A cylindrical rod of copper (E = 110 GPa, 16 × 106 psi) having a yield strength of 240 MPa (35,000 psi) is to be subjected to a load of 6640 N (1493 lbf). If the length of the rod is 370 mm (14.57 in.), what must be the diameter to allow an elongation of 0.53 mm (0.02087 in.)?

Answer:

a. T = 119.68 N.m

b. r = 140 rev

Explanation:

first we know that:

∑T = Iα

where ∑T is the sumatory of the torques, I is the moment of inertia and α is the angular aceleration.

so, if the prop goes from rest to 400 rpm in 14 seconds we can find the α of the system:

1. change the 400 rpm to radians as:

                W = 400*2π/60

                W = 41.888 rad /s

2. Then, using the next equation, we find the α as:

                w = αt

solving for α

               α = [tex]\frac{w}{t}[/tex]

note: t is the time, so:

               α = [tex]\frac{41.888}{14}[/tex]

               α = 2.992 rad/s^2

Now using the first equation, we get:

T = Iα

T = 40(2.992)

T = 119.68 N.m

On the other hand, for know the the number of revolutions turned as the prop achieves operating speed, we use the following equation:

θ = wt +[tex]\frac{1}{2}[/tex]α[tex]t^2[/tex]

Where w =  41.888 rad /s, α = 2.992 rad/s^2, t is the time and θ give as the number of radians that the prop made in the fisrt 14 seconds, so:

θ = (41.888)(14)+[tex]\frac{1}{2}[/tex](2.992)(14[tex])^2[/tex]

θ = 879.648 rad

and that divided by 2π give us the number of revolutions r, so:

r = 140 rev

Answer:

Explanation:

A )

[tex]L_{max} = \sqrt{l(l+1)}[/tex]ℏ

where l is orbital quantum number

l = n-1 where n is principal quantum no

Given n = 7

l = 7 - 1 = 6

[tex]L_{max} = \sqrt{6(6+1)}[/tex]ℏ

= 6.48ℏ

B)

Here

n = 26

l = 26 - 1

= 25

[tex]L_{max} = \sqrt{25(25+1)}[/tex]ℏ

= 25.49ℏ

= 25.5ℏ

C )

n = 191

l = 191 - 1

190

[tex]L_{max} = \sqrt{190(190+1)}[/tex]

= 190.499ℏ

= 191ℏ

The magnitude of the maximum orbital angular momentum Lmax for an electron in a hydrogen atom can be calculated using the formula Lmax = √(l*(l + 1)), where l is the orbital quantum number, which for maximum Lmax is n - 1 for principal quantum number n. The values for n = 7, 26, and 191 are approximately 16.9706 ℏ, 81.6144 ℏ, and 675.3918 ℏ, respectively.

The magnitude of the maximum orbital angular momentum for an electron in a hydrogen atom can be expressed in terms of the principal quantum number n. The maximum value of the orbital quantum number, l, is n - 1.

The magnitude of the maximum orbital angular momentum Lmax is then given by the square root of l*(l + 1). Therefore:

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Answer:

The time period of the solar system's orbit is [tex]2.05\times10^{8}\ year[/tex]

Explanation:

Given that,

Distance = 25000 light year

Speed of light [tex]c=3\times10^{8}\ m/s[/tex]

Speed of astronomers = 230 m/s

Suppose the orbit is circular, we need to find the time period of the solar system's orbit.

We need to calculate the time period

Using formula of time period

[tex]t = \dfrac{d}{v}[/tex]

[tex]t = \dfrac{2\pi r}{v}[/tex]

r = distance

t = time

v = speed

Put the value into the formula

[tex]t=\dfrac{2\pi\times25000\times9.46\times10^{15}}{230\times10^{3}}[/tex]

[tex]t=6.460\times10^{15}\ sec[/tex]

Time in years

[tex]t=\dfrac{6.460\times10^{15}}{3.15\times10^{7}}[/tex]

[tex]t=2.05\times10^{8}\ year[/tex]

Hence, The time period of the solar system's orbit is [tex]2.05\times10^{8}\ year[/tex]

The solar system is about 25,000 light years away from the centre of the Milky Way galaxy. It orbits the galaxy at a speed of 230 km/s, which translates to a galactic year of approximately 225 million Earth years. The galaxy's mass, including dark matter, influences the orbital velocities of all celestial bodies.

The solar system, orbiting at a speed of 230 km/s, is located approximately 25,000 light years from the Milky Way galaxy's center. A light year is defined as the distance light travels in one year with a speed of 3.0×108m/s. Understanding this concept helps us comprehend the vast size of our galaxy. Additionally, the solar system's movement around the Milky Way is referred to as a galactic year, which lasts approximately 225 million years in Earth's time.

The sun, and all stars in the galaxy, orbit the galactic center in a nearly circular path lying in the galaxy's disk. Moreover, the total mass of our Galaxy can be determined by measuring the orbital velocities of stars and interstellar matter, including the Sun. This vast mass, around 2 × 1012 Msun, includes a significant part composed of dark matter that emits no electromagnetic radiation.

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To solve this problem it is necessary to apply the concepts related to the Power depending on the temperature and the heat transferred.

By definition the power can be expressed as

[tex]P = \frac{\Delta T}{\Delta Q}[/tex]

Where,

[tex]\Delta T = T_m - T_a =[/tex] Change at the temperature, i.e, the maximum acceptable die temperature ([tex]T_m[/tex]) with the allowable temperature in chassis ([tex]T_A[/tex])

[tex]\Delta Q = Q_A-Q_D =[/tex] Change in the thermal resistance to ambient ([tex]Q_A[/tex]) and the Thermal resistance from die to package ([tex]Q_D[/tex])

Our values are given as,

[tex]T_m=110\°C[/tex]

[tex]T_a= 50\°C[/tex]

[tex]Q_A= 8\°C/W[/tex]

[tex]Q_D= 2\°C/W[/tex]

Replacing we have,

[tex]P = \frac{110-50}{8-2}[/tex]

[tex]P = 10W[/tex]

The power that can dissipate the chip is 10W

a. The value of vₓ = 33.77 km/s

b. The value of Vy = 81.05 km/s

The magnetic force, F on the proton moving in the uniform magnetic field is given by

F = qv × B where

Re-writing the force in matrix form, we have

[tex]\left[\begin{array}{ccc}F_{x} &F_{y} &F_{z} \\\end{array}\right] = q\left[\begin{array}{ccc}i&j&k\\v_{x} &v_{y}&v_{z}\\B_{x}&B_{y}&B_{z}\end{array}\right][/tex]

Taking the determinant, we have

[tex]F_{x}i + F_{y}j + F_{z}k = q[(v_{y}B_{z} - v_{z}B_{y})]i + q[(v_{x}B_{z} - v_{z}B_{x})]j + q[(v_{x}B_{y} - v_{y}B_{x})]k[/tex]

Equating the components of the force, we have

[tex]F_{x} = q[(v_{y}B_{z} - v_{z}B_{y})] (1)\\F_{y} = q[(v_{x}B_{z} - v_{z}B_{x})] (2)\\F_{z} = q[(v_{x}B_{y} - v_{y}B_{x})] (3)[/tex]

[tex]F_{x}/q = [(v_{y}B_{z} - v_{z}B_{y})] (4)\\F_{y}/q = [(v_{x}B_{z} - v_{z}B_{x})] (5)\\F_{z}/q = [(v_{x}B_{y} - v_{y}B_{x})] (6)[/tex]

Since F = (3.82 × 10⁻¹⁷ N)l + (1.59 × 10⁻¹⁷ N)j + 0k, equation (5) and (6) become

[tex]1.59 X 10^{-17} /1.602 X 10^{-19} = [(v_{x}B_{z} - v_{z}B_{x})] (5)\\993 = [(v_{x}B_{z} - v_{z}B_{x})] (7)\\Also\\0/q = [(v_{x}B_{y} - v_{y}B_{x})] (6)\\0 = (v_{x}B_{y} - v_{y}B_{x}) \\v_{x}B_{y} = v_{y}B_{x}\\v_{y} = \frac{v_{x}B_{y}}{B_{x}} (8)[/tex]

The value of vₓ = 33.77 km/s

Since [tex]B_{x}[/tex] = 10 × 10 ⁻³ T, [tex]B_{z}[/tex] = 30 × 10 ⁻³ T and [tex]v_{z}[/tex] = 2000 m/s, substituting the values of the variables into equation (7), we have

[tex]v_{x}B_{z} - v_{z}B_{x} = 993 (7)[/tex]

vₓ(30 × 10 ⁻³ T) - 2000 m/s × 10 × 10 ⁻³ T = 993 N

vₓ(30 × 10 ⁻³ T) - 20 m/sT = 993 N

vₓ(30 × 10 ⁻³ T) = 993 N + 20 m/sT

vₓ(30 × 10 ⁻³ T) = 1013 N

vₓ = 1013 N/(30 × 10 ⁻³ T)

vₓ = 33.77 × 10³ m/s

vₓ = 33.77 km/s

So, the value of vₓ = 33.77 km/s

The value of Vy = 81.05 km/s

Since [tex]B_{x}[/tex] = 10 × 10 ⁻³ T, [tex]B_{y}[/tex] = 24.0 × 10 ⁻³ T and [tex]v_{x}[/tex] = 33.77 × 10³ m/s, substituting the values of the variables into equation (8), we have

[tex]v_{y} = \frac{v_{x}B_{y}}{B_{x}} (8)[/tex]

Vy = 33.77 × 10³ m/s × 24.0 × 10 ⁻³ T/10 × 10 ⁻³ T

Vy = 810.48 m/sT/10 × 10 ⁻³ T

Vy = 81.048 × 10³  m/s

Vy ≅ 81.05 km/s

So, the value of Vy = 81.05 km/s

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Answer:

a) True. The speed of the wave in the string does not depend on the temperature, but the speed of the wave the air depends on the Temperatue.

Explanation:

Let's analyze the vibration of the string, when the string is touched a transverse wave is produced whose speed is determined by the tension and density of the string

        V = √T/μ

This wave advances and bounces at one of the ends forming a standing wave that induces a vibration in the surrounding air that therefore also produces a longitudinal wave whose velocity is a function of time

       v = 331 √( 1+ T/273)

With this information we can review the statements given

a) True. The speed of the wave in the string does not depend on the temperature, but the speed of the wave the air depends on the Temperatue.

b) False. The wave in the string is transverse, but the wave in the air is longitudinal

c) False. The wave in the string is transverse

d) False. It's the other way around

Answer:

A

Explanation:

Just got it right on edge

Since a light string is wrapped around the outer rim of a solid uniform cylinder, the mass of the cylinder is equal to 18.4 kilograms.

Given the following data:

To calculate the mass of the cylinder:

First of all, we would determine the acceleration of the stone by using the the third equation of motion;

[tex]V^2 = U^2 + 2aS[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]3.4^2 = 0^2 + 2a(2.40)\\\\11.56 = 0 + 4.8a\\\\11.56 = 4.8a\\\\a = \frac{11.56}{4.8}[/tex]

Acceleration, a = 2.41 [tex]m/s^2[/tex]

Next, we would solve for the torque by using the formula:

[tex]T = m(g - a)\\\\T = 3(9.8 - 2.41)\\\\T = 3 \times 7.39[/tex]

Torque, T = 22.17 Newton.

In rotational motion, torque is given by the formula:

[tex]T = I\alpha \\\\Tr = \frac{Mr^2}{2} \times \frac{a}{r} \\\\Tr = \frac{Mra}{2}\\\\2Tr = Mra\\\\M = \frac{2T}{a}\\\\M = \frac{2\times 22.17}{2.41}\\\\M = \frac{44.34}{2.41}[/tex]

Mass = 18.4 kilograms

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The system described is based on rotational dynamics and energy conservation principles in physics. The initial potential energy of the stone is converted into kinetic energy, and the stone's tensional force is equated with the cylinder's net torque. By systematically resolving these equations, we find the mass of the cylinder to be about 66.16 kg.

The subject of your question pertains to rotational dynamics in Physics. In this particular system, there are two main things to consider. First, we need to acknowledge the law of conservation of energy. The entire mechanical energy of the system would remain constant because there is neither any frictional force present nor any external force doing work. Second, we have to consider the moment of inertia for the cylinder.

For the stone, the initial potential energy (mgh) is converted into kinetic energy (0.5*m*v^2). So, 3*9.81*2.4 = 0.5*3*(3.4)^2, which solves to be accurate.

The tension in the string is equal to the force exerted by the stone, so T = m*g = 3*9.81 = 29.43 N.

As the string unwinds, the cylinder spins faster so, we equate the tensional force with the net torque, τ_net = I*α. Here, α = tangential acceleration (which equals g) divided by radius, hence, α = g/radius.

By simplifying these equations, we find that the mass of the cylinder is approximately 66.16 kg.

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Answer:

I = 21.94 A

Explanation:

Given that

B= 0.15 T

Number of turns N= 3100

Length L = 57 cm

Diameter ,d= 7 cm

We know that magnetic field in the solenoid given as

[tex]B=\dfrac{\mu_0IN}{L}[/tex]

[tex]I=\dfrac{BL}{\mu_0N}[/tex]

Now by putting the values

[tex]I=\dfrac{0.15\times 0.57}{4\pi \times 10^{-7}\times 3100}[/tex]

I = 21.94 A

Therefore current need to produce 0.15 T magnetic filed is 21.94 A.

To find the relative distance from one point to another it is necessary to apply the Relativity equations.

Under the concept of relativity the distance measured from a spatial object is given by the equation

[tex]l = l_0 \sqrt{1-\frac{v^2}{c^2}}[/tex]

Where

[tex]l_0[/tex]= Relative length

v = Velocity of the spaceship

c = Speed of light

Replacing with our values we have that

[tex]l = l_0 \sqrt{1-\frac{v^2}{c^2}}[/tex]

[tex]l = 1.2*10^{11} \sqrt{1-\frac{0.8c^2}{c^2}}[/tex]

[tex]l = 1.2*10^{11} \sqrt{1-0.8^2}[/tex]

[tex]l = 7.2*10^{10}m[/tex]

Therefore the distance between Mars and Venus measured by the Martin is [tex]7.2*10^{10}m[/tex]

Answer:

[tex]E=477.92\ W.m^{-2}[/tex]

Explanation:

Given that:

Absolute temperature of the body, [tex]T=273+30=303\ K[/tex]

Using Stefan Boltzmann Law of thermal radiation:

[tex]E=\epsilon. \sigma.T^4[/tex]

where:

[tex]\sigma =5.67\times 10^{-8}\ W.m^{-2}.K^{-4}[/tex]   (Stefan Boltzmann constant)

Now putting the respective values:

[tex]E=1\times 5.67\times 10^{-8}\times 303^4[/tex]

[tex]E=477.92\ W.m^{-2}[/tex]

Answer:

The flow rate is [tex]9.7\times 10^{- 11}\ m^{3}/s[/tex]

Solution:

As per the question:

Pressure drop,  = 1.45 kPa = 1450 Pa

Radius of the artery, R = [tex]2.50\times 10^{- 5}\ m[/tex]

length of the artery, L = [tex]1.10\times 10^{- 3}\ m[/tex]

Temperature, T = [tex]37^{\circ}C[/tex]

Viscosity, [tex]\eta = 2.084\times 10^{- 3}\ Pa.s[/tex]

Now,

The flow rate is given by:

[tex]Q = \frac{\pi R^{4}P}{8\eta L}[/tex]

[tex]Q = \frac{\pi (2.50\times 10^{- 5})^{4}\times 1450}{8\times 2.084\times 10^{- 3}\times 1.10\times 10^{- 3}} = 9.7\times 10^{- 11}\ m^{3}/s[/tex]

The flow rate through the artery can be calculated using Poiseuille's law, which takes into account the pressure difference, radius, length, and viscosity of the fluid. Using the given values, the flow rate is approximately 0.00686 m^3/s.

The flow rate through an artery can be calculated using Poiseuille's law, which states that the flow rate is directly proportional to the pressure difference and the fourth power of the radius, and inversely proportional to the length and viscosity of the fluid.

Using the given values, we can calculate the flow rate as follows:

Flow rate = (pressure difference * pi * radius^4) / (8 * viscosity * length)

Substituting the given values into the equation gives:

Flow rate = (1.45 * 10^3 * 3.14 * (2.5 * 10^-5)^4) / (8 * 2.084 * 10^-3 * 1.10 * 10^-3) = 0.00686 m^3/s

Therefore, the flow rate through the artery is approximately 0.00686 m^3/s.

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To determine the maximum amplitude of the piston's oscillation without the bolt losing contact, we need to set the gravitational force equal to the spring force, and solve for the amplitude. Using the given values, including the mass of the bolt, the acceleration due to gravity, and the frequency of oscillation, we can calculate the maximum amplitude.

In order for the bolt to stay in contact with the piston's surface, the maximum amplitude of the piston's oscillation needs to be determined.

For simple harmonic motion, the restoring force is proportional to the displacement and acts opposite to the direction of motion. In this case, the restoring force is provided by the weight of the bolt (mg) and the force exerted by the piston (kx), where x is the displacement from equilibrium and k is the spring constant of the piston.

At maximum amplitude, the net force acting on the bolt is zero, so we can set the gravitational force equal to the spring force: mg = kA. Rearranging this equation gives us the maximum amplitude A = mg/k.

Now we can calculate the maximum amplitude using the given values: mass of bolt = 33.0 g = 0.033 kg, acceleration due to gravity g = 9.81 m/s^2, and frequency of oscillation f = 3.05 Hz. The period of oscillation is T = 1/f.

Using the equation T = 2*pi*sqrt(m/k) for the period of a mass-spring system, we can solve for the spring constant k: k = (2*pi/T)^2*m = (2*pi/1/f)^2*0.033 = (2*pi*f)^2*0.033. Plugging in the given value for f, we find k = (2*pi*3.05)^2*0.033.

Finally, we can calculate the maximum amplitude A = mg/k = 0.033*9.81/(2*pi*3.05)^2*0.033. Evaluating this expression gives us the maximum amplitude of the piston's oscillation.

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Answer:

The wavelength of the light is 530 nm.

Explanation:

Given that,

Distance D= 1.0 m

Distance between slits d= 0.30 mm

Number of fringe = 9

Width = 1.6 cm

We need to calculate the angle

Using formula of angle

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

[tex]tan\theta=\dfrac{1.6\times10^{-2}}{1.0}[/tex]

[tex]\theta=\tan^{-1}(\dfrac{1.6\times10^{-2}}{1.0})[/tex]

[tex]\theta=0.91^{\circ}[/tex]

We need to calculate the wavelength of the light

Using formula of wavelength

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

[tex]\lambda=\dfrac{d\sin\theta}{m}[/tex]

Put the value into the formula

[tex]\lambda= \dfrac{0.30\times10^{-3}\times\sin0.91}{9}[/tex]

[tex]\lambda=5.29\times10^{-7}\ m[/tex]

[tex]\lambda=530\ nm[/tex]

Hence, The wavelength of the light is 530 nm.

Answer:

530 nm

Explanation:

Screen distance, D = 1 m

slit distance, d = 0.3 mm

n = 9 th bright

y = 1.6 cm

Let λ be the wavelength of light used.

y = n x D x λ / d

1.6 x 10^-2 = 9 x 1 x λ / (0.3 x 10^-3)

λ = 5.3333 x 10^-7 m

λ = 533.33 nm

λ = 530 nm ( by rounding off)

Answer: [tex]1.955(10)^{13} \frac{Pa.s}{m^{3}}[/tex]

Explanation:

This can be solved by the Poiseuille’s law for a laminar flow:

[tex]R=\frac{8 \eta L}{\pi r^{4}}[/tex]

Where:

[tex]R[/tex] is the resistance of the arteriole

[tex]\eta=3(10)^{-3} Pa.s[/tex] is the viscosity of blood

[tex]L=1000 \mu m=1000(10)^{-6}m[/tex] is the length of the arteriole

[tex]r=25 \mu m=25(10)^{-6}m[/tex] is the radius of the arteriole

[tex]R=\frac{8 (3(10)^{-3} Pa.s)(1000(10)^{-6}m)}{\pi (25(10)^{-6}m)^{4}}[/tex]

[tex]R=1.955(10)^{13} \frac{Pa.s}{m^{3}}[/tex]

The resistance of the arteriole is calculated using the Hagen-Poiseuille equation, resulting in an approximate value of 1.96 × 10¹⁵ Pa·s·m⁻⁴. The formula determines resistance by varying viscosity, length, and radius values, which is essential for understanding blood flow dynamics in physiological settings.

We can solve this problem using the formula for fluid resistance in a cylindrical tube (Hagen-Poiseuille equation):

[tex]R = (8 \times \eta \times L) / (\pi \times r_4)[/tex]

Where:

Substituting the values:

[tex]R = (8 \times 3 \times 10^{-3} Pa\cdot s \times 1000 \times 10^{-6} m) / (\pi \times (25 \times 10^{-6} m))[/tex]

Calculating the values inside the formula:

So, