When a comet comes close to the Sun, what happens to it that makes it bright and easier to see?

Answer:

One characteristic of mass wasting processes is that they move materials relatively short distances compared to streams. - d.

Answer:

Work, W = 846.72 Joules

Explanation:

Given that,

Mass of the watermelon, m = 4.8 kg

It is dropped from rest from the roof of 18 m building. We need to find the work done by the gravity on the watermelon from the roof to the ground. It is same as gravitational potential energy i.e.

W = mgh

[tex]W=4.8\ kg\times 9.8\ m/s^2\times 18\ m[/tex]

W = 846.72 Joules

So, the work done by the gravity on the watermelon is 846.72 Joules. Hence, this is the required solution.

The energy stored by the electric field between two plates, known as a parallel-plate capacitor, is calculated using the respective formulas for voltage, capacitance and energy in a capacitor, factoring in the charge of the plates, surface area, and separation distance.

The energy stored by the electric field between two parallel plates (known as a parallel-plate capacitor) can be calculated using the formula for the energy stored in a capacitor, which is U = 0.5 * C * V^2. Here, V is the voltage across the plates and C is the capacitance of the capacitor.

To calculate V, we use the relation V = E * d, where E is the electric field, which can be found using the formula E = Q/A (Charge per Area), and d is the distance separating the plates. In this case, the charge Q is 14 nC, the area A is (8.5 cm)^2 and the distance d is 2.5 mm.

Finally, to find C, we use the formula C = permittivity * (A/d), where the permittivity of free space is approximately 8.85 × 10^-12 F/m. With these variables, C, V and U can be calculated accordingly.

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The energy stored in the electric field of a parallel-plate capacitor can be calculated using the dimensions of the plates, their separation, and charges. For this particular scenario, the energy stored would be approximately 0.44 μJ.

In this scenario, we have a parallel-plate capacitor where each plate is a square with side 8.5 cm and has a charge of 14 nC. The plates are separated by a 2.5mm gap. Using this information, we can calculate the energy stored in the capacitor.

Firstly, the surface area A of the plates can be calculated using the formula A = s², where s is the side of the square. Converting 8.5 cm to meters, we get s = 0.085 m. So, A = (0.085 m)² = 0.007225 m².

Next, we calculate the electric field E between the plates using the formula E = Q/ε₀A, where Q is the charge on one plate and ε₀ is the permittivity of free space (8.85 x 10^-12 F/m). Converting 14 nC to coulombs gives Q = 14 x 10^-9 C. Substituting these values into the formula, we find E ≈ 224.717 kV/m.

The energy U stored in an electric field can be calculated using the formula U = 0.5ε₀EA². Substituting the earlier calculated values, we find U ≈ 0.44 μJ.

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Answer:

The distance between adjacent antinodes of the standing wave is 1.25 m.

Explanation:

Given that,

Frequency f= 120 MHz

We need to calculate the distance between adjacent antinodes of the standing wave

Using formula of distance

[tex]\Delta x=\dfrac{\lambda}{2}[/tex].....(I)

We know that,

[tex]\lambda=\dfrac{c}{f}[/tex]

Put the value of [tex]\lambda[/tex] in to the equation (I)

[tex]\Delta x=\dfrac{c}{2f}[/tex]

Where, c = speed of light

f = frequency

Put the all value into the formula

[tex]\Delta x=\dfrac{3\times10^{8}}{2\times120\times10^{6}}[/tex]

[tex]\Delta x=1.25\ m[/tex]

Hence, The distance between adjacent antinodes of the standing wave is 1.25 m.

Answer:

Total energy, [tex]TE=3.006\times 10^{-10}\ J[/tex]

Explanation:

It is given that, a proton in a certain particle accelerator has a kinetic energy that is equal to its rest energy.  Let KE is the kinetic energy of the proton and E₀ is its rest energy. So,

[tex]KE=E_o[/tex]              

The total energy of the proton is equal to the sun of kinetic energy and the rest mass energy.

