A wheel that was initially spinning is accelerated at a constant angular acceleration of 5.0 rad/s^2. After 8.0 s, the wheel is found to have made an angular displacement of 400 radians. (a) How fast was the wheel spinning initially? (b) What is the final angular velocity of the wheel?

Answer:

Energy, [tex]E=9\times 10^{13}\ J[/tex]

Explanation:

It is given that,

Mass of matter, m = 1 g = 0.001 Kg

If this matter is completely converted into energy, we need to find the yield. The mass of an object can be converted to energy as :

[tex]E=mc^2[/tex]

c = speed of light

[tex]E=0.001\ kg\times (3\times 10^8\ m/s)^2[/tex]

[tex]E=9\times 10^{13}\ J[/tex]

So, the energy yield will be [tex]9\times 10^{13}\ J[/tex]. Hence, this is the required solution.

Answer:

The terminal speed of his is 137.68 m/s.

Explanation:

Given that,

Mass of skydiver = 78 kg

Area of box[tex]A =24\times35=840\ cm[/tex]

Drag coefficient = 0.80

Density of air [tex]\rho= 1.2\times kg/m^3[/tex]

We need to calculate the terminal velocity

Using formula of drag force

[tex]F_{d} = \dfrac{1}{2}\rho v^2Ac[/tex]

Where,

[tex]\rho[/tex] = density of air

A = area

C= coefficient of drag

Put the value into the formula

[tex]78\times9.8=\dfrac{1}{2}\times1.2\times v^2\times24\times10^{-2}\times35\times10^{-2}\times0.80[/tex]

[tex]v^2=\dfrac{2\times78\times9.8}{1.2\times24\times10^{-2}\times35\times10^{-2}\times0.80}[/tex]

[tex]v=\sqrt{\dfrac{2\times78\times9.8}{1.2\times24\times10^{-2}\times35\times10^{-2}\times0.80}}[/tex]

[tex]v=137.68\ m/s[/tex]

Hence, The terminal speed of his is 137.68 m/s.

The terminal speed of the skydiver with dimensions as a rectangular box as he falls feet first is 137.68 m/s.

Terminal speed of a body is the maximum speed, which is achieved by the object when it fall through a fluid.

In the case of terminal velocity, the force of gravity becomes equal to the sum of the drag force and buoyancy force due to fluid on body.

Terminal velocity can be find out as,

[tex]v=\sqrt{\dfrac{2mg}{\rho AC_d}}[/tex]

Here, (m) is the mass, (g) is gravitational force, ([tex]\rho[/tex]) is the density of fluid, (A) is the project area and ([tex]C_d[/tex]) is the drag coefficients.

It is given that, the mass of the skydiver is 78 kg The dimensions of the skydiver is s 24 cm × 35 cm × 170 cm.

The coefficient of drag is 0.80 and the density of air is 1.2 kg/m³.

Put the values in the above formula as,

[tex]v=\sqrt{\dfrac{2\times78\times9.8}{1.2\times0.24\times0.35\times 0.8}}\\v=137.68\rm m/s[/tex]

Thus the terminal speed of the skydiver as he falls feet first is 137.68 m/s.

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Answer:

The ball doesn't strike the building because it strikes the ground at d=1.62 meters.

Explanation:

V= 5 m/s < 70º

Vx= 1.71 m/s

Vy= 4.69 m/s

h= Vy * t - g * t²/2

clearing t for the flying time of the ball:

t= 0.95 s

d= Vx * t

d= 1.62 m

Answer:

v = 0.025 m/s

Explanation:

Given that the voltage is

[tex]V = 5000 x^2[/tex]

now at x = 0

[tex]V_1 = 0 Volts[/tex]

also we have at x = 8 cm

[tex]V_2 = 5000(0.08)^2 = 32 Volts[/tex]

now change in potential energy of the charge is given as

[tex]\Delta U = q\Delta V[/tex]

[tex]\Delta U = (10 \times 10^{-9})(32 - 0)[/tex]

now by mechanical energy conservation law

[tex]\frac{1}{2}mv^2 - 0 = 3.2 \times 10^{-7}[/tex]

[tex]\frac{1}{2}(1 \times 10^{-3})v^2 = 3.2 \times 10^{-7}[/tex]

[tex]v = 0.025 m/s[/tex]

The maximum speed of the particle in arrangement of source charges produces 32 volts electric potential is 0.025 meter per second.

