

When the pendulum falls freely the net acceleration due to gravity is zero.

As we know that the time period of simple pendulum is inversely proportional to the square root of acceleration due to gravity, thus the time period becomes infinity.

In freefall, the pendulum's effective acceleration due to gravity becomes zero, causing the pendulum to not swing, and its period becomes theoretically infinite and immeasurable.

Effect of Freefall on a Pendulum's Period

When considering simple pendulum motion in an elevator under normal conditions, we can determine its periodic time (T) using the formula T = 2π√(L/g), where L is the length of the pendulum and g is the acceleration due to gravity. This equation illustrates that the period of the pendulum (T) is affected by two variables: the length of the pendulum (L) and the acceleration due to gravity (g).

When the elevator is in free fall, the effective acceleration g becomes zero because the elevator and the pendulum are both in a state of free fall with the same acceleration due to gravity. Therefore, in this scenario, the pendulum would experience weightlessness and would not oscillate, resulting in an infinite theoretical oscillation period, making the concept of a period inapplicable.

The period is normally independent of mass or amplitude for small angles, but since freefall changes the acceleration experienced by the pendulum to zero, it significantly affects the pendulum's oscillation, negating the normal conditions for calculating a pendulum's period.

Answer:

Weight in planet X = 0.413 N

Explanation:

Weight = Mass x Acceleration due to gravity.

W = mg

Mass, m = 100 g = 0.1 kg

We have equation of motion s = ut + 0.5 at²

Displacement, s = 10 m

Initial velocity, u = 0 m/s

Time, t = 2.2 s

Substituting

        s = ut + 0.5 at²

        10 = 0 x 2.2 + 0.5 x a x 2.2²        

        a = 4.13 m/s²

Acceleration due to gravity, a = 4.13 m/s²

W = mg = 0.1 x 4.13 = 0.413 N

Weight in planet X = 0.413 N

In Planet X, a 100-g ball is released from rest from a height of 10.0 m and it takes 2.2 s for it to reach the ground. The weight of the ball on the surface of Planet X is 0.41 N.

In physics, gravitational acceleration is the acceleration of an object in free fall within a vacuum (and thus without experiencing drag).

A 100-g (m) ball is released from rest from a height of 10.0 m (s) and it takes 2.2 s (t) for it to reach the ground. We can calculate the gravitational acceleration using the following kinematic equation.

s = 1/2 × g × t²

g = 2 s / t² = 2 (10.0 m) / (2.2 s)² = 4.1 m/s²

We will use Newton´s second law of motion.

w = m × g = 0.100 kg × 4.1 m/s² = 0.41 N

In Planet X, a 100-g ball is released from rest from a height of 10.0 m and it takes 2.2 s for it to reach the ground. The weight of the ball on the surface of Planet X is 0.41 N.

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Answer:

The amplifier gain is 30 dB.

(c) is correct option.

Explanation:

Given that,

Output power = 100 mW

Input power = 0.1 mW

We need to calculate the amplifier gain in dB

Using formula of power gain

[tex]a_{p}=10\ log_{10}(A_{p})[/tex]....(I)

We calculate the [tex]A_{p}[/tex]

[tex]A_{p}=\dfrac{P_{o}}{P_{i}}[/tex]

[tex]A_{p}=\dfrac{100}{0.1}[/tex]

[tex]A_{p}=1000[/tex]

Now, put the value of  [tex]A_{p}[/tex] in equation (I)

[tex]a_{p}=10\ log_{10}(1000)[/tex]

[tex]a_{p}=10\times log_{10}10^{3}[/tex]

[tex]a_{p}=10\times 3log_{10}10[/tex]

[tex]a_{p}=30\ dB[/tex]

Hence, The amplifier gain is 30 dB.

The gain of an amplifier, given a power output of 100 mW and an input power of 0.1 mW, can be calculated using the gain formula in decibels, which results in a gain of 30 dB.

In this context, the gain of the amplifier can be calculated using the formula for Gain in decibels (dB), which is 10 times the log base 10 of the output power divided by the input power. Therefore, we first divide 100 mW by 0.1 mW, which gives us 1000. Taking the log base 10 of 1000 returns 3, and multiplying 3 by 10 gives us a gain of 30 dB.

So the correct answer to your question: 'An amplifier has a power output of 100 mW when the input power is 0.1 mW, what is the amplifier gain?' is option c which states that the gain is 30 dB.

