Two disks are rotating about the same axis. Disk A has a moment of inertia of 3.3 kg · m2 and an angular velocity of +6.6 rad/s. Disk B is rotating with an angular velocity of -9.3 rad/s. The two disks are then linked together without the aid of any external torques, so that they rotate as a single unit with an angular velocity of -2.1 rad/s. The axis of rotation for this unit is the same as that for the separate disks. What is the moment of inertia of disk B?

Answer:

39.08 m

Explanation:

θ = 31.6 degree

u = 29.3 m/s

Let it hits at the maximum height h. It takes half of the time of flight to cover the distance.

T = 2 u Sinθ / g

t = T / 2 = u Sinθ / g = (29.3 x Sin 31.6) / 9.8 = 1.566 s

The horizontal distance covered in this time is

d = u Cosθ x t = 29.3 x Cos 31.6 x 1.566 = 39.08 m

Thus, the fire hose be located at a distance of 39.08 m from the building.

Answer:

Not Changed

Explanation:

To know what happened with the volume you need to know the Ideal gas Law

[tex]\frac{P_{1}V_{1}}{T1} =\frac{P_{2}V_{2} }{T2}[/tex]

This law is a combination of the other four laws: Boyles's, Charles's, Avogadro's, and Guy-Lussac's.

The initial state is represented by P1, V1, T1 and the final by P2, V2, T2.

In this case:

[tex]T_{2} =3T_{1} \\P_{2} =3P_{1}[/tex]

Replacing on the equation

[tex]\frac{P_{1}V_{1}}{T1} =\frac{3P_{1}V_{2}}{3T_{1} }[/tex]

If we clear from the equation V2

[tex]\frac{P_{1}V_{1}3T_{1}}{T_{1} 3P_{1}} ={V_{2}}[/tex]

Then cancel both P1 and T1

[tex]\frac{3V_{1}}{3} =V_{2}[/tex]

You will found that

[tex]V_{1} =V_{2}[/tex]

Answer:

The normal force between the motorcycle and track at the top of the loop = 2343.75 N

Explanation:

At the top of the loop the track is directed tangential to the motion, the normal to the tangent is along radius of circle. that is here we need to find centripetal force.

We have expression for centripetal force

                  [tex]F=\frac{mv^2}{r}[/tex]

Here mass, m = 175 kg

        Velocity, v = 7.5 m/s

        Diameter, d = 8.4 m

        Radius, r = 0.5d = 0.5 x 8.4 = 4.2 m

Substituting

        [tex]F=\frac{mv^2}{r}=\frac{175\times 7.5^2}{4.2}=2343.75N[/tex]

The normal force between the motorcycle and track at the top of the loop = 2343.75 N

Answer:

0.9

Explanation:

h = 400 mm, h' = 325 mm

Let the coefficient of restitution be e.

h' = e^2 x h

325 = e^2 x 400

e^2 = 0.8125

e = 0.9

The coefficient of restitution between the ball and the surface is 0.9.

The coefficient of restitution (e) between the ball and the surface can be determined using the formula:

[tex]\[ e = \sqrt{\frac{\text{height after collision}}{\text{height before collision}}} \][/tex]

First, we need to convert the heights from millimeters to meters for consistency, since the standard units for height in physics are meters.

The initial height (h_i) before the collision is 400 mm, which is equivalent to 0.4 m (since 1 m = 1000 mm).

The final height (h_f) after the collision is 325 mm, which is equivalent to 0.325 m.

Now, we can plug these values into the formula for the coefficient of restitution:

[tex]\[ e = \sqrt{\frac{0.325 \text{ m}}{0.4 \text{ m}}} \][/tex]

[tex]\[ e = \sqrt{\frac{325}{400}} \][/tex]

[tex]\[ e = \sqrt{0.8125} \][/tex]

[tex]\[ e = 0.9 \][/tex]

Therefore, the coefficient of restitution between the ball and the surface is 0.9. This means that the collision is relatively elastic, with the ball retaining a significant portion of its initial kinetic energy after the rebound.