Electric potential energy is the energy which is required to move a unit charge from a point to another point in the electric field.

It can be given as,

[tex]U=qV[/tex]

Here, (q) is the charge and (V) is the electric potential difference.

Given infroamtion-

The electric potential producers by the arrangement of source charges is given by,

[tex]V=5000x^2[/tex]

The mass of the particle is 1.0 gram.

The charge of the particles 10 nC.

As, the electric potential producers by the arrangement of source charges is given by,

[tex]V=5000x^2[/tex]

At the x equal to 8 cm or 0.08 m, the equation become,

[tex]V=5000(0.08)^2\\V=32\rm V[/tex]

Thus the potential difference at the is 32 volts.

The electric potential energy of the particle is,

[tex]U=10\times10^{-9}\times32\\U=3.2\times10^{-7}\rm j[/tex]

Now the electric potential energy is equal to the kinetic energy of the particle. Thus,

[tex]\dfrac{1}{2}\times0.001\times v^2=3.2\times10^{-7}\\v=0.025\rm m/s[/tex]

Thus the maximum speed of particle is 0.025 meter per second.

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Answer:

Magnetic field, B = 1.84 T

Explanation:

It is given that,

Charge on alpha particle, q = +2e = [tex]3.2\times 10^{-19}\ C[/tex]

Mass of alpha particle, [tex]m=6.6\times 10^{-27}\ kg[/tex]

Speed of alpha particles, [tex]v=1.6\times 10^7\ m/s[/tex]

We need to find the magnetic field strength required to bend them into a circular path of radius, r = 0.18 m

So, [tex]F_m=F_c[/tex]

[tex]F_m\ and\ F_c[/tex] are magnetic force and centripetal force respectively

[tex]qvB=\dfrac{mv^2}{r}[/tex]

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

[tex]B=\dfrac{6.6\times 10^{-27}\ kg\times 1.6\times 10^7\ m/s}{3.2\times 10^{-19}\ C\times 0.18\ m}[/tex]

B = 1.84 T

So, the value of magnetic field is 1.84 T. Hence, this is the required solution.

The magnetic field strength required to bend them into a circular path is 1.83 T.

The magnetic force on the emitted charge is given as;

F = qvB

The centripetal force of the emitted charge is given as;

F = mv²/r

The magnetic field strength required to bend them into a circular path is calculated as follows;

qvB = mv²/r

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

[tex]B = \frac{6.6 \times 10^{-27} \times 1.6 \times 10^7}{(2\times 1.6 \times 10^{-19} ) \times 0.18} \\\\B = 1.83 \ T[/tex]

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Answer:

The specific heat of the substance is c= 512.04 J/kg K

Explanation:

ΔQ= 2205 J

m= 0.03434 kg

ΔT= 125.4 ºC

ΔQ= m * c * ΔT

c= ΔQ / (m * ΔT)

c= 512.04 J/Kg K

Answer:

0.032 m

Explanation:

Consider the forces acting on the block

m = mass of the block = 1.50 kg

[tex]f_{s}[/tex] = Static frictional force

[tex]F_{n}[/tex] = Normal force on the block from the wall

[tex]F_{s}[/tex] = Spring force due to compression of spring

[tex]F_{g}[/tex] = Force of gravity on the block = mg = 1.50 x 9.8 = 14.7 N

k = spring constant = 860 N/m

μ = Coefficient of static friction between the block and wall = 0.54

x = compression of the spring

Spring force is given as

[tex]F_{s}[/tex] = kx

From the force diagram of the block, Using equilibrium of force along the horizontal direction, we get the force equation as  

[tex]F_{n}[/tex] = [tex]F_{s}[/tex]

[tex]F_{n}[/tex] = kx                                             eq-1

Static frictional force is given as

[tex]f_{s}[/tex] = μ [tex]F_{n}[/tex]

Using eq-1

[tex]f_{s}[/tex] = μ k x                                                eq-2

From the force diagram of the block, Using equilibrium of force along the vertical direction, we get the force equation as

[tex]f_{s}[/tex] = [tex]F_{g}[/tex]

Using eq-2

μ k x = 14.7

(0.54) (860) x = 14.7

x = 0.032 m

Answer:

The speed of meteoroid is 21.61 km/s in south-east.