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Answer:

[tex]p_2 = 664081 N/m^{2}[/tex]

Explanation:

from the ideal gas law we have

PV = mRT

[tex]P = \rho RT[/tex]

[tex]\rho = \frac{P}{RT}[/tex]

HERE  R is gas constant for dry air  =  287  J K^{-1} kg^{-1}

[tex]\rho = \frac{7.00 10^{5}}{287(18+273)}[/tex]

[tex]\rho = 8.38 kg/m^{3}[/tex]

We know by ideal gas law

[tex]\rho = \frac{m_1}{V_1}[/tex]

[tex]m_1 = \rho V_1 = 8.38 *2*10^{-3}[/tex]

[tex]m_1 = 0.0167 kg[/tex]

for m_2

[tex]m_2 = \rho V_i - V_removed[/tex]

[tex]m_2 = 8.38*(.002 - 10^{-4})[/tex]

[tex]m_2 = 0.0159 kg[/tex]

WE KNOW

PV = mRT

V, R and T are constant therefore we have

P is directly proportional to mass

[tex]\frac{p_2}{p_1}=\frac{m_2}{m_1}[/tex]

[tex]p_2 = p_1 * \frac{m_2}{m_1}[/tex]

[tex]p_2 =7*10^{5} * \frac {.0159}{0.0167}[/tex]

[tex]p_2 = 664081 N/m^{2}[/tex]

This problem can be solved using Boyle's Law, which relates the pressure and volume of a gas at a constant temperature. The question asks for the new pressure of a bicycle tire after letting out a certain volume of air. The answer is approximately 7.37 x 10⁵ Pa.

The subject of this question is gas laws, specifically Boyle's Law which states that the pressure and volume of a gas have an inverse relationship when the temperature is kept constant. Assuming the temperature and volume of the tire remain constant before and after you let out the air, when a volume of 100 cm³ (which we will convert to 0.1 L for consistency) of air is let out, the new total volume of the gas is 1.9 L.

According to Boyle's Law, P1*V1 = P2*V2, where P1 and V1 represent the initial pressure and volume, and P2 and V2 represent the final pressure and volume. Plugging the values into this equation, we get:

(7.00 x 10⁵ Pa)(2.00 L) = P2 * (1.9 L)

Which gives us:

P2 = (7.00 x 10⁵ Pa * 2.00 L) / 1.9 L

Therefore, the pressure in the bike tire after letting out 100 cm³ of gas is approximately 7.37 x 10⁵ Pa.

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Answer:

The wavelength of the light is 633 nm.

Explanation:

Given that,

Distance between the two slits d= 0.025 cm

Distance between the screen and slits D = 120 cm

Distance between the slits y= 1.52 cm

We need to calculate the angle

Using formula of double slit

[tex]\tan\theta=\dfrac{y}{D}[/tex]

Where, y = Distance between the slits

D = Distance between the screen and slits

Put the value into the formula

[tex]\tan\theta=\dfrac{1.52}{120}[/tex]

[tex]\theta=\tan^{-1}\dfrac{1.52}{120}[/tex]

[tex]\theta=0.725[/tex]

We need to calculate the wavelength

Using formula of wavelength

[tex]d\sin\theta=n\lambda[/tex]

Put the value into the formula

[tex]0.025\times\sin0.725=5\times\lambda[/tex]

[tex]\lambda=\dfrac{0.025\times10^{-2}\times\sin0.725}{5}[/tex]

[tex]\lambda=6.326\times10^{-7}\ m[/tex]

[tex]\lambda=633\ nm[/tex]

Hence, The wavelength of the light is 633 nm.

Answer:

115 m/s, 414 km/hr

Explanation:

There are two forces acting on a skydiver: gravity and air resistance (drag).  At terminal velocity, the two forces are equal and opposite.

∑F = ma

D − mg = 0

D = mg

Drag force is defined as:

D = ½ ρ v² C A

where ρ is the fluid density,

v is the velocity,

C is the drag coefficient,

and A is the cross sectional surface area.

Substituting and solving for v:

½ ρ v² C A = mg

v² = 2mg / (ρCA)

v = √(2mg / (ρCA))

We're given values for m and A, and we know the value of g.  We need to look up ρ and C.

Density of air depends on pressure and temperature (which vary with elevation), but we can estimate ρ ≈ 1.21 kg/m³.

For a skydiver falling headfirst, C ≈ 0.7.

Substituting all values:

v = √(2 × 80.0 kg × 9.8 m/s² / (1.21 kg/m³ × 0.7 × 0.140 m²))

v = 115 m/s

v = 115 m/s × (1 km / 1000 m) × (3600 s / hr)

v = 414 km/hr

The terminal velocity of the skydiver in m/s and km/h is;  115m/s  and  414 km/h

Using Given data :

mass of skydiver ( M ) = 80 kg

Cross sectional surface area ( A ) = 0.14 m^2

p ( fluid density ) ≈ 1.21 kg/m³.

