A string of length L= 1.2 m and mass m = 20g is under 400 N of tension, its two ends are fixed. a. How many nodes will you see in the 5th harmonic.
i. 4
ii. 5
iii. 6
iv. 7
v. None of the above.

Final answer:

The longer wire will have twice the resistance of the shorter wire.

Explanation:

The resistance of a wire depends on both its length and cross-sectional area. In this case, the longer wire has twice the length and twice the cross-sectional area of the shorter wire. The resistance of a wire is directly proportional to its length and inversely proportional to its cross-sectional area. Therefore, the longer wire will have twice the resistance of the shorter wire.

Answer:

Factor by which the intrnal energy changes is 66.63

Explanation:

Given:

When the is motor operated normally, the current through the motor,I₁ =0.750A

Motor winding resistance, R = 19.6Ω

Thus, the Power dissipated at Normal condition,(P₁):

P₁=I₁²×R

P₁=(0.750)²×19.6= 11.025 W

Also,

Power = Rate of change of energy = [tex]\frac{dE}{dt}[/tex]

Now,

when the motor is connected to 120 V power supply, the current through the  motor (I₂) =[tex]\frac{V}{R}[/tex]

or

I₂ =[tex]\frac{120}{19.6}A[/tex]

or

I₂ =6.122 A

Thus, the power dissipated due the current I₂ will be

P₂ = I₂² × R

⇒ P₂ = (6.122)² × 19.6 = 734.693 W

Hence, the factor by which the internal energy rate in resistor winding increases is = [tex]\frac{P_2}{P_1}=\frac{734.693}{11.025}=[/tex]66.63

The solution involves using Ohm's Law and the formula for power dissipation to determine the increase in the rate of change of internal energy in the motor windings when the rotor seizes and compare it to normal operating conditions.

The student's question pertains to a direct current motor and involves calculating the rate of change of internal energy, back electromotive force (emf), and current under various scenarios. To address the student's question, one would need to apply Ohm's Law (V = IR, where V is voltage, I is current, R is resistance) and the power dissipation formula (P = I²R, where P is power).

Explanation of Solution

Initially, the motor's normal operation can be described by the values given: I = 0.750 A, V = 120 V, R = 19.6 Ω. The power dissipated (P) by the motor windings when the motor is operating normally would be P = I²R = (0.750 A)²(19.6 Ω).

However, if the rotor seizes, the motor will essentially become a static resistor and the current through it will increase. This is because without the back emf generated by the rotating rotor, the voltage across the motor windings will be the full supply voltage. The current in this case (let's call it I') can be calculated using Ohm's Law: I' = V/R = 120 V / 19.6 Ω. The new power dissipation (P') when the rotor seizes would be P' = I'²R. To find by what factor the rate of change of internal energy increases, we would need to compare P' to the original power dissipation P by calculating P'/P.

The maximum speed at which the truck can briefly move around the curve without making the toolbox slide can be found by setting the sliding acceleration equal to the centripetal acceleration (v²/r) and solving for v. Factors like tire friction and flat terrain are also taken into account.

To determine how fast the truck can go before the toolbox starts to slide, we need to calculate the maximum static friction (the force that prevents sliding).

The formula to identify the maximum speed that didn't result in sliding, we would use centripetal acceleration, which is the product of tangential speed squared divided by the curve's radius.

The acceleration of the crate on the truck bed must equal centripetal acceleration to prevent sliding. Given that the sliding acceleration is 2.06 m/s², setting this value to equal centripetal acceleration (v²/r where r is 22 m) and solving for v (velocity/speed), would provide the maximum speed at which the truck can move without the toolbox sliding.Tire friction also plays a role, as it allows vehicles to move at higher speeds without sliding off the road.

This also assumes that the road is relatively flat and the truck moves uniformly, similar to the situation illustrated in Figure 24.7 with a motorist moving in a straight direction.

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The speed at which a toolbox in a truck will start to slide when the truck is turning around a curve depends on the balance of the centripetal force and static friction force. These depend on the truck's speed, the curve's radius, the toolbox's mass and the friction coefficient. Without specific values, a numerical answer can't be given.