Explanation:

Given that,

A meteoroid is traveling through the atmosphere at 18.3 km/s. while descending at a rate of 11.5 km/s it means 11.5 km/s in south.

We need to draw a diagram

Using Pythagorean theorem

[tex]AC^2=AB^2+BC^2[/tex]

[tex]AC^2=(18.3)^3+(11.5)^2[/tex]

[tex]AC=\sqrt{(18.3)^2+(11.5)^2}[/tex]

[tex]AC=21.61\ km/s[/tex]

Hence, The speed of meteoroid is 21.61 km/s in south-east.

The speed of the meteoroid is calculated using the Pythagorean theorem and is approximately 21.62 km/s.

To calculate the speed (magnitude of the velocity), the equation is: speed = √(horizontal velocity)² + (vertical velocity)².

Thus, the speed = √(18.3 km/s)² + (11.5 km/s)² = √(335.29 + 132.25) km²/s² = √467.54 km²/s² = 21.62 km/s.

The meteoroid's speed through the atmosphere is approximately 21.62 km/s.

Answer:

B= [tex]\sqrt{65}[/tex] ≅8.06

Explanation:

Using the Pythagorean theorem:

[tex]C^{2}[/tex]= [tex]A^{2}[/tex] + [tex]B^{2}[/tex]

where C represents the length of the hypotenuse and A and B the lengths of the triangle's other two sides, we can find out the lenght of B assuming the value of the hypotenuse being 9 and A being 4.

[tex]9^{2}[/tex]=[tex]4^{2}[/tex] + [tex]B^{2}[/tex]

81= 16+ [tex]B^{2}[/tex]

81-16= [tex]B^{2}[/tex]

B= [tex]\sqrt{65}[/tex] ≅8.06

The length of B is equal to 8.06 units

Data given;

To solve this question, we have to use the formula of finding resultant vectors

Since it's a right-angle triangle, let's use Pythagoras' theorem

[tex]C^2=A^2 + B^2\\9^2 = 4^2 + B^2\\b^2 = 9^2 - 4^2\\b^2 = 65\\b = \sqrt{65}\\b = 8.06[/tex]

From the calculation above, the length of B is equal to 8.06.

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You haven't told us anything about the detectors being used.  We don't know how the sensitivity of the detector is related to the total number of photons absorbed, and we don't even know whether you and your friend are both using the same type of detector.

All we can do, in desperation, is ASSUME that the minimum time required to just detect a star is inversely proportional to the total number of its photons that strike the detector.  That is, assume . . .

(double the number of photons) ===> (detect the source in half the time) .

-- The intensity of light delivered to the prime focus of a telescope is directly proportional to the AREA of its objective lens or mirror, which in turn is proportional to the square of its radius or diameter.

So your telescope gathers (0.18/0.05)² = 12.96 times as much light as your friends telescope does.

-- So we'd expect your instrument to detect the same star in

(119.5 min) / (12.96) = 9.22 minutes  .

We're simply comparing the performance of two different telescopes as they observe the same object, so the star's magnitude doesn't matter.

If your telescope has diameter 0.18 metres, for 9.22 minutes, you need to expose for.

An optical telescope is one that collects and sharply concentrates light. Mostly from the visible parts of the spectrum. That is to make a magnified image for close inspection, to take a picture, or you might say to get information from an electronic sensor image.

A reflecting telescope, sometimes called a reflector, is one that creates images by reflecting light off either a single curved mirror or a group of mirrors. Sir Isaac Newton created the reflecting telescope inside the 17th century as a replacement for the refracting one.

your telescope gathers (0.18/0.05)² = 12.96 times as much light as your friends telescope does.

(119.5 min) / (12.96) = 9.22 minutes

Therefore, for 9.22 minutes, you need to expose for.

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Answer:

a) There are 100 centimeters in 1 meter.

b) [tex]\texttt{A cm is equal to }\frac{1}{2.54}\texttt{ inch}[/tex]

Explanation:

a) We have the conversion

         1 m = 100 cm

   So there are 100 centimeters in 1 meter.

b) 1 inch = 2.54 cm

    [tex]1cm=\frac{1}{2.54}inch[/tex]

   [tex]\texttt{A cm is equal to }\frac{1}{2.54}\texttt{ inch}[/tex]

Answer:

Force, F = 16814.95 N

Explanation:

It is given that,

Mass of space station, m = 2010 kg

Altitude, [tex]d=5.35\times 10^5\ m[/tex]

Mass of earth, [tex]m=5.98\times 10^{24} kg[/tex]

Mean radius of earth, [tex]r=6.37\times 10^6\ m[/tex]

Magnitude of force is given by :

[tex]F=G\dfrac{Mm}{R^2}[/tex]

R = r + d

[tex]R=6.37\times 10^6\ m+5.35\times 10^5\ m=6905000\ m[/tex]

[tex]F=6.67\times 10^{-11}\times \dfrac{2010\ kg\times 5.98\times 10^{24} kg}{(6905000\ m)^2}[/tex]

F = 16814.95 N

So, the force between the space station and the Earth is 16814.95 N. Hence, this is the required solution.

Answer:

EMF = 1.20 V

Explanation:

It is given that,

Magnetic field used by the screwdriver, B = 13.5 T

Length of screwdriver, l = 10.5 cm = 0.105  m

Speed with which it is moving. v = 0.85 m/s

We need to find the maximum EMF induced in the screwdriver. It is given by :

[tex]\epsilon=BLv[/tex]

[tex]\epsilon=13.5\ T \times 0.105\ m \times 0.85\ m/s[/tex]

[tex]\epsilon=1.20\ V[/tex]

So, the maximum emf of the screwdriver is 1.20 V. Hence, this is the required solution.

Answer:

21828 m

Explanation:

[tex]v_{ground}[/tex] = speed of sound through ground = 6 km/s = 6000 m/s

[tex]v_{air}[/tex] = speed of sound through air = 343 m/s

t = time taken for the vibrations to arrive

t' = time taken for sound to arrive = t + 60

d = distance of the point of explosion

time taken for the vibrations to arrive is given as

[tex]t = \frac{d}{v_{ground}}[/tex]                            eq-1

time taken for the sound to arrive is given as

[tex]t' = \frac{d}{v_{air}}[/tex]

[tex]t + 60 = \frac{d}{v_{air}}[/tex]

using eq-1

[tex]\frac{d}{v_{ground}} + 60 = \frac{d}{v_{air}}[/tex]

[tex]\frac{d}{6000} + 60 = \frac{d}{343}[/tex]

d = 21828 m

Answer:

(a) 2.23 m/s

(b) 0.9 m

Explanation:

h = 1 m, t = 0.1 second

horizontal component of initial velocity, ux = 2 m/s

vertical component of initial velocity, uy = 0  

(a) Let v be the velocity after 0.1 seconds. Its vertical component is vy and horizontal component is vx.

The horizontal component of velocity remains constant as in this direction, acceleration is zero.

vx = ux = 2 m/s

Use first equation of motion in Y axis direction.

vy = uy + g t

vy = 0 + 9.8 x 0.1 = 0.98 m/s

Resultant velocity after 0.1 second

v^2 = vx^2 + vy^2

v^2 = 2^2 + 0.98^2

v = 2.23 m/s

(b) Let it takes time t to land.

Use second equation of motion along Y axis

h = uy t + 1/2 g t^2

1 = 0 + 1/2 x 9.8 x t^2

t = 0.45 second

Let it lan at a distance x.

so, x = ux x t

x = 2 x 0.45 = 0.9 m

Answer:

[tex]\omega_O = 0.16 rad /sec[/tex]

Q = 0.24

Explanation:

given data:

resonant  angular frequency is given as  \omega_O = \frac{1}{\sqrt{LC}}

where L is inductor = 150 mH

C is capacitor = 0.25\mu F

[tex]\omega = \frac{1}{\sqrt{150*10^{6}*0.25*10^{-6}}}}[/tex]

[tex]\omega_O = 0.16 rad /sec[/tex]

QUALITY FACTOR is given as

[tex]Q = \frac{1}{R}{\sqrt\frac{L}{C}}[/tex]

Putting all value to get quality factor value

Q =[tex] \frac{1}{1000}{\sqrt\frac{150*10^{6}}{0.25*10^{-6}}}[/tex]

Q = 0.24

Final answer:

The maximum power dissipation in the resistor occurs at a frequency of approximately 1175.5 Hz in an RLC series circuit with the given values of R, L, and C. The quality factor of the circuit is approximately 57.74.