Answer:

A) 0.660 g/ml

B) 1.297 ml

C) 0.272 g

Explanation:

Every substance, body or material has mass and volume, however the mass of different substances occupy different volumes.  This is where density [tex]D[/tex] appears as a  physical characteristic property of matter that establishes a relationship between the mass [tex]m[/tex] of a body or substance and the volume [tex]V[/tex] it occupies:

[tex]D=\frac{m}{V}[/tex] (1)

Knowing this, let's begin with the answers:

Here the mass is [tex]m=0.155g[/tex] and th volume [tex]V=0.000235L=0.235mL[/tex]

Solving (1) with these values:

[tex]D=\frac{0.155g}{0.235mL}[/tex] (2)

[tex]D=0.660g/mL[/tex] (3)

In this case the mass of a sample is [tex]m=4.71g[/tex] and its density is [tex]D=3.63g/mL[/tex].

Isolating [tex]V[/tex] from (1):

[tex]V=\frac{m}{D}[/tex] (4)

[tex]V=\frac{4.71g}{3.63g/mL}[/tex] (5)

[tex]V=1.297mL[/tex] (5)

In this case the volume of a sample is [tex]V=0.293mL[/tex] and its density is [tex]D=0.930g/mL[/tex].

Isolating [tex]m[/tex] from (1):

[tex]m=D.V[/tex] (6)

[tex]m=(0.930g/mL)(0.293mL)[/tex] (7)

[tex]m=0.272g[/tex] (8)

To compute density, volume or mass, utilize the density formula (Density = Mass / Volume). For problem A, the liquid's density is 0.660 g/mL. For B, the volume of the sample is 1.30 mL, and for C, the mass of the sample is 0.272 g.

The formula to calculate density is Density = Mass / Volume. A. To calculate the density in g/mL of a liquid with a mass of 0.155 g and a volume of 0.000235 L, we first need to convert the volume from liters to milliliters: 0.000235 L is equal to 0.235 mL. Then, divide the mass by the volume to get the density: Density = 0.155 g / 0.235 mL = 0.660 g/mL.

B. To calculate the volume in milliliters of a 4.71-g sample of a solid with a density of 3.63 g/mL, divide the mass by the density: Volume = 4.71 g / 3.63 g/mL = 1.30 mL.

C. To calculate the mass of a 0.293-mL sample of a liquid with a density of 0.930 g/mL, multiply the volume by the density: Mass = 0.293 mL * 0.930 g/mL = 0.272 g.

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Answer:

current = 1.51 A

Explanation:

Initially the capacitor without any dielectric is connected across AC source

so the capacitive reactance of that capacitor is given as

[tex]x_c = \frac{1}{\omega c}[/tex]

now we have

[tex]i = \frac{V_{rms}}{x_c}[/tex]

here we know that

[tex]i = 0.29 A[/tex]

now other capacitor with dielectric of 4.2 is connected in parallel with the first capacitor

so here net capacitance is given as

[tex]c_{eq} = 4.2c + c = 5.2c[/tex]

now the equivalent capacitive reactance is given as

[tex]x_c' = \frac{1}{\omega(5.2c)}[/tex]

[tex]x_c' = \frac{x_c}{5.2}[/tex]

so here we have new current in that circuit is given as

[tex]i' = \frac{V_{rms}}{x_c'}[/tex]

[tex]i' = 5.2 (i) = 5.2(0.29)[/tex]

[tex]i' = 1.51 A[/tex]

Answer:

option (a)

Explanation:

Wavelength is defined as the distance traveled by the wave in one complete oscillation.

The number of oscillations completed in one second is called frequency.

The relation for the wave velocity is given by

wave velocity = frequency x wavelength

Answer : The time taken for the concentration will be, 7.98 seconds

Explanation :

First order reaction : A reaction is said to be of first order if the rate is depend on the concentration of the reactants, that means the rate depends linearly on one reactant concentration.

Expression for rate law for first order kinetics is given by :

[tex]k=\frac{2.303}{t}\log\frac{[A]_o}{[A]}[/tex]

where,

k = rate constant  = [tex]0.150s^{-1}[/tex]

t = time taken for the process  = ?

[tex][A]_o[/tex] = initial concentration = 0.860 M

[tex][A][/tex] = concentration after time 't' = 0.260 M

Now put all the given values in above equation, we get:

[tex]0.150s^{-1}=\frac{2.303}{t}\log\frac{0.860}{0.260}[/tex]

[tex]t=7.98s[/tex]

Therefore, the time taken for the concentration will be, 7.98 seconds

To determine the time for the concentration of substance A to decrease in a first-order reaction, the formula t = (1/k) * ln([A]0/[A]t) is used with the known rate constant and initial and final concentrations.