The question is asking about the speed at which the toolbox in the truck would start sliding due to the forces acting upon it when the truck turns around a curve. This is related to centripetal force and frictional force.

When the truck turns, the toolbox experiences a centripetal force which pushes it towards the center of the curve. At the same time, friction between the toolbox and the truck bed is resisting this pushing force, keeping the box in place. At some point, if the truck is going too fast, the centripetal force will overcome friction, and the toolbox will start to slide.

We can use the formula for centripetal force, which is F = mv²/r, where m is the mass of the toolbox, v is the velocity of the truck and r is the radius of the curve. And we know that the maximum static friction force (F_max) is f_s * m * g, where f_s is the static friction coefficient and g is the acceleration due to gravity.

The toolbox starts to slide when F = F_max, so we can find the maximum speed (v_max) before sliding happens by making these two equations equal to each other and solving for v. Without the specific values for the mass of the toolbox and the static friction coefficient, we cannot calculate a numerical answer.

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Answer:

(a) 10.52 A

(b) 11 ohm

(c) 1.77 x 10^7 J

Explanation:

P = 1210 W, V = 115 V,

Let i be the current.

(a) Use the formula of power

P = V i

1210 = 115 x i

i = 10.52 A

(b) Let the resistance of heating coil is R.

Use the formula given below

V = i x R

R = V / i = 115 / 10.52 = 11 ohm

(c) Use the formula for energy

E = V x i x t  

E = 115 x 10.52 x 4.07 x 60 x 60 = 1.77 x 10^7 J

Answer:

Part a)

E = 4812.5 N/C

Part b)

Two field must be perpendicular to each other as well as perpendicular to the velocity of charge so that net force on the moving charge will be zero

Explanation:

For uniform electric and magnetic field we will have a charge particle moving through it such that net force on it is zero

so here we have

[tex]qE = qvB[/tex]

magnetic force on moving charge will balanced by electrostatic force on moving charge

[tex]v = \frac{E}{B}[/tex]

now we have

[tex]v = 8.75 km/s[/tex]

B = 0.550 T

now we have

[tex]E = (8.75 \times 10^3)(0.550)[/tex]

[tex]E = 4812.5 N/C[/tex]

Part b)

Two field must be perpendicular to each other as well as perpendicular to the velocity of charge so that net force on the moving charge will be zero

Answer:

energy dissipated = 132 kJ

Explanation:

mass = 61 kg

height drop = 227 m

velocity = 11 m/s

potential energy due to height drop from top to bottom

                         P.E. =  m g h

                         P.E. =  61× 9.8× 227

                         P.E. = 135,700 J

kinetic energy = [tex]\frac{1}{2}mv^2[/tex]

                        = [tex]\frac{1}{2}\times 61 \times 11^2[/tex]

                        = 3690.5 J

energy dissipated = P.E - K.E.

                              = 135,700 J -3690.5 J

                              =132,009.5 J = 132 kJ

Explanation:

1. Force applied on an object is given by :

F = W = mg

(a) A 160 lb human being, F = 160 lb

g = acceleration due to gravity, g = 32 ft/s²

[tex]m=\dfrac{F}{g}[/tex]

[tex]m=\dfrac{160\ lb}{32\ ft/s^2}[/tex]

m = 5 kg

(b) A 1.9 lb cockatoo, F = 1.9 lb

[tex]m=\dfrac{F}{g}[/tex]

[tex]m=\dfrac{1.9\ lb}{32\ ft/s^2}[/tex]

m = 0.059 kg

2. (a) A 2300 kg rhinoceros, m = 2300 kg

[tex]W=2300\ kg\times 32\ ft/s^2=73600\ lb[/tex]

(b) A 22 g song sparrow, m = 22 g = 0.022 kg

[tex]W=0.022\ kg\times 32\ ft/s^2=0.704\ lb[/tex]

Hence, this is the required solution.