[tex]TE=KE+E_o[/tex]

[tex]TE=2E_o[/tex]

[tex]TE=2m_{proton}c^2[/tex]

[tex]TE=2\times 1.67\times 10^{-27}\ kg\times (3\times 10^8\ m/s)^2[/tex]

[tex]TE=3.006\times 10^{-10}\ J[/tex]

So, the total energy of the proton as measured by a physicist working with the accelerator is [tex]3.006\times 10^{-10}\ J[/tex]

Answer:

[tex]\omega' = 0.89\omega[/tex]

Explanation:

Rotational inertia of uniform solid sphere is given as

[tex]I = \frac{2}{5}MR^2[/tex]

now we have its angular speed given as

angular speed = [tex]\omega[/tex]

now we have its final rotational kinetic energy as

[tex]KE = \frac{1}{2}(\frac{2}{5}MR^2)\omega^2[/tex]

now the rotational inertia of solid cylinder about its axis is given by

[tex]I = \frac{1}{2}MR^2[/tex]

now let say its angular speed is given as

angular speed = [tex]\omega'[/tex]

now its rotational kinetic energy is given by

[tex]KE = \frac{1}{2}(\frac{1}{2}MR^2)\omega'^2[/tex]

now if rotational kinetic energy of solid sphere is same as rotational kinetic energy of solid sphere then

[tex]\frac{1}{2}(\frac{2}{5}MR^2)\omega^2 = \frac{1}{2}(\frac{1}{2}MR^2)\omega'^2[/tex]

[tex]\frac{2}{5}\omega^2 = \frac{1}{2}\omega'^2[/tex]

[tex]\omega' = 0.89\omega[/tex]

Answer:

w_cyl = ±√(4/5) ω

Explanation:

Kinetic energy

E = (1/2)Iw²

where I is the moment of inertia and w the angular frequency of rotation.

The moment of inertia of a solid sphere of mass M and radius R is:

I = (2/5)MR²,  

Solid cylinder is of mass M and radius R

I = (1/2)MR²

Equate the energies through

(1/2)×(2/5) M R²× (w_sphere)² = (1/2)× (1/2) MR² × (w_cyl)²

(w_cyl)² = (4/5)(w_sphere)²

w_cyl = ±√(4/5) ω

The energy of rotation is independent of the direction of rotation

Answer:

v = 895 m/s

Explanation:

Time period is given as 39 Earth Days

[tex]T = 39 days \times 24 hr \times 3600 s[/tex]

[tex]T = 3369600 s[/tex]

now the radius of the orbit is given as

[tex]r = 4.8 \times 10^8 m[/tex]

so the total path length is given as

[tex] L = 2 \pi r[/tex]

[tex]L = 2\pi (4.8 \times 10^8)[/tex]

[tex]L = 3.015 \times 10^9 [/tex]

now the speed will be given as

[tex]v = \frac{L}{T}[/tex]

[tex]v = \frac{3.015 \times 10^9}{3369600} [/tex]

[tex]v = 895 m/s[/tex]

Answer:

Explanation:

As we know that magnetic flux is given by

[tex]\phi = B.A[/tex]

[tex]\phi = B.\pi r^2[/tex]

now from Faraday's law

[tex]EMF = \frac{d\phi}{dt}[/tex]

[tex]EMF = \frac{d(B. \pi r^2)}{dt}[/tex]

[tex]EMF = 2\pi r B \frac{dr}{dt}[/tex]

now we have

[tex]r = 40/2 = 20 cm[/tex]

B = 12 T

[tex]\frac{dr}{dt} = 5 \times 10^{-3} m/s[/tex]

Part a)

now at t = 1 s

r = 20 + 0.5 = 20.5 cm

[tex]EMF = (2\pi (0.205))(12)(5 \times 10^{-3})[/tex]

[tex]EMF = 0.077 Volts[/tex]

Part b)

now at t = 10 s

r = 20 + 0.5(10) = 25 cm

[tex]EMF = (2\pi (0.25))(12)(5 \times 10^{-3})[/tex]