The question asks how long it would take for the concentration of substance A to decrease from 0.860 M to 0.260 M given a first-order reaction with a rate constant of 0.150 s−¹ at 400 °C.

For a first-order reaction, the time (t) it takes for the concentration of a reactant to change can be found using the formula:

t = (1/k) * ln([A]0/[A]t)

Where:

In this case, we can substitute the given values into the equation:

t = (1/0.150 s−¹) * ln(0.860 M / 0.260 M) = (1/0.150 s−¹) * ln(3.3077)

The time can be calculated by completing the computation for ln(3.3077) and dividing by the rate constant.

Answer:

(a) 85.64°F

(b) 78.28°C

(c)  - 268.8°C  

Explanation:

(a) 29.8°C into °F

To convert °C into °F, we use the formula given below

C / 100 = (F - 32) / 180

C / 5 = (F - 32) / 9

9 x 29.8 / 5 = F - 32

F = 85.64°F

(b) 172.9°F into °C

C / 5 = (F - 32) / 9

C = (172.9 - 32) x 5 / 9

C = 78.28°C

(c) 4.2 K into °C

C = K - 273

C = 4.2 - 273 = - 268.8°C

Answer:

23

Explanation:

First, we need to convert the hose diameter from inches to meters.

0.75 in × (2.54 cm / in) × (1 m / 100 cm) = 0.0191 m

Calculate the flow rate given the velocity and hose diameter:

Q = vA

Q = v (¼ π d²)

Q = (0.30 m/s) (¼ π (0.0191 m)²)

Q = 8.55×10⁻⁵ m³/s

Find the volume of the pool:

V = π r² h

V = π (1.5 m)² (1.0 m)

V = 7.07 m³

Find the time:

t = V / Q

t = (7.07 m³) / (8.55×10⁻⁵ m³/s)

t = 82700 s

t = 23 hr

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:

Speed, v = 6.91 m/s

Explanation:

Given that,

Spring constant, k = 594 N/m

It is attached to a table and is compressed down by 0.196 m, x = 0.196 m

We need to find the speed of the spring when it is released. Here, the elastic potential energy is balanced by the kinetic energy of the spring such that,

[tex]\dfrac{1}{2}kx^2=\dfrac{1}{2}mv^2[/tex]

[tex]v=\sqrt{\dfrac{kx^2}{m}}[/tex]

[tex]v=\sqrt{\dfrac{594\ N/m\times (0.196\ m)^2}{0.477\ kg}}[/tex]

v = 6.91 m/s

So, the speed of the ball is 6.91 m/s. Hence, this is the required solution.

Answer:

The beat frequency is 33.33 Hz.

Explanation:

Given that,

Length of first pipe =60 cm

Length of other pipe = 68 cm

Speed of sound = 340 m/s

We need to calculate the frequency

We know that,

When they operate at fundamental frequency then the length is given by,

[tex]L=\dfrac{\lambda}{2}[/tex]

The wavelength is given by

[tex]\lambda=2L[/tex]

For first organ pipe,

Using formula of frequency

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

[tex]f_{1}=\dfrac{v}{2L_{1}}[/tex]...(I)

Put the value into the formula

[tex]f_{1}=\dfrac{340}{2\times60\times10^{-2}}[/tex]

[tex]f_{1}=283.33\ Hz[/tex]

For second organ pipe,

[tex]f_{2}=\dfrac{v}{2L_{2}}[/tex]...(II)

Put the value in the equation (II)

[tex]f_{2}=\dfrac{340}{2\times68\times10^{-2}}[/tex]

[tex]f_{2}=250\ Hz[/tex]

Therefore the beat frequency

[tex]\Delta f=f_{1}-f_{2}[/tex]

[tex]\Delta f=283.33-250[/tex]

[tex]\Delta f=33.33\ Hz[/tex]

Hence,  The beat frequency is 33.33 Hz.

The beat frequency heard when two organ pipes are sounded together at their fundamental frequencies, where one pipe is 60 cm long and the other is 68 cm long, is 33.33 Hz.

To find the beat frequency heard when two organ pipes are sounded together at their fundamental frequencies, we need to calculate the frequencies of each pipe. The formula for frequency is f = v/λ, where v is the speed of sound and λ is the wavelength. Since the pipes are open at both ends, the wavelength is twice the length of the pipe. So, the frequency of the 60 cm pipe would be f₁ = v/(2L₁), and the frequency of the 68 cm pipe would be f₂ = v/(2L₂). Now, to calculate the beat frequency, we subtract the frequencies: beat frequency = |f₁ - f₂|.

Using the given speed of sound (340 m/s), we can substitute the lengths of the pipes into the frequency formula to find the frequencies:

f₁ = 340/(2 x 0.6) Hz = 283.33 Hz

f₂ = 340/(2 x 0.68) Hz = 250 Hz

Now, we can calculate the beat frequency:

beat frequency = |283.33 - 250| Hz = 33.33 Hz

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The close were the centers of the objects at closest approach 2.7 x 10^-14 m

E =8.8 MeV = 8.8 x 1.6 x 10^-13 J

q = 2 e = 2 x 1.6 x 10^-19 C

Q = 82 e = 82 x 1.6 x 10^-19 C

Let d be the distance of closest approach

E = k Q q / d

Where, K = 9 x 10^9 Nm^2 / C^2

d = k Q q / E

d = (9 x 10^9 x 82 x 1.6 x 10^-19 x 2 x 1.6 x 10^-19) / (8.8 x 1.6 x 0^-13)

d = 2.7 x 10^-14 m

Answer:

4 m/s

Explanation:

M = mass of the hammer = 8.91 kg

m = mass of the metal piece = 0.411 kg

h = height gained by the metal piece = 3.88 m

Potential energy gained by the metal piece is given as

PE = mgh

PE = (0.411) (9.8) (3.88)

PE = 15.6 J

KE = Kinetic energy of the hammer

Given that :

Potential energy of metal piece = (0.219) Kinetic energy of the hammer

PE = (0.219) KE

15.6 = (0.219) KE

KE = 71.2 J

v = speed of hammer

Kinetic energy of hammer is given as

KE = (0.5) M v²

71.2 = (0.5) (8.91) v²

v = 4 m/s

Answer:

volumme =0.36 ml

Explanation:

total heat required can be obtained by using following formula

[tex]q= mC \Delta T[/tex].......(1)

where,

m - mass of water,

C - specific heat capacity of water and = 4.184 j g^{-1} °C

[tex]\Delta T[/tex] - total change in temperature.  = 10°C

The density of water is 1 g/cc. hence, 200 mL of water is equal to 200 g

putting all value in the above equation (1)

q = 200*4.184* 10 ° = 8368 J.

Therefore total number of moles of ethanol required to supply 8368 J of heat is

[tex]\frac {8368}{1368000} = 0.006117 moles.[/tex]

The molar mass of ethanol is 46 g/mol.

The mass of ethanol required is 46* 0.006117 = 0.28138 g

The density of ethanol is 0.78 g/ml.

The volume of ethanol required is

[tex]\frac {0.28138}{0.78} = 0.36 ml[/tex]

To heat 200 mL of water by 10 °C, you would need approximately 0.00785 mL of ethanol. This is calculated by converting the volume of water to its mass, calculating the heat energy using the specific heat capacity of water, and then converting that energy to the amount of ethanol required using the heat of combustion of ethanol.

To calculate the amount of ethanol required to heat 200 mL of water by 10 °C, we need to use the formula: Q = mcΔT, where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

First, we need to convert the volume of water to its mass. Since the density of water is 1 g/mL, the mass of 200 mL of water is 200 g.

Next, we can calculate the heat energy using the formula: Q = mcΔT. The specific heat capacity of water is 4.184 J/g °C, the mass of water is 200 g, and the change in temperature is 10 °C. Plugging in these values, we get: Q = (200 g)(4.184 J/g °C)(10 °C) = 8376 J.

Finally, we can convert this heat energy to the amount of ethanol required using the heat of combustion of ethanol. The heat of combustion of ethanol is -1368 kJ/mol. To convert from J to kJ, we divide the heat energy by 1000 to get: 8376 J / 1000 = 8.376 kJ.

Now we can calculate the volume of ethanol. Since the density of ethanol is 0.78 g/mL, we can use the formula: V = m/ρ, where V is the volume, m is the mass, and ρ is the density. The mass of ethanol can be calculated by rearranging the formula: m = Q/ΔH, where Q is the heat energy and ΔH is the heat of combustion. Plugging in the values, we get: m = 8.376 kJ / -1368 kJ/mol = -0.00611 mol.

Now we can calculate the volume of ethanol using the formula: V = m/ρ. The mass of ethanol is -0.00611 mol and the density of ethanol is 0.78 g/mL. Plugging in these values, we get: V = -0.00611 mol / (0.78 g/mL) = -0.00785 mL.

Since volume cannot be negative, the volume of ethanol required to heat 200 mL of water by 10 °C is 0.00785 mL.

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Answer:

The total work done will be zero.

Explanation:

Given that,

Mass = 100 kg

Force = 392 N

Velocity = 20 m/s

Distance s= 10 m

We need to calculate the work done

Using balance equation

The net force will be

[tex]F'=F-\mu mg[/tex]

[tex]F'=392-0.4\times100\times9.8[/tex]

[tex]F'=0[/tex]

The net force is zero.

Hence, The total work done will be zero by all forces on the object.

Answer:

Power, P = 722.96 watts

Explanation:

It is given that,

Voltage, V = 120 V

Length of nichrome wire, l = 8.9 m

Diameter of wire, d = 0.86 mm

Radius of wire, r = 0.43 mm = 0.00043 m

Resistivity of wire, [tex]\rho=1.3\times 10^{-6}\ \Omega-m[/tex]

We need to find the power drawn by this heater. Power is given by :

[tex]P=\dfrac{V^2}{R}[/tex]

And, [tex]R=\rho\dfrac{l}{A}[/tex]

[tex]P=\dfrac{V^2\times A}{\rho\times l}[/tex]

[tex]P=\dfrac{120^2\times \pi (0.00043)^2}{1.3\times 10^{-6}\times 8.9}[/tex]

P = 722.96 watts

So, the power drawn by this heater element is 722.96 watts. Hence, this is the required solution.    

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.

Explanation:

Force between two point changes, F₁ = 1 N

Distance between them, r₁ = 2 cm = 0.02 m

We know that the electrostatic force is given by :

[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]

i.e.

[tex]F\propto \dfrac{1}{r^2}[/tex]

i.e

[tex]\dfrac{F_1}{F_2}=(\dfrac{r_2}{r_1})^2[/tex]

Let F₂ is the force when the distance between the charges is 8 cm, r₂ = 0.08 m

[tex]F_2=\dfrac{F_1\times r_1^2}{r_2^2}[/tex]

[tex]F_2=\dfrac{1\ N\times (0.02\ m)^2}{(0.08\ m)^2}[/tex]

F₂ = 0.0625 N

So, the distance between the sphere is 8 N, the new force is equal to 0.0625 N. Hence, this is the required solution.

By using steam tables or similar thermodynamic property charts, you can determine the initial and final temperature and enthalpy of R-134a under specified pressure conditions in a rigid container. As the pressure changes with heating, you can find the corresponding changes in temperature and enthalpy.

The subject matter of this question falls under the domain of thermodynamics in physics. It involves an element of heating, changes in pressure and the resultant alterations in temperature. Given the specifics, we can't directly provide a numerical solution in this setting as it requires the usage of refprop or other similar thermodynamic property database or the proper specific enthalpy and temperature charts for R-134a. However, conceptually, it can be handled via using steam tables (or more specifically refrigerant tables) for the substance R-134a. Initially, you lookup the properties using the given pressure and known mass and volume to calculate the initial temperature and enthalpy. Similarly, when the container is heated and the pressure raises to 600 kPa, you can again refer to the steam tables with this new pressure (and the volume which remains unchanged as the container is rigid) to determine the new temperature and enthalpy.

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Answer:

364.4 J

Explanation:

I = Moment of inertia of the forearm = 0.550 kgm²

v = linear velocity of the ball relative to elbow joint = 17.1 m/s

r = distance from the joint = 0.470 m

w = angular velocity

Using the equation

v = r w

17.1 = (0.470) w

w = 36.4 rad/s

Rotational kinetic energy of the forearm is given as

RKE = (0.5) I w²

RKE = (0.5) (0.550) (36.4)²

RKE = 364.4 J

Answer:

363.96 J

Explanation:

v = 17.1 m/s

r = 0.47 m

I = 0.55 kgm^2

Let ω be the angular velocity

ω = v / r = 17.1 / 0.47 = 36.38 rad/s

The kinetic energy of rotation is

K = 1/2 I ω^2 = 0.5 x 0.55 x 36.38 x 36.38 = 363.96 J

Explanation:

Charge on proton, q₁ = e

Charge on alpha particles, q₂ = 2e

The magnetic force is given by :

[tex]F=qvB\ sin\theta[/tex]

Here, [tex]\theta=90=sin(90) = 1[/tex]

For proton, [tex]F_p=ev_pB[/tex]..........(1)

For alpha particle, [tex]F_a=2ev_aB[/tex]..........(2)

Since, a proton (charge e), traveling perpendicular to a magnetic field, experiences the same force as an alpha particle. So,

[tex]ev_pB=2ev_aB[/tex]

[tex]\dfrac{v_p}{v_a}=\dfrac{2}{1}[/tex]

So, the ratio of the speed of proton to the alpha particle is 2 : 1 .Hence, this is the required solution.

If a proton and an alpha particle experience the same force in a magnetic field, the proton must be traveling at twice the speed of the alpha particle. This is because the force exerted by a magnetic field on a moving charge depends on the charge of the particle, the speed of the particle, and the strength of the magnetic field.

The force exerted by a magnetic field on a moving charge depends on the charge of the particle, the speed of the particle, and the strength of the magnetic field. Given that a proton (charge e) and alpha particle (charge 2e) experience the same force in the same magnetic field, we can create an equation to solve for their speed ratio.

The force on a particle due to a magnetic field is given by F = qvB where q is the charge, v is the speed, and B is the magnetic field. Since the force on the proton and alpha particle are the same, we can set their force equations equal to each other.

This means that e * v_proton * B = 2e * v_alpha * B. Simplifying, the ratio v_proton/v_alpha = 2.

Therefore, the proton is moving twice as fast as the alpha particle.

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Answer:

(kg⋅m/s)

Explanation:

The unit of momentum is the product of the units of mass and velocity. In SI units, if the mass is in kilograms and the velocity is in meters per second then the momentum is in kilogram meters per second (kg⋅m/s)

This question is dealing with fundamental SI units. Thus, let's list the seven basic SI Units upon which other units are expressed;

Mass - kilogram (kg)

Length - meter (m)

Time - second (s)

Amount of substance - mole (mol)

Electric current - ampere (A)

Thermodynamic temperature - kelvin (K)

Luminous intensity - candela (cd)

Fundamental SI unit of momentum is Kg.m/s

Now, we want to write the SI Unit of momentum.

From the question, we are told that Physicists often measure the momentum of subatomic particles using the formula;

MeV/c, where c is the speed of light, and 1 MeV = 1.6 ✕ 10−13 kg.m²/s²

Now, we know that unit of speed is in m/s.

Thus, in units, momentum = MeV/c = (kg.m²/s²)/(m/s)

Simplifying this gives; Kg.m/s

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Answer:

The speed of the car and the stopping distance are 50.13 m/s and 279.22 m.

Explanation:

Given that,

Initial speed = 10.0 m/s

Acceleration = 3.00 m/s^2

Distance [tex]d = \dfrac{1}{4}\times1.609[/tex]

[tex]d = 0.40225\ km=0.40225\times10^{3}\ m[/tex]

We need to calculate the speed of the car,

Using equation of motion

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

Where, u = initial velocity

v = final velocity

a = acceleration

Put the value in the equation

[tex]v^2=(10.0)^2+2\times3.00\times0.40225\times10^{3}[/tex]

[tex]v^2=2513.5[/tex]

[tex]v=50.13\ m/s[/tex]

(B). We need to calculate the stopping distance of the car,

Using equation of motion again

[tex]v^2=u^2+2as[/tex]

Here,initial velocity = 50.13 m/s

Final velocity = 0

Acceleration = -4.50 m/s²

Put the value in the equation

[tex]0=(50.13)^2-2\times4.50\times s[/tex]

[tex]s=\dfrac{(50.13)^2}{2\times4.50}[/tex]

[tex]s=279.22\ m[/tex]

Hence, The speed of the car and the stopping distance are 50.13 m/s and 279.22 m.