Final answer:

The mass of a 160 lb human being is 72.576 kg, and the mass of a 1.9 lb cockatoo is 0.86184 kg. A 2300 kg rhinoceros weighs 5060 lbs in English units, while a 22 g song sparrow weighs 0.0484 lbs.

Explanation:

To calculate the mass (in SI units) of a 160 lb human being, we use the conversion factor 1 lb = 0.4536 kg. So, the calculation is:160 lb x 0.4536 kg/lb = 72.576 kg

(a) The mass of a 160 lb human being in SI units is 72.576 kg.

For a 1.9 lb cockatoo, the conversion to kilograms is:1.9 lb x 0.4536 kg/lb
 = 0.86184 kg

(b) The mass of a 1.9 lb cockatoo in SI units is 0.86184 kg

To convert the mass of a 2300 kg rhinoceros to weight in English units, knowing that 1 kg = 2.2 lbs (where weight in pounds is considered the gravitational force on the mass), the weight is calculated as:

2300 kg x 2.2 lbs/kg= 5060 lbs

(c) The weight of a 2300 kg rhinoceros in English units is 5060 lbs.

Finally, for converting a 22 g song sparrow to weight in English units:

22 g x (1 kg/1000 g) x (2.2 lbs/kg)= 0.0484 lbs

(d) The weight of a 22 g song sparrow in English units is 0.0484 lbs.

Answer:

0.2447 m

Explanation:

Amplitude, A = 0.25 m, T = 5.67 s, t = 29.6 s, y = ?

The general equation of SHM is given by

y = A Sin wt

y = 0.25 Sin (2 x 3.14 t /5.67)

Put t = 29.6 s

y = 0.25 Sin (2 x 3.14 x 29.6 / 5.67)

y = 0.2447 m

Answer:

The wavelength of electron is [tex]6.99\times 10^{-22}\ m[/tex]

Explanation:

The kinetic energy of the electron is, [tex]E=0.5\ MeV=0.5\times 10^6\ eV[/tex]

We need to find the wavelength of this electron. It can be calculated using the concept of DE-broglie wavelength as :

[tex]\lambda=\dfrac{h}{\sqrt{2mE} }[/tex]

h is Plank's constant

m is the mass of electron

[tex]\lambda=\dfrac{6.67\times 10^{-34}\ J-s}{\sqrt{2\times 9.1\times 10^{-31}\ kg\times 0.5\times 10^6\ eV} }[/tex]        

[tex]\lambda=6.99\times 10^{-22}\ m[/tex]

So, the wavelength of electron is [tex]6.99\times 10^{-22}\ m[/tex]. Hence, this is the required solution.

The wavelength of an electron with a kinetic energy of 0.50 MeV (relativistic) is approximately 7.28 x 10^-12 m.

The wavelength of an electron with a kinetic energy of 0.50 MeV can be calculated using the relativistic de Broglie equation:

λ = h/(m*c)

Where λ is the wavelength, h is Planck's constant (6.63 x 10^-34 Js), m is the mass of the electron (9.11 x 10^-31 kg), and c is the speed of light (3.00 x 10^8 m/s).

Substituting the values:

λ = (6.63 x 10^-34 Js)/((9.11 x 10^-31 kg)*(3.00 x 10^8 m/s))

λ ≈ 7.28 x 10^-12 m

Therefore, the wavelength of the electron is approximately 7.28 x 10^-12 m.

Answer:

The ratio of the young's modulus of steel and copper is [tex]1.79\times10^{-2}[/tex]

(c) is correct option.

Explanation:

Given that,

Length of steel wire = 4.7 m

Cross section[tex]A = 3\times10^{-3}\ m^2[/tex]

Length of copper wire = 3.5 m

Cross section[tex]A = 4\times10^{-5}\ m^2[/tex]

We need to calculate the ratio of young's modulus of steel and copper

Using formula of young's modulus for steel wire

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

[tex]Y_{s}=\dfrac{Fl_{s}}{A_{s}\Delta l}[/tex]....(I)

The young's modulus for copper wire

[tex]Y_{c}=\dfrac{Fl_{c}}{A_{c}\Delta l}[/tex]....(II)

From equation (I) and (II)

The ratio of the young's modulus of steel and copper

[tex]\dfrac{Y_{s}}{Y_{c}}=\dfrac{\dfrac{Fl_{s}}{A_{s}\Delta l}}{\dfrac{Fl_{c}}{A_{c}\Delta l}}[/tex]

[tex]\dfrac{Y_{s}}{Y_{c}}=\dfrac{A_{c}\times l_{s}}{A_{s}\times l_{c}}[/tex]

[tex]\dfrac{Y_{s}}{Y_{c}}=\dfrac{4\times10^{-5}\times4.7}{3\times10^{-3}\times3.5}[/tex]

[tex]\dfrac{Y_{s}}{Y_{c}}=1.79\times10^{-2}[/tex]

Hence, The ratio of the young's modulus of steel and copper is [tex]1.79\times10^{-2}[/tex]

Answer:

99 dB

Explanation:

We have given that sound intensity with a damaged muffler =8× [tex]10^{-3}[/tex]

we have to find the intensity level of the this sound in decibels

for calculating in decibels we have to use the formula

β=[tex]10log\frac{I}{I_0}[/tex]

  =[tex]10log\frac{.008}{10^{-12}}[/tex]

   =10 log8+10 log[tex]10^{9}[/tex]

   =10 log8+90 log10

   =10×0.9030+90

   =99 dB

Final answer:

To calculate the intensity level of a sound in decibels, use the formula β = 10 log10(I/I0). Given the intensity of [tex]8.0 \times 10^{-3[/tex] threshold of hearing of [tex]10^{-12[/tex] the sound intensity level is approximately 99 decibels.

Explanation:

The student asks about calculating the intensity level of sound in decibels, given the sound intensity at the ear of a passenger in a car with a damaged muffler (8.0 × 10-3 W/m2). To find the intensity level in decibels (dB), we use the formula:

β = 10 log10(I/I0)

where:

Plugging in the values:

β = 10 log10(8.0 × 10-3 / 10-12)

β = 10 log10(8.0 × 109)

β = 10 (log10(8) + log10(109))

β = 10 (0.903 + 9)

β = 10 × 9.903

β = 99.03 dB

The intensity level of the sound in the car with the damaged muffler is approximately 99 decibels.

Answer:

Frequency, [tex]f=1.2\times 10^{10}\ Hz[/tex]

Explanation:

It is given that,

Magnetic field, B = 0.43 T

We need to find the frequency the electron traverse a circular path. It is also known as cyclotron frequency. It is given by :

[tex]f=\dfrac{qB}{2\pi m}[/tex]

[tex]f=\dfrac{1.6\times 10^{-19}\ C\times 0.43\ T}{2\pi \times 9.11\times 10^{-31}\ Kg}[/tex]

[tex]f=1.2\times 10^{10}\ Hz[/tex]

Hence, this is the required solution.

Answer:

[tex]\dot{W} = 339.84 W[/tex]

Explanation:

given data:

flow Q = 9 m^{3}/s

velocity = 8 m/s

density of air = 1.18 kg/m^{3}

minimum power required to supplied to the fan is equal to the POWER POTENTIAL of the kinetic energy and it is given as

[tex]\dot{W} =\dot{m}\frac{V^{2}}{2}[/tex]

here [tex]\dot{m}[/tex]is mass flow rate and given as

[tex]\dot{m} = \rho*Q[/tex]

[tex]\dot{W} =\rho*Q\frac{V^{2}}{2}[/tex]

Putting all value to get minimum power

[tex]\dot{W} =1.18*9*\frac{8^{2}}{2}[/tex]

[tex]\dot{W} = 339.84 W[/tex]

The minimum power required to accelerate air to a velocity of 8 m/s at a rate of 9 m^3/s with a density of 1.18 kg/m^3 is roughly 339.84 Watts.

The subject of this question is physics, specifically the topic of power in mechanical systems. To find the minimum power required, we use the formula Power = 0.5 * density * volume flow rate * velocity^2. Plugging in the given values: Power = 0.5 * 1.18 kg/m^3 * 9 m^3/s * (8 m/s)^2, we get approximately 339.84 W. Therefore, the minimum power that must be supplied to the fan is roughly 339.84 Watts.

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Answer:

(a) 1.042 J

b) 0.108 m

Explanation:

(a)

W = work done to stretch the spring = 2 J

k = spring constant

L₀ = initial length = 30 cm = 0.30 m

L = final length = 42 cm = 0.42 m

x = stretch of the spring = L - L₀ = 0.42 - 0.30 = 0.12 m

Work done to stretch the spring is given as

W = (0.5) k x²

2 = (0.5) k (0.12)²

k = 277.78 N/m

x₀ = initial stretch of the spring = 35 - 30 = 5 cm = 0.05 m

x = final stretch of the spring = 40 - 30 = 10 cm = 0.10 m

W' = work needed to stretch the spring from 35 cm to 40 cm

Work needed to stretch the spring from 35 cm to 40 cm is given as

W' = (0.5) k (x² - x₀²)

W' = (0.5) (277.78) (0.10² - 0.05²)

W' = 1.042 J

b)

x = Stretch of the spring beyond natural length

F = force = 30 N

Spring force is given as

F = k x

30 = (277.78) x

x = 0.108 m

(a) The work is needed to stretch the spring from 35 cm to 40 cm is 1.042 J.

(b) Beyond the length of 0.108 m  force of 30 N keep the spring stretched.

In order to stretch or compress the spring some amount of work has to be done. Which is given as,

          [tex]\rm{W=\frac{1}{2} Kx^{2} }[/tex]  

where k = spring constant of the spring

          x = compression of spring

          W= Work required or spring work

Initial length is given by 30 cm = 0.30 m

final length is given by  42 cm = 0.42 m

When spring is stretched change in the length occurs denoted by x

x = final length - initial length =0.42 - 0.30 = 0.12 m

[tex]\rm{W=\frac{1}{2} Kx^{2} }[/tex]

[tex]\rm{2=\frac{1}{2} K(0.12)^{2} }[/tex]

K= 277.78 N/m

(a) work is needed to stretch the spring from 35 cm to 40 cm

stretching length from30cm to 35 is 5 cm which is 0.05 m

stretching length from cm to 30 is 40 cm is 10 cm which is 0.1m

x = final stretch of the spring = 40 - 30 = 10 cm = 0.10 m

Work needed to stretch the spring from 35 cm to 40 cm is given

W = (0.5)K([0.1]² - [0.5]²)

W = (0.5) (277.78) (0.10² - 0.05²)

W = 1.042 J

x is stretch of the spring beyond natural length

F = force = 30 N

Spring force is given as

F = k x

30 = (277.78) x

x = 0.108 m

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The angular momentum of a rotation object is the product of its moment of inertia and its angular velocity:

L = Iω

L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.

Apply the conservation of angular momentum. The total angular momentum before disks A and B are joined is:

L_{before} = (3.3)(6.6) + B(-9.3)

L_{before} = -9.3B+21.78

where B is the moment of inertia of disk B.

The total angular momentum after the disks are joined is:

L_{after} = (3.3+B)(-2.1)

L_{after} = -2.1B-6.93

L_{before} = L_{after}

-9.3B + 21.78 = -2.1B - 6.93

B = 4.0kg·m²

The moment of inertia of disk B is 4.0kg·m²

Answer:

72 m/s

Explanation:

D1 = 3 cm, v1 = 2 m/s

D2 = 0.5 cm,

Let the velocity at narrow end be v2.

By use of equation of continuity

A1 v1 = A2 v2

3.14 × 3 × 3 × 2 = 3.14 × 0.5 ×0.5 × v2

v2 = 72 m/s

Answer:

Option C is the correct answer.

Explanation:

We equation for elongation

      [tex]\Delta L=\frac{PL}{AE}[/tex]

Here we need to find load required,

We need to double the wire, that is ΔL = 2L - L = L

A = 5 x 10⁻⁵ m²

E = 2 x 10¹¹ N/m²

Substituting

     [tex]L=\frac{PL}{5\times 10^{-5}\times 2\times 10^{11}}\\\\P=10^7N[/tex]

Option C is the correct answer.