[tex]EMF = 0.094 Volts[/tex]

Answer:

v = 2917.35 m/s

Explanation:

let Fc be the centripetal force avting on the satelite , Fg is the gravitational force between mars and the satelite, m is the mass of the satelite and M is the mass of mars.

at any point in the orbit  the forces acting on the satelite are balanced such that:

Fc = Fg

mv^2/r = GmM/r^2

v^2 = GM/r

   v = \sqrt{GM/r}  

      = \sqrt{(6.6708×10^-11)(6.38×10^23)/(3.38×10^6 + 1.62×10^6)}

      = 2917.35 m/s

Therefore, the orbital velocity of the satelite orbiting mars is 2917.35 m/s.

Answer:

0.45 m

Explanation:

m = 1.23 kg, k = 37.4 N/m, vmax = 2.48 m/s

velocity is maximum when it passes through the mean position.

[tex]T = 2\pi \sqrt{\frac{m}{k}}[/tex]

[tex]T = 2\pi \sqrt{\frac{1.23}{37.4}}[/tex]

T = 1.139 sec

w = 2 π / T

w = 2 x 3.14 / 1.139

w = 5.51 rad / s

Vmax = w A

Where, A be the amplitude

2.48 = 5.51 A

A = 2.48 / 5.51 = 0.45 m

Final answer:

The amplitude of oscillation for the given spring oscillator is 0.58 m.

Explanation:

The amplitude of oscillation can be calculated using the equation:

A = vmax/ω

where A is the amplitude, vmax is the maximum speed, and ω is the angular frequency.

The angular frequency can be calculated using the equation:

ω = sqrt(k/m)

where k is the spring constant and m is the mass of the oscillator.

Substituting the given values into the equations:

ω = sqrt(37.4 N/m / 1.23 kg) = 4.29 rad/s

A = 2.48 m/s / 4.29 rad/s = 0.58 m

Therefore, the amplitude of oscillation is 0.58 m.

Answer:

5.4 x 10⁶ Pa

Explanation:

h = depth to which penguins dive under seawater = 530 m

P₀ = Atmospheric pressure = 1.013 x 10⁵ pa

ρ = density of seawater = 1025 kg/m³

P = absolute pressure experienced by penguin at that depth

Absolute pressure is given as

P = P₀ + ρgh

Inserting the values

P = 1.013 x 10⁵ + (1025) (9.8) (530)

P = 5.4 x 10⁶ Pa

A. 409 Hz

The fundamental frequency of a string is given by:

[tex]f_1=\frac{1}{2L}\sqrt{\frac{T}{m/L}}[/tex]

where

L is the length of the wire

T is the tension in the wire

m is the mass of the wire

For the piano wire in this problem,

L = 0.400 m

T = 1070 N

m = 4.00 g = 0.004 kg

So the fundamental frequency is

[tex]f_1=\frac{1}{2(0.400)}\sqrt{\frac{1070}{(0.004)/(0.400)}}=409 Hz[/tex]

B. 24

For this part, we need to analyze the different harmonics of the piano wire. The nth-harmonic of a string is given by

[tex]f_n = nf_1[/tex]

where [tex]f_1[/tex] is the fundamental frequency.

Here in this case

[tex]f_1 = 409 Hz[/tex]

A person is capable to hear frequencies up to

[tex]f = 1.00 \cdot 10^4 Hz[/tex]

So the highest harmonics that can be heard by a human can be found as follows:

[tex]f=nf_1\\n= \frac{f}{f_1}=\frac{1.00\cdot 10^4}{409}=24.5 \sim 24[/tex]

The fundamental frequency of the piano wire is approximately 409 Hz, based on the given mass, length, and tension. The highest harmonic that could be heard by a person capable of hearing frequencies up to 10,000 Hz would be the 24th harmonic.