Answer:

[tex]A = 0.2875 m^2[/tex]

Explanation:

As we know that

[tex]Q = 4.6 \mu C[/tex]

E = 1.8 kV/mm

now we know that electric field between the plated of capacitor is given as

[tex]E = \frac{\sigma}{\epsilon_0}[/tex]

now we will have

[tex]1.8 \times 10^6 = \frac{\sigma}{\epsilon_0}[/tex]

[tex]\sigma = (1.8 \times 10^6)(8.85 \times 10^{-12})[/tex]

[tex]\sigma = 1.6 \times 10^{-5} C/m^2[/tex]

now we have

[tex]\sigma = \frac{Q}{A}[/tex]

now we have area of the plates of capacitor

[tex]A = \frac{Q}{\sigma}[/tex]

[tex]A = \frac{4.6 \times 10^{-6}}{1.6 \times 10^{-5}}[/tex]

[tex]A = 0.2875 m^2[/tex]

Answer:

If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

Explanation:

R₁ = Resistance of first resistor

R₂ = Resistance of second resistor

V = Voltage of battery = 12 V

I = Current = 0.33 A (series)

I = Current = 1.6 A (parallel)

In series

[tex]\text{Equivalent resistance}=R_{eq}=R_1+R_2\\\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{0.33}\\\Rightarrow R_1+R_2=36.36\\ Also\ R_1=36.36-R_2[/tex]

In parallel

[tex]\text{Equivalent resistance}=\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}\\\Rightarrow {R_{eq}=\frac{R_1R_2}{R_1+R_2}[/tex]

[tex]\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{1.6}\\\Rightarrow \frac{R_1R_2}{R_1+R_2}=7.5\\\Rightarrow \frac{R_1R_2}{36.36}=7.5\\\Rightarrow R_1R_2=272.72\\\Rightarrow(36.36-R_2)R_2=272.72\\\Rightarrow R_2^2-36.36R_2+272.72=0[/tex]

Solving the above quadratic equation

[tex]\Rightarrow R_2=\frac{36.36\pm \sqrt{36.36^2-4\times 272.72}}{2}[/tex]

[tex]\Rightarrow R_2=25.78\ or\ 10.57\\ If\ R_2=25.78\ then\ R_1=36.36-25.78=10.58\ \Omega\\ If\ R_2=10.57\ then\ R_1=36.36-10.57=25.79\Omega[/tex]

∴ If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

The two resistors in question, when wired in a series, have a combined resistance of about 36.36 Ohms. When wired in parallel, they share a resistance of roughly 7.5 Ohms. The two resistor values would then most likely be around 28.86 Ohms and 7.5 Ohms.

The nature of your question indicates a focus on the properties and calculations associated with electrical resistors in a circuit. These resistors can either be configured in a series or parallel connection, which will drastically change their behavior and derived readings.

From the context of your question, it is evident that a 12V battery is being used in conjunction with two resistors. In a series connection, the sum of the voltages across each resistor will equal the total voltage, effectively segmenting the 12V battery's output. In a parallel connection, each resistor would experience the full 12V impact, leading to larger current readings. Your values indicate a 0.33A current in a series scenario and a 1.60A current in a parallel situation.

Knowing this, we can apply Ohm's Law, which states Voltage (V) equals Current (I) times Resistance (R). In the series connection, the total resistance can be calculated as 12V divided by 0.33A equals approximately 36.36 Ohms. For the parallel connection, the total resistance would be 12V divided by 1.60A equals approximately 7.5 Ohms. Subtracting these two values, we can find that one of the resistors is around 28.86 Ohms while the other is roughly 7.5 Ohms. This would satisfy both the series and parallel conditions outlined in your question.

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Answer:

The inductance of the inductor is 35.8 mH

Explanation:

Given that,

Voltage = 120-V

Frequency = 1000 Hz

Capacitor [tex]C= 2.00\mu F[/tex]

Current = 0.680 A

We need to calculate the inductance of the inductor

Using formula of current

[tex]I = \dfrac{V}{Z}[/tex]

[tex]Z=\sqrt{R^2+(L\omega-\dfrac{1}{C\omega})^2}[/tex]

Put the value of Z into the formula

[tex]I=\dfrac{V}{\sqrt{R^2+(L\omega-\dfrac{1}{C\omega})^2}}[/tex]

Put the value into the formula

[tex]0.680=\dfrac{120}{\sqrt{(100)^2+(L\times2\pi\times1000-\dfrac{1}{2\times10^{-6}\times2\pi\times1000})^2}}[/tex]

[tex]L=35.8\ mH[/tex]

Hence, The inductance of the inductor is 35.8 mH

Answer:

Inductance,L:

"The property of the conductor or the solenoid to generate the electromotive force,emf due to the flow of current,I."

Unit:  henry,H as it is equivalent to, kg.m².sec⁻².A⁻².

Explanation:

Data:

Solution:

We need to calculate the inductance, L of the solenoid inside a circuit,

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