C ( drag coefficient ) = 0.7

Determine the terminal velocity of the skydiver

At terminal velocity drag force and gravity is equal and opposite therefore canceling out each other

∑ F = ma

Drag force - Mg = 0

therefore;  D = Mg ----- ( 1 )

where D ( drag force ) = 1/2 pv² C A ---- ( 2 )

p = fluid density , C = drag coefficient , A = cross sectional area

v = velocity

Back to equations 1 and 2  ( equating them )

1/2 pv² CA = Mg ---- ( 3 )

v² = 2mg / ( p C A )

∴ V = √ ( 2mg / (p C A ))

V = √ ( 2 * 80 * 9.8 ) / ( 1.21 * 0.7 * 0.140 ))

    = 115 m/s  

also  V = 414 km/h

Hence we can conclude that the terminal velocity of the skydiver is in m/s and km/h are 115m/s and  414 km/h

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Answer:

10042.6 ohm

Explanation:

f = 10 kHz = 10000 Hz, L = 36 mH = 0.036 H, R = 10 kilo Ohm = 10000 ohm

C = 5 nF = 5 x 10^-9 F

XL = 2 x π x f x L

XL = 2 x 3.14 x 10000 x 0.036 = 2260.8 ohm

Xc = 1 / ( 2 x π x f x C) = 1 / ( 2 x 3.14 x 10000 x 5 x 10^-9)

Xc = 3184.7 ohm

Total impedance is Z.

Z^2 = R^2  + (XL - Xc)^2

Z^2 = 10000^2 + ( 2260.8 - 3184.7 )^2

Z = 10042.6 ohm

Answer:

Final velocity, v = 1.74 m/s

Explanation:

Given that,

Mass of car 1, m₁ = 22509 kg

Velocity of car 1, v₁ = 4.21 m/s

Mass of car 2, m₂ = 31647 kg

It is stationary, v₂ = 0

Let v be the velocity of the two cars after they collide and attach. It can be calculated using law of conservation of momentum as :

[tex]m_1v_1+m_2v_2=(m_1+m_2)v[/tex]

[tex]v=\dfrac{m_1v_1+m_2v_2}{(m_1+m_2)}[/tex]

[tex]v=\dfrac{22509\ kg\times 4.21\ m/s+0}{22509\ kg+31647\ kg}[/tex]

v = 1.74 m/s

So, the velocity of two cars after the collision is 1.74 m/s. Hence, this is the required solution.

Answer:

Spongebob: Bye Mr. Krabs! Bye Squidward! BYE SQUIDWARD!

Patrick: (clearly triggered) Why'd you say "bye squidward" twice?

Spongebob: I LiKe SqUiDwArD

The International Space Station's operational weight in orbit is effectively zero due to its state of continuous free-fall around Earth, even though the gravitational force at its altitude is not significantly less than on Earth's surface.

The question relates to the weight of the International Space Station (ISS) when in orbit. Weight in physics is defined as the force exerted on an object due to gravity, calculated as the product of mass and gravitational acceleration (g). On Earth's surface, g is approximately 9.81 m/s2, but this value decreases with altitude due to the equation g = GME / r2, where G is the gravitational constant, ME is Earth's mass, and r is the distance from Earth's center. At the ISS's altitude (> 350 km), g is about 8.75 m/s2. However, it's crucial to understand that while the ISS has a significant mass, leading to a large weight calculation on Earth, its apparent weight in orbit is effectively zero due to it being in a continuous free-fall state around Earth, experiencing microgravity. This explains why astronauts appear weightless, even though the actual gravitational force at that altitude is not much less than on Earth's surface. Therefore, while the ISS has a calculable weight based on its mass and Earth's gravitational pull at its altitude, its operational weight in orbit, in terms of the experience within it, is zero.

Answer:

49.85 V

Explanation:

u = 0, s = 5.62 cm, t = 1.15 x 10^-6 s

Let the electric field is E and voltage is V.

Use second equation of motion

s = ut + 1/2 a t^2

5.62 x 10^-2 = 0 + 0.5 a x (1.15 x 10^-6)^2

a = 8.5 x 10^10 m/s^2

m x a = q x E

E = m x a / q

E = (1.67 x 10^-27 x 8.5 x 10^10) / (1.6 x 10^-19)

E = 887.19 V/m

V = E x s

V = 887.19 x 5.62 x 10^-2 = 49.85 V