To find the fundamental frequency (f1) of the piano wire, we can use the formula for the fundamental frequency of a stretched string, which is f1 = (1/2L) * sqrt(T/μ), where L is the length of the string, T is the tension, and μ is the mass per unit length (linear mass density).

First, we need to calculate the linear mass density of the piano wire:

Now, we plug in the values into the formula:

f1 = (1 / (2 * 0.400 m)) * sqrt(1070 N / 0.010 kg/m) = (1 / 0.8 m) * sqrt(107000 N/m) = (1 / 0.8) * 327.16 Hz ≈ 409 Hz.

For part B, we need to determine the highest harmonic that can be heard if the upper limit of human hearing is 10,000 Hz. Since the fundamental frequency is 409 Hz, the harmonics will be integer multiples of this value. The highest harmonic number (n) will be the largest integer such that:

n * f1 ≤ 10,000 Hz

Doing the division gives us n ≤ 10,000 / 409, and we round down to the nearest whole number because we cannot have a fraction of a harmonic. So, the highest harmonic heard would be:

n = floor(10,000/409) = 24 (since 24 * 409 ≈ 9816 Hz which is below the 10,000 Hz threshold).

Answer:

The weight of the astronaut is 0.4802  N.

Explanation:

Gravitational potential energy, [tex]U=2.94\times 10^6\ J[/tex]

Distance above earth, [tex]d=4\times 10^5\ m[/tex]

The gravitational potential energy is given by :

[tex]U=\dfrac{GMm}{R}[/tex]

G is universal gravitational constant

M is the mass of Earth, [tex]M=5.97\times 10^{24}\ kg[/tex]

m is mass of astronaut

R is the radius of earth, R = R + d

[tex]R=6.37\times 10^6\ m+4\times 10^5\ m=6770000\ m[/tex]

[tex]m=\dfrac{U(R+d)^2}{GM}[/tex]

[tex]m=\dfrac{2.94\times 10^6\ J\times (6770000\ m)}{6.67\times 10^{-11}\times 5.97\times 10^{24}\ kg}[/tex]

m = 0.049 kg

The weight of the astronaut is given by :

W = mg

[tex]W=0.049\ kg\times 9.8\ m/s^2[/tex]

W = 0.4802  N

So, the weight of the astronaut when he returns to the earth surface is 0.4802 N. Hence, this is the required solution.

Final answer:

The weight of the astronaut when he returns to the Earth's surface is approximately 73.32 N.

Explanation:

To calculate the weight of the astronaut when he returns to the Earth's surface, we can use the formula for gravitational potential energy:

PE = mgh

where PE is the gravitational potential energy, m is the mass of the astronaut, g is the acceleration due to gravity, and h is the altitude of the astronaut.

Given that the gravitational potential energy is 2.94 x 10^6 J and the altitude is 4.00 x 10^5 m, we can rearrange the formula to solve for m:

m = PE / (gh)

Substituting the values, we get:

m = (2.94 x 10^6 J) / ((9.80 m/s^2) * (4.00 x 10^5 m))

Calculating this, we find that the mass of the astronaut is approximately 7.49 kg.

Now, to find the weight of the astronaut when he returns to the Earth's surface, we can use the formula:

Weight = mg

Substituting the mass we just calculated, we get:

Weight = (7.49 kg) * (9.80 m/s^2)

Calculating this, we find that the weight of the astronaut when he returns to the Earth's surface is approximately 73.32 N.

In Mathematics and Physics, scalar is a quantity or a single number that shows the measurement of a medium in magnitude only (It does not include direction as vectors do); examples of scalars are voltage, mass, temperature, speed, volume, power, energy, and time.

Two examples of scalars I have used recently are Degrees Celsius to measure the temperature of my living room and Cubic Feet to measure the volume of my mug.

A scalar is a quantity represented only by a magnitude and devoid of direction. Examples of scalars include temperature and energy, which are significant in physics and everyday measurements. Unlike vectors, scalars do not change with coordinate system rotations.

Definition of Scalar

A scalar is a physical quantity that is represented only by a magnitude (or numerical value) but does not involve any direction. In contrast to vectors, which have both magnitude and direction, scalars are not affected by coordinate system rotations or translations. Scalars come in handy when representing physical quantities where direction is non-applicable.

Two examples of scalars that I have used recently are:

When dealing with physics problems or everyday measurements, it's crucial to understand whether you're working with a scalar or a vector to correctly interpret the quantity in question.

Answer:

112.17 m/s

56.427 years

Explanation:

h = 3.18 x 10^10 m

R = 6.4 x 10^6 m

r = R + h = 3.18064 x 10^10 m

M = 6 x 10^24 kg

The formula for the orbital velocity is given by

[tex]v = \sqrt{\frac{G M }{r}}[/tex]

[tex]v = \sqrt{\frac{6.67 \times 10^{-11}\times 6\times 10^{24}  }{3.18064\times 10^{10}}}[/tex]

v = 112.17 m/s

Orbital period, T = 2 x 3.14 x 3.18064 x 10^10 / 112.17

T = 0.178 x 10^10 s

T = 56.427 years

Answer:

option (d)

Explanation:

The relation between the rms velocity and the molecular mass is given by

v   proportional to  \frac{1}{\sqrt{M}} keeping the temperature constant

So for two gases

[tex]\frac{v_{A}}{v_{B}}=\sqrt{\frac{M_{B}}{M_{A}}}[/tex]

[tex]\frac{2v_{B}}{v_{B}}=\sqrt{\frac{M_{B}}{M_{A}}}[/tex]

[tex]{\frac{M_{B}}{M_{A}}} = 4[/tex]

[tex]{\frac{M_{B}}{4}} = M_{A}[/tex]

If the RMS speed of the molecules of gas A is twice that of gas B then the molecular mass of A compared to that of B is one-fourth that of B.

We know that the RMS velocity of the molecule is given as,

[tex]V = \dfrac{1}{\sqrt{M}}[/tex]

Given to us

RMS speed of the molecules of gas A is twice that of gas B, therefore, [tex]V_A = 2 V_B[/tex]

Substitute the value of RMS in the equation [tex]V_A = 2 V_B[/tex],

[tex]\dfrac{1}{\sqrt{M_A}} = \dfrac{2}{\sqrt{M_B}}\\\\\\\sqrt{\dfrac{M_B}{M_A}} = 2\\\\\\\dfrac{M_B}{M_A} = 4[/tex]

Hence, If the RMS speed of the molecules of gas A is twice that of gas B then the molecular mass of A compared to that of B is one-fourth that of B.

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For a DC circuit, the following equation relates the voltage, resistance, and current:

V = IR

V is the total voltage supplied, I is the total current, and R is the total resistance.

Given values:

V = 14.4V

R = 11.0Ω

Plug in the values and solve for I:

14.4 = I×11.0

I = 1.309A

Since one electron carries 1.6×10⁻¹⁹C of charge, divide the current by this number.

1.309/(1.6×10⁻¹⁹) = 8.182×10¹⁸

Round this value to 2 significant figures:

8.2×10¹⁸ electrons per second.

Answer:

[tex]R=16.8[/tex]Ω

Explanation:

l=1000m

a=1mm2

p=1.68x10^-8

Electrical Resistivity Equation

[tex]R=p(\frac{L}{A} )\\[/tex]

where

p= proportional constant ρ (the Greek letter “rho”) is known as Resistivity.

L = is the length in metres (m)

A= is the area in square metres (m2),

R=1.68×10^-8×(1000/1×10^-6)

[tex]R=16.8[/tex]Ω

Answer:

The width of the slit is 0.167 mm

Explanation:

Wavelength of light, [tex]\lambda=608\ nm=608\times 10^{-9}\ m[/tex]

Distance from screen to slit, D = 88.5 cm = 0.885 m

The distance on the screen between the fifth order minimum and the central maximum is 1.61 cm, y = 1.61 cm = 0.0161 m

We need to find the width of the slit. The formula for the distance on the screen between the fifth order minimum and the central maximum is :

[tex]y=\dfrac{mD\lambda}{a}[/tex]

where

a = width of the slit

[tex]a=\dfrac{mD\lambda}{y}[/tex]

[tex]a=\dfrac{5\times 0.885\ m\times 608\times 10^{-9}\ m}{0.0161\ m}[/tex]

a = 0.000167 m

[tex]a=1.67\times 10^{-4}\ m[/tex]

a = 0.167 mm

So, the width of the slit is 0.167 mm. Hence, this is the required solution.

Answer:

3272.4 Hz

Explanation:

L = 31 mH

XL = 637 ohm

XL = 2 π f L

f = XL / (2 π L)

f = 637 / ( 2 x 3.14 x 31 x 10^-3)

f = 3272.4 Hz

Answer:

7.6 m/s

Explanation:

m = 2.3 kg, h = 4.6 m, x = 25 cm = 0.25 m

Use the conservation of energy

Potential energy of the block = Elastic potential energy of the spring

m g h = 1/2 k x^2

Where, k be the spring constant

2.3 x 9.8 x 4.6 = 0.5 x k x (0.25)^2

k = 3317.88 N/m

Now, let v be the velocity of the block, when the compression is 15 cm.

Again use the conservation of energy

Potential energy of the block = Kinetic energy of block + Elastic potential  

                                                                                         energy of the spring

m g h = 1/2 m v^2 + 1/2 k x^2

2.3 x 9.8 x 4.6 = 0.5 x 2.3 x v^2 + 0.5 x 3317.88 x (0.15)^2

103.684 = 1.15 v^2 + 37.33

v = 7.6 m/s

Answer:  The gauge pressure of the air in the tires is 179.5 kPa.

Solution :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]

where,

[tex]P_1[/tex] = initial pressure of gas = Atmospheric pressure + gauge pressure = 101 kPa + 165 kPa = 266 kPa  

[tex]P_2[/tex] = final pressure of gas = ?

[tex]V_1[/tex] = initial volume of gas = [tex]1.63\times 10^{-2}m^3[/tex]

[tex]V_2[/tex] = final volume of gas = [tex]1.70\times 10^{-2}m^3[/tex]

[tex]T_1[/tex] = initial temperature of gas = [tex]0^oC=273+0=273K[/tex]

[tex]T_2[/tex] = final temperature of gas = [tex]27.3^oC=273+27.3=300.3K[/tex]

Now put all the given values in the above equation, we get the final pressure of gas.

[tex]\frac{266\times 1.63\times 10^{-2}}{273K}=\frac{P_2\times 1.70\times 10^{-2}}{300.3K}[/tex]

[tex]P_2=280.5kPa[/tex]

Gauge pressure = Absolute pressure - atmospheric pressure  = (280.5 - 101) kPa= 179.5 kPa

Therefore, the gauge pressure of the air in the tires is 179.5 kPa.

Answer:

[tex]11304 \frac{in^{3}}{s}[/tex]

Explanation:

r = radius of right circular cone = 150 in

h = height of right circular cone = 144 in

[tex]\frac{dr}{dt}[/tex] = rate at which radius increase = 1.5 in/s

[tex]\frac{dh}{dt}[/tex] = rate at which height decrease = - 2.4 in/s

Volume of the right circular cone is given as

[tex]V = \frac{\pi r^{2}h}{3}[/tex]

Taking derivative both side relative to "t"

[tex]\frac{dV}{dt} = \frac{\pi }{3}\left ( r^{2}\frac{dh}{dt} \right + 2rh\frac{dr}{dt})[/tex]

[tex]\frac{dV}{dt} = \frac{3.14 }{3}\left ( (150)^{2}(- 2.4) + 2(150)(144)(1.5))[/tex]

[tex]\frac{dV}{dt} = 11304 \frac{in^{3}}{s}[/tex]

Answer:

Length of Eiffel tower, when the temperature is 35 degrees = 300.21 m

Explanation:

Thermal expansion is given by the expression

[tex]\Delta L=L\alpha \Delta T \\ [/tex]

Here length of Eiffel tower, L = 300 m

Coefficient of thermal expansion, α = 0.000012 per degree Celsius

Change in temperature, = 35 - (-24) = 59degrees Celsius

Substituting

[tex]\Delta L=L\alpha \Delta T= 300\times 0.000012\times 59=0.2124m \\ [/tex]

Length of Eiffel tower, when the temperature is 35 degrees = 300 + 0.2124 = 300.21 m

Answer:

a) Train's nose will be present 9.19 seconds in the station.

b) The nose leaves the station at 21.62 m/s.

c) Train's end will leave after 15.33 seconds from station.

d) The end leaves the station at 20.70 m/s.

Explanation:

a) We have equation of motion s = ut + 0.5at²

   Here u = 23 m/s, a = -0.15 m/s², s = 205 m

   Substituting

        205 = 23t + 0.5 x (-0.15) x t²

        0.075t² -23 t +205 = 0

  We will get t = 9.19 or t = 297.47

  We have to consider the minimum time

  So train's nose will be present 9.19 seconds in the station.

b) We have equation of motion v= u + at

   Here u = 23 m/s, a = -0.15 m/s², t = 9.19

   Substituting

        v= 23 - 0.15 x 9.19 = 21.62 m/s

   The nose leaves the station at 21.62 m/s.

c) We have equation of motion s = ut + 0.5at²

   Here u = 23 m/s, a = -0.15 m/s², s = 205 + 130 = 335 m

   Substituting

        335 = 23t + 0.5 x (-0.15) x t²

        0.075t² -23 t +335 = 0

  We will get t = 15.33 or t = 291.33

  We have to consider the minimum time

  So train's end will leave after 15.33 seconds from station.      

d) We have equation of motion v= u + at

   Here u = 23 m/s, a = -0.15 m/s², t = 15.33

   Substituting

        v= 23 - 0.15 x 15.33 = 20.70 m/s

   The end leaves the station at 20.70 m/s.

Answer:

- 2.7 rad/s^2

Explanation:

f0 =  + 5.85 rotations per second (counter clockwise)

t = 21.3 s

f =  - 3.31 rotations per second (clockwise)

w0 = 2 x 3.14 x 5.85 = + 36.738 rad/s

w = - 2 x 3.14 x 3.31 = - 20.79 rad/s

Let α be teh angular acceleration.

α = (w - w0) / t

α = (-20.79 - 36.738) / 21.3

α = - 2.7 rad/s^2

The flywheel's average angular acceleration is -0.43 rad/s².

The average angular acceleration of the flywheel is determined by applying the following kinematic equation as shown below;

α = (ωf - ωi)/t

where;

α = (-3.31 - 5.85)/21.3

α = -0.43 rad/s²

Thus, the flywheel's average angular acceleration is -0.43 rad/s².

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Answer:

(a) 92 %

(b) 6.76 %

Explanation:

length, l = 4 m, height, h = 5 m, width, w = 3 m, density of water = 1000 kg/m^3

density of ice = 920 kg/m^3, density of mercury = 13600 kg/m^3

(a) Let v be the volume of ice below water surface.

By the principle of flotation

Buoyant force = weight of ice block

Volume immersed x density of water x g = Total volume of ice block x density

                                                                      of ice x g

v x 1000 x g = V x 920 x g

v / V = 0.92

% of volume immersed in water = v/V x 100 = 0.92 x 100 = 92 %

(b) Let v be the volume of ice below the mercury.

By the principle of flotation

Buoyant force = weight of ice block

Volume immersed x density of mercury x g = Total volume of ice block x  

                                                                      density of ice x g

v x 13600 x g = V x 920 x g

v / V = 0.0676

% of volume immersed in water = v/V x 100 = 0.0676 x 100 = 6.76 %

Answer:

At time 10.28 s after A is fired bullet B passes A.

Passing of B occurs at 4108.31 height.

Explanation:

Let h be the height at which this occurs and t be the time after second bullet fires.

Distance traveled by first bullet can be calculated using equation of motion

[tex]s=ut+0.5at^2 \\ [/tex]

Here s = h,u = 450m/s a = -g and t = t+3

Substituting

[tex]h=450(t+3)-0.5\times 9.81\times (t+3)^2=450t+1350-4.9t^2-29.4t-44.1\\\\h=420.6t-4.9t^2+1305.9 [/tex]

Distance traveled by second bullet

Here s = h,u = 600m/s a = -g and t = t

Substituting

[tex]h=600t-0.5\times 9.81\times t^2=600t-4.9t^2\\\\h=600t-4.9t^2 \\ [/tex]

Solving both equations

[tex]600t-4.9t^2=420.6t-4.9t^2+1305.9\\\\179.4t=1305.9\\\\t=7.28s \\ [/tex]

So at time 10.28 s after A is fired bullet B passes A.

Height at t = 7.28 s

[tex]h=600\times 7.28-4.9\times 7.28^2\\\\h=4108.31m \\ [/tex]

Passing of B occurs at 4108.31 height.

Answer:

The change in length is 3.4 mm.

Explanation:

Given that,

Length = 4 m

Diameter = 2.3 cm

Load = 70 kN

Modulus of elasticity = 200 GPa

We need to calculate the change in length

Using formula of modulus of elasticity

[tex]E=\dfrac{\dfrac{F}{A}}{\dfrac{\Delta l}{l}}[/tex]

[tex]\Delta l=\dfrac{Fl}{AE}[/tex]

Where, F = force

A = area

L = length

E = modulus elasticity

Put the value into the formula

[tex]\Delta l=\dfrac{70\times10^{3}\times4}{\pi\times(1.15\times10^{-2})^2\times200\times10^{9}}[/tex]

[tex]\Delta l=0.00336\ m[/tex]

[tex]\Delta l=3.4\ mm[/tex]

Hence, The change in length is 3.4 mm.

Given:

R = 13 Ω

L = 46 mH

V = 240 V(rms)

f = 60 Hz

Formula used:

[tex]I_{L} = \frac{V}{R + jX_{L} }[/tex]

[tex]X_{L} = 2\pi fL[/tex]

Solution:

Now, using the above formula for [tex]X_{L}[/tex]:

[tex]X_{L} = 2\pi\times 60\times 46\times 10^{-3} [/tex] = 17.34 Ω

From the above formula for  [tex]I_{L}[/tex]:

[tex]I_{L} = \frac{240\angle0}{13 + j17.34 }[/tex]

[tex]I_{L}[/tex] = (6.64 - j8.86) A = [tex]11.07\angle-53.14^{\circ}[/tex] A

[tex]i_{L}(t)[/tex] = [tex]\sqrt{2}\times 11.07cos(2\pi \times 60t - 53.14)[/tex] A

[tex]i_{L}(t)[/tex] = 15.65cos(376.99t - 53.14)A

Load current in the given AC circuit can be obtained by using the concept of Impedance and Ohm's law in its AC variant where current, I = V/Z. Impedance, Z is calculated taking into account both Resistance and Reactance.

In the case of the mentioned electric dryer, we are dealing with an AC circuit that contains both resistance and inductance. In such cases, we should leverage the concept of Impedance, which is the effective resistance in an AC circuit resulting from combined effect of resistors, inductors and capacitors.

Impedance is defined by the relation Z = √ (R^2 + (XL)^2), where R is the resistance (13 Ω) and XL = 2πfL is the Reactance, f is the frequency (60Hz), and L the inductance (46mmH). However, when calculating the load current, we use Ohm's law in its AC version, I = V/Z, where I denotes the current, V the voltage supplied (240V), and Z the impedance defined earlier.

By substitizing all the values and calculating we can get the desired load current.

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