Final answer:

The angular magnification produced by a +4.00 cm focal length lens with an object at 3.75 cm distance is approximately 7.25 times.

Explanation:

The question presented is a problem in optics, more specifically in the subtopic of lens magnification, which is a part of Physics. To calculate the angular magnification for a lens, we can use the thin lens formula and the magnification formula. Given that the focal length of the lens (f) is +4.00 cm and the object distance (do) is 3.75 cm, we first determine if an image is formed by using the lens equation 1/f = 1/do + 1/di. However, since this equation will not yield a real value for di (image distance) because do is smaller than f, we can't directly use it to find the angular magnification.

Instead, the angular magnification (M) of a simple lens while viewing a close object is approximately given by M = 1 + (25 cm / f), where 25 cm is the near point of a normal human eye. Thus, for this lens:

M = 1 + (25 cm / 4.00 cm)

M = 1 + 6.25

M = 7.25

This means the lens would produce an angular magnification of approximately 7.25 times when the object is placed at 3.75 cm from the lens.

Answer:

2.87 Volt

Explanation:

N = 19

change in magnetic field = 5.6 T

Time, dt = 744 ms = 0.744 s

Induced emf,

e = N × change in flux / time

e = 19 × 3.14 × 0.08 × 0.08 ×5.6/0.744

e = 2.87 Volt

Answer:

The observer experienced a change I'm frequency of 6971.42Hertz

Explanation:

In waves, the frequency of a wave is directly proportional to its velocity i.e F ∝ V

F = kV

k = F/V

F1/V1 = F2/V2 = k... 1

Given F1 = 800Hz at V1 = 35m/s

F2 = ? when V2 = 340m/s

Substituting this given datas into equation 1 to get the new frequency F2:

800/35 = F2/340

35F2 = 800×340

F2 = 800×340/35

F2 = 7771.42Hertz

Change of frequency = F2-F1

Change of frequency = 7771.42-800

Change of frequency = 6971.42Hertzz

The change of frequency does the observer hear due to police car siren is 6971.42 Hz.

Frequency of wave is the number of waves, which is passed thorough a particular point at a unit time.

For the cars approaching to the observer, the Doppler formula can be given as

[tex]\dfrac{f_1}{v_1}=\dfrac{f_2}{v_s}[/tex]

Here, [tex](v_s)[/tex] is the speed of the sound.

A police car has an 800-Hz siren. It is traveling at 35.0 m/s on a day when the speed of sound through air is 340 m/s.

As, the car approaches and passes an observer. Thus by the above formula,

[tex]\dfrac{800}{35}=\dfrac{f_2}{340}\\f_2=7771.41\rm Hz[/tex]

As the frequency of police car siren is 800-Hz. Thus, the change of the frequency observed by the observer is,

[tex]\Delta f=7771.42-8000\\\Delta f=6971.42\rm Hz[/tex]

Thus, change of frequency does the observer hear due to police car siren is 6971.42 Hz.

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Explanation:

A batter hits a baseball into the air. The formula that models the baseball’s height above the ground, y, in feet, x seconds after it is hit is given by :

[tex]y=-16x^2+64x+5[/tex].........(1)

(a) We need to find the maximum height reached by the baseball. The maximum height reached is calculated by differentiating equation (1) wrt x as :

[tex]\dfrac{dy}{dx}=0[/tex]

[tex]\dfrac{d(-16x^2+64x+5)}{dx}=0[/tex]

[tex]-32x+64=0[/tex]

x = 2 seconds

(b) The height of the ball is given by equation (1) and putting the value of x in it as :

[tex]y=-16(2)^2+64(2)+5[/tex]

y = 69 feets

So, at 2 seconds the baseball reaches its maximum height and its maximum height is 69 feets.

Answer:

wow

Explanation:

Answer:

A)

0.8 c

B)

0.87 c

Explanation:

A)

L₀ = Original length of the meter stick = 1 m

L = Length observed = 0.6 m

[tex]v[/tex] = speed of the meter stick

Using the equation

[tex]L = L_{o} \sqrt{1 - \left ( \frac{v}{c} \right )^{2}}[/tex]

[tex]0.6 = 1 \sqrt{1 - \left ( \frac{v}{c} \right )^{2}}[/tex]

[tex]0.36 = 1 - \left ( \frac{v}{c} \right )^{2}[/tex]

[tex]v[/tex] = 0.8 c

B)

T₀ = Time of the clock at rest = t

T = Time of the clock at motion = (0.5) t

[tex]v[/tex] = speed of the clock

Using the equation

[tex]T = T_{o} \sqrt{1 - \left ( \frac{v}{c} \right )^{2}}[/tex]

[tex]0.5 t = t \sqrt{1 - \left ( \frac{v}{c} \right )^{2}}[/tex]

[tex]0.5  =  \sqrt{1 - \left ( \frac{v}{c} \right )^{2}}[/tex]

v = 0.87 c

Answer:

23.98 rpm

Explanation:

d = diameter of merry-go-round = 2.4 m

r = radius of merry-go-round = (0.5) d = (0.5) (2.4) = 1.2 m

m = mass of merry-go-round = 270 kg

I = moment of inertia of merry-go-round

Moment of inertia of merry-go-round is given as  

I = (0.5) m r² = (0.5) (270) (1.2)² = 194.4 kgm²

M = mass of john = 34 kg

Moment of inertia of merry-go-round and john together after jump is given as

  I' = (0.5) m r² + M r² = 194.4 + (34) (1.2)² = 243.36 kgm²

w = final angular speed

w₀ = initial angular speed of merry-go-round = 20 rpm = 2.093 rad/s

v = speed of john before jump

using conservation of angular momentum

Mvr + I w₀ = I' w

(34) (5) (1.2) + (194.4) (2.093) = (243.36) w

w = 2.51 rad/s

w = 23.98 rpm

Answer:

111 N

Explanation:

weight (downwards) = 600 N

reaction force (upwards) = 900 N

t = 0.37 s

Net force

F = reaction force - weight = 900 - 600 = 300 N

Impulse = force x time = 300 x 0.37 = 111 N

Explanation:

The number of charges flowing per second inside the conductor is called the current through the conductor. Mathematically, it is given by :

[tex]I=\dfrac{Q}{t}[/tex]

Where

Q is the charge

t is time

Since, Q = ne

n is number of electrons

e is the charge on one electron

So, to find the current through the conductor we must know the number of electrons moving per second through it. Hence, this is the required solution.

Answer:

Electric field, E = 45.19 N/C

Explanation:

It is given that,

Surface charge density of first surface, [tex]\sigma_1=0.2\ nC/m^2=0.2\times 10^{-9}\ C/m^2[/tex]

Surface charge density of second surface, [tex]\sigma=-0.6\ nC/m^2=-0.6\times 10^{-9}\ C/m^2[/tex]

The electric field at a point between the two surfaces is given by :

[tex]E=\dfrac{\sigma}{2\epsilon_o}[/tex]

[tex]E=\dfrac{\sigma1-\sigma_2}{2\epsilon_o}[/tex]

[tex]E=\dfrac{0.2\times 10^{-9}-(-0.6\times 10^{-9})}{2\times 8.85\times 10^{-12}}[/tex]

E = 45.19 N/C

So, the magnitude of the electric field at a point between the two surfaces is 45.19 N/C. Hence, this is the required solution.

Answer:

2954.6 N/C, 46.36 degree from positive  axis

Explanation:

E1 = 1300 N/C, θ1 = 35 degree

E2 = 1700 N/C, θ2 = 55 degree

Now write the electric fields in vector form

E1 = 1300 ( Cos 35 i + Sin 35 j) = 1064.9 i + 745.6 j

E2 = 1700 ( Cos 55 i + Sin 55 j) = 975.08 i + 1392.6 j

Resultant electric field

E = E1 + E2

E =  1064.9 i + 745.6 j + 975.08 i + 1392.6 j

E = 2039.08 i + 2138.2 j

Magnitude of E

E = sqrt (2039.08^2 + 2138.2^2)

E = 2954.6 N/C

Let it makes an angle Φ from X axis

tan Φ = 2138.2 / 2039.08 = 1.049

Φ = 46.36 degree from positive X axis.

This question is about calculating the overall electric field from two individual fields using vector addition. This involves resolving each field into its components, adding the respective components, and then using Pythagoras' theorem and the tangent inverse relation to determine the overall magnitude and direction of the resultant field.

The total electric field can be determined by using vector addition to add together the individual electric fields due to each of the two components mentioned in your question. Firstly, we resolve each field into its x and y components. The x-component and y-component for E1 are E1*cos(θ1) and E1*sin(θ1), respectively. Similarly, the x and y components for E2 are E2*cos(θ2) and E2*sin(θ2), respectively.

We then add together the respective x and y components: E(x total) = E1*cos(θ1) + E2*cos(θ2) and E(y total) = E1*sin(θ1) + E2*sin(θ2).

The overall magnitude of the resultant electric field can be calculated using Pythagoras' theorem, √[(E(x total))^2 + (E(y total))^2]. The direction of the total electric field can be evaluated using the tangent inverse relation θ total = tan^-1 [E(y total)/E(x total)].

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The center of mass of these objects is located at approximately 12.652 meters on the y-axis.

We can find the center of mass (center of gravity) of these objects by calculating the weighted average of their positions along the y-axis. Here's how:

Consider Each Object:

We have four objects with masses m1, m2, m3, and m4 at positions y1, y2, y3, and y4 respectively.

Center of Mass Formula:

The center of mass coordinate (y_cm) is calculated as:

y_cm = (Σ(mi * yi)) / Σ(mi)

Σ (sigma) represents summation over all objects (i = 1 to 4 in this case).

mi is the mass of the i-th object.

yi is the y-axis position of the i-th object.

Apply the formula to our case:

y_cm = [(2.00 kg * 13.00 m) + (3.00 kg * 12.50 m) + (2.50 kg * 0.00 m) + (4.00 kg * 20.500 m)] / (2.00 kg + 3.00 kg + 2.50 kg + 4.00 kg)

Calculate:

y_cm = [26.00 kgm + 37.50 kgm + 0.00 kgm + 82.00 kgm] / 11.50 kg

y_cm ≈ 145.50 kg*m / 11.50 kg ≈ 12.652 m (rounded to four decimal places)

Therefore, the center of mass of these objects is located at approximately 12.652 meters on the y-axis.

Answer:

speed, distance  and direction of motion of the object can be determined by analyzing the radio wave.

Explanation:

We know that radar operates by transmitting radio waves to a destination and these waves are comes back to the receiver station. By Considering this transmission and receiver process, we can measure the distance, velocity and path of an object's movement.

Distance can be assessed by taking following consideration,  the velocity of the waves is V. It can help to assess the time made for the waves to be emitted by the radar and felt by the receiver, let the time be t.

Therefore  distance can be determine as D= v*t/2,

here 2 signifies that the distance travelled by the wave in either direction ( from transmitter to receiver and vice verse)

Using the source wave frequency, speed can be computed. In a specific frequency, the radar starts sending out the frequencies and the reflected wave will have a distinct frequency. The velocity can be determine by

[tex]v= (\Delta f/f)(c/2),[/tex]

where[tex]\Delta f[/tex]is the change in frequency and

c is the speed of light (the wave).

Direction can be determine by applying above principle. change in frequency is used to determine the direction in the following way:

When the frequency transition is very low, the object moves away from the radar and vice verse.

Answer:

(A) 667.5 N/m

(B)

Explanation:

(A) Let the spring constant be k.

Using the formula F = kx

k = 251 / 0.376

K = 667.5 N/m

(B)

Work done

W = 0.5 × kx^2

W = 0.5 × 667.5 × 0.376 × 0.376

W = 47.2 J