Explanation:

In an RLC series circuit, the maximum power dissipation in the resistor is achieved at the resonant frequency, which is given by the formula:

fr = 1 / (2π √LC)

Substituting the given values:

R = 100Ω, L = 150mH (or 0.15H), and C = 0.25μF (or 0.00000025F), we can calculate the resonant frequency:

fr = 1 / (2π √(0.15 x 0.00000025))

fr ≈ 1175.5 Hz

Therefore, the maximum power dissipation in the resistor occurs at a frequency of approximately 1175.5 Hz.

The quality factor (Q) of the circuit is a measure of its damping ability. It is given by the formula:

Q = R √C / L

Substituting the given values:

R = 100Ω, L = 150mH (or 0.15H), and C = 0.25μF (or 0.00000025F), we can calculate the quality factor:

Q = 100 √(0.00000025) / 0.15

Q ≈ 57.74

Therefore, the quality factor of the circuit is approximately 57.74.

Answer:

The amplitude of oscillation for the oscillating mass is 0.28 m.

Explanation:

Given that,

Mass = 0.14 kg

Equation of simple harmonic motion

[tex]x(t)=(0.28\ m)\cos[(8\ rad/s)t][/tex]....(I)

We need to calculate the amplitude

Using general equation of simple harmonic equation

[tex]y=A\omega \cos\omega t[/tex]

Compare the equation (I) from general equation

The amplitude is 0.28 m.

Hence, The amplitude of oscillation for the oscillating mass is 0.28 m.

Answer:

Here, for parallel resisitors

Option D) all of the other choices are are true

is correct.

Explanation:

In parallel connection:

1) Voltage across each element connected in parallel remain same.

2) Kirchhoff's Current Law (KCL), sum of the current entering and leaving the junction will be zero

3) The equivalent resistance of the elements connected in parallel is always less than the individual resistance of any resistor in the circuit.

The equivalent resistance in a parallel circuit is given by:

[tex]\frac{1}{R_{eq}} = \frac{1}{R_1} +\frac{1}{R_2} +......+ \frac{1}{R_n}[/tex]

Final answer:

When resistors are connected in parallel, the voltage across each resistor is the same, the total current flowing from the battery equals the sum of the currents flowing through each resistor, and the equivalent resistance of the combination is less than the resistance of any one of the resistors.

Explanation:

When two or more resistors are connected in parallel to a battery:

The voltage across each resistor is the same. This is because in a parallel circuit, all the resistors have the same potential difference across them.

The total current flowing from the battery equals the sum of the currents flowing through each resistor. In a parallel circuit, the total current is divided among the different resistors.

The equivalent resistance of the combination is less than the resistance of any one of the resistors. This is because adding resistors in parallel decreases the overall resistance of the circuit.

Explanation:

It is given that,

Mass of first ball, m₁ = 2 kg

Mass of other ball, m₂ = 3.5 kg

Velocities of both balls, u = 0.9 m/s

(1) The lighter ball rebounds opposite its initial direction, with speed 0.90 m/s. We need to find the final velocity of second ball. Applying the conservation of momentum as :

[tex]2\ kg\times 0.9-3.5\ kg\times 0.9\ m/s=-2\ kg\times 0.9\ m/s+3.5v[/tex]

v is the final velocity of heavier ball.

v = 0.128 m/s

or

v = 0.13 m/s

Initial kinetic energy, [tex]E_i=\dfrac{1}{2}\times (2\ kg+3.5\ kg)\times (0.9\ m/s)^2=2.23\ J[/tex]

Final kinetic energy, [tex]E_f=\dfrac{1}{2}\times 2\ kg\times (0.9\ m/s)^2+\dfrac{1}{2}\times 3.5\ kg\times (0.13\ m/s)^2=0.84\ J[/tex]

Lost in kinetic energy, [tex]\Delta KE=0.84-2.23=-1.39\ J[/tex]

Hence, this is the required solution.

Answer:

A) 5.2 x 10³ N

B) 8.8 x 10³ N

Explanation:

Part A)

[tex]F_{g}[/tex] = weight of the craft in downward direction = tension force in the cable when stationary = 7000 N

[tex]T[/tex] = Tension force in upward direction

[tex]F_{d}[/tex] = Drag force in upward direction = 1800 N

Force equation for the motion of craft is given as

[tex]F_{g}[/tex] - [tex]F_{d}[/tex] - [tex]T[/tex] = 0

7000 - 1800 - [tex]T[/tex] = 0

[tex]T[/tex] = 5200 N

[tex]T[/tex] = 5.2 x 10³ N

Part B)

[tex]F_{g}[/tex] = weight of the craft in downward direction = tension force in the cable when stationary = 7000 N

[tex]T[/tex] = Tension force in upward direction

[tex]F_{d}[/tex] = Drag force in downward direction = 1800 N

Force equation for the motion of craft is given as

[tex]T[/tex]  - [tex]F_{g}[/tex] - [tex]F_{d}[/tex] = 0

[tex]T[/tex] - 7000 - 1800  = 0

[tex]T[/tex] = 8800 N

[tex]T[/tex] = 8.8 x 10³ N

Explanation:

It is given that,

Speed of the baseball, u = 44 m/s

Speed of the baseball, v = 53 m/s

Mass of the ball, m = 145 g = 0.145 kg

Time of contact between the ball and the bat, t = 2.2 ms = 0.0022 s

[tex]F=ma[/tex]

[tex]F=\dfrac{mv}{t}[/tex]

[tex]F_1=\dfrac{0.145\ kg\times 44\ m/s}{0.0022\ s}[/tex]

F₁ = 2900 N...........(1)

[tex]F=ma[/tex]

[tex]F=\dfrac{mv}{t}[/tex]

[tex]F_2=\dfrac{0.145\ kg\times 53\ m/s}{0.0022\ s}[/tex]

F₂ = 3493.18 N.........(2)

In average vector form force is given by :

[tex]F=F_1+F_2[/tex]

[tex]F=(2900i+(-3493.18)\ N[/tex]

[tex]F=(2900i-3493.18j)\ N[/tex]

Hence, this is the required solution.

Answer:

0.0667 m

Explanation:

λ = wavelength of light = 400 nm = 400 x 10⁻⁹ m

D = screen distance = 2.5 m

d = slit width = 15 x 10⁻⁶ m

n = order = 1

θ = angle = ?

Using the equation

d Sinθ = n λ

(15 x 10⁻⁶) Sinθ = (1) (400 x 10⁻⁹)

Sinθ = 26.67 x 10⁻³

y = position of first minimum

Using the equation for small angles

tanθ = Sinθ = y/D

26.67 x 10⁻³ = y/2.5

y = 0.0667 m

The first minimum in a single-slit diffraction pattern of light with a wavelength of 400 nm incident on a single slit of width 15 microns, 2.5 m from the screen, is approximately 66.67 mm from the central maximum.

The question asks for the distance of the first minimum from the central maximum in a single-slit diffraction pattern.

The distance to the first minimum from the central maximum in a single-slit diffraction pattern is calculated using the formula: [tex]y = \frac{\lambda X D}{d}[/tex]

where y - is the distance from the central maximum to the first minimum on the screen (meters), λ (lambda) - is the wavelength of light (meters), D - is the distance between the slit and the screen (meters), and d - the width of the slit (meters)

Given λ (lambda) = 400 nm = [tex]400 X 10^{-9} m[/tex], D = 2.5 m, d = 15 microns = [tex]15 X 10^{-6} m[/tex]

Calculation:

[tex]y = \frac{(400 X 10^{-9} m) X (2.5 m)}{(15 X 10^{-6} m)}\\y = 0.0667 \ meters (or 66.7 \ millimeters)[/tex]

The first minimum is 66.7 millimeters from the central maximum.

Answer:

3260.33 J

Explanation:

[tex]n [/tex]  = number of rods = 5

[tex]L [/tex]  = length of each rod = 0.715 m

[tex]m [/tex]  = mass of rod = 2.51 kg

[tex]I [/tex]  = total moment of inertia

Total moment of inertia is given as

[tex]I = \frac{nmL^{2}}{3}[/tex]

[tex]I = \frac{(5)(2.51)(0.715)^{2}}{3}[/tex]

[tex]I [/tex] = 2.14 kgm²

[tex]w [/tex]  = angular speed = 527 rpm = 55.2 rad/s

Rotational kinetic energy is given as

E = (0.5) [tex]I [/tex]  ([tex]w [/tex] )²

E = (0.5) (2.14) (55.2)²

E = 3260.33 J

Answer:

c) Elastic Modulus

Explanation

As we know that when deformation is under elastic limit then stress applied to the given material is proportional to the strain developed in it

So here we can say that since they both are directly proportional to each other so the proportionality constant here is known as Modulus of elasticity.

So we can say it is given as

[tex]stress = E (strain)[/tex]

so now if we draw a graph between between stress and strain then it must be a straight line and the the slope of this straight line is given as

[tex]Slope = \frac{Stress}{Strain}[/tex]

So here correct answer will be

c) Elastic Modulus

Final answer:

The slope of the stress-strain curve in the elastic deformation region is known as the Elastic Modulus or Young's Modulus, which measures the stiffness of the material. The correct answer to the given question is 'c. Elastic Modulus'.

Explanation:

The slope of the stress-strain curve in the elastic deformation region represents the relationship between stress and strain under elastic conditions. This slope is in fact the Elastic Modulus, which is also known as Young's Modulus when it's in tension or compression. It serves as a measure of the stiffness or rigidity of the material, indicating how much stress is required to achieve a certain amount of strain.

In the context of a stress-strain curve, the plastic modulus is associated with plastic deformation, not the elastic region. Poisson's ratio is another material property that describes the ratio of transverse strain to axial strain, and is not the slope of the curve. Hence, the correct answer to the question is 'c. Elastic Modulus'.

Answer:

-423 m³/s

Explanation:

Volume of a cone is:

V = ⅓ π r² h

Given r = 2h:

V = ⅓ π (2h)² h

V = ⁴/₃ π h³

Taking derivative with respect to time:

dV/dt = 4π h² dh/dt

Given h = 1010 cm and dh/dt = 33 cm/s:

dV/dt = 4π (1010 cm)² (33 cm/s)

dV/dt ≈ 4.23×10⁸ cm³/s

dV/dt ≈ 423 m³/s

The pile is growing at 423 m³/s, so the bin is draining at -423 m³/s.

The rate at which the sand is leaving the bin at that instant is [tex]423\times 10^{6} cm^{3}/s[/tex].

It is given that the radius of a conical bin is two times its height and at the instant when the height of the bin is [tex]1010cm[/tex], the height of the pile increases at a rate of [tex]33 cm/s[/tex].

The formula for the volume of a cone is given as,

[tex]V=\frac{1}{3}\pi r^{2}h[/tex]

Substitute [tex]r=2h[/tex] as per the question

[tex]V=\frac{4}{3}\pi h^{3}[/tex]

This is the volume of the conical bin.

To find the rate of change in the volume of the bin, differentiate the expression for volume w.r.t. time using the chain rule as follows,

[tex]\frac{dV}{dt}=\frac{dV}{dh}\times \frac{dh}{dt}[/tex]

[tex]\frac{dV}{dt}=\frac{4}{3}\pi (3h^{2}) \times \frac{dh}{dt}\\\\\frac{dV}{dt}=4\pi h^{2} \times \frac{dh}{dt}\\[/tex]

Now, according to the question, at [tex]h=1010cm[/tex], [tex]\frac{dh}{dt}=33[/tex].

Substituting these values, the rate at which the sand is leaving the bin is,

[tex]\frac{dV}{dt}=4\pi (1010)^{2} \times 33\\\frac{dV}{dt}=423,025,503.90235\\\frac{dV}{dt}=423\times10^{6}cm^{3}/s[/tex]

So, the rate at which the sand leaves the conical bin at the given instant is [tex]423\times10^{6}cm^{3}/s[/tex]

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Answer:

option (b)

Explanation:

Let the electric field is given by E.

mass of proton = mp

mass of electron = me

acceleration of proton = ap

acceleration of electron = ae

Charge on both the particle is same but opposite in nature.

The force on proton = q E

The force on electron = - q E

acceleration of proton, ap = q E / mp

acceleration of electron, ae = - q E / me

We observe that the force is same in magnitude but opposite in direction, acceleration is also different and opposite in direction.

Final answer:

The electron and proton experience the same magnitude of force but different accelerations due to their mass difference, and move in opposite directions because of their opposite charges.

Explanation:

When a proton and an electron are placed in a uniform electric field and released, they both experience the same magnitude of force, because they have equal and opposite charges of the same magnitude. However, their accelerations differ due to their masses. The electron has a much smaller mass compared to the proton, and according to Newton's second law (F = ma), a given force will produce a larger acceleration on an object with a smaller mass.

Therefore, while the magnitudes of the forces are the same, the electron will experience a greater magnitude of acceleration than the proton. Finally, they move in opposite directions because the electric field exerts a force in the direction of the field on positive charges and in the opposite direction on negative charges. Therefore, the electron moves in the opposite direction to the proton when released in the same electric field.

Answer:

The speed of the ambulance is 4.66 m/s.

Explanation:

Given that,

The siren emitting a whine at 1570 Hz

The cyclist pedaling a bike at 2.45 m/s

The cyclist hears a frequency of 1560 Hz

We know that,

Speed of sound wave

[tex]v = 343\ m/s[/tex]

We calculate the speed of the ambulance

Using Doppler effect,

[tex]f'=f\times\dfrac{v+v_{o}}{v+v_{s}}[/tex]

Where,

[tex]f' [/tex]= frequency of ambulance siren

[tex]f [/tex]= cyclist hears the frequency

[tex]v_{s}[/tex]=speed of source

[tex]v_{v}[/tex]= speed of observer

Put the value in to the formula

[tex]v_{s}=f\times\dfrac{v+v_{o}}{f'}-v[/tex]

[tex]v_{s}=1570\times\dfrac{343-2.45}{1560}-343[/tex]

[tex]v_{s}=4.66\ m/s[/tex]

Hence, The speed of the ambulance is 4.66 m/s.

Answer:

(a) 6.42 x 10^-18 N

(b) 9.73 x 10^8 m/s^2

Explanation:

v = 760 m/s, B = 0.034 T, m = 6.6 x 10^-27 kg, q = 3.2 x 10^-19 C, theta = 51 degree

(a) F = q v B Sin theta

F = 3.2 x 10^-19 x 760 x 0.034 x Sin 51

F = 6.42 x 10^-18 N

(b) Acceleration, a = Force / mass

a = (6.42 x 10^-18) / (6.6 x 10^-27)

a = 0.973 x 10^9

a = 9.73 x 10^8 m/s^2

Answer:

Magnetic field, B = 0.0045 T            

Explanation:

It is given that,

Speed of the proton, [tex]v=1.5\times 10^7\ m/s[/tex]

Acceleration of the proton, [tex]a=0.66\times 10^{13}\ m/s^2[/tex]

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

The magnetic force is balanced by the force due to the acceleration of the proton as :

[tex]qvB=ma[/tex]

[tex]B=\dfrac{ma}{qv}[/tex]

[tex]B=\dfrac{1.67\times 10^{-27}\ kg\times 0.66\times 10^{13}\ m/s^2}{1.6\times 10^{-19}\ C\times 1.5\times 10^7\ m/s}[/tex]

B = 0.0045 T

So, the magnitude of magnetic field on the proton is 0.0045 T. Hence, this is the required solution.

Answer:

A) Apparent Weight = 590 N

Explanation:

As we know that frequency is given as

[tex]f = \frac{1}{30}[/tex]

[tex]f = 0.033 Hz[/tex]

now the angular speed is given as

[tex]\omega = 2\pi f[/tex]

[tex]\omega = 2\pi(0.033) = 0.21 rad/s[/tex]

now at the top position we will have

[tex]mg - N = m\omega^2 R[/tex]

[tex]N = mg - m\omega^2 R[/tex]

[tex]N = 600 - \frac{600}{9.8}(0.21)^2(4.0)[/tex]

[tex]N = 590 N[/tex]

The woman's apparent weight at the top of the Ferris wheel is less than her actual weight due to the centripetal force experienced during the uniform circular motion. This apparent weight can be calculated by subtracting the centripetal force from the gravitational force, yielding an answer of 590 N.

This problem revolves around the concepts of centripetal force and apparent weight in the context of uniform circular motion. The woman's apparent weight at the top of the Ferris wheel is less than her actual weight because of the centripetal force directed towards the center of the Ferris wheel.

To calculate the apparent weight, we should subtract the centripetal force from the gravitational force. The gravitational force is her actual weight, and since weight = mass * gravity, her mass equals 600N/9.8 m/s2 ~= 61.2 kg. The angular velocity of the Ferris wheel (ω) is 2π rad/30s since it makes one revolution every 30s. Using the formula centripetal force = m*ω2*r, we find that the centripetal force equals 61.2 kg * (2π rad/30s)2 * 4m = 10 N approximately.

Finally, the woman's apparent weight at the top is the gravitational force minus the centripetal force, or 600 N - 10 N, which equals 590 N. Therefore, the correct answer is A) 590 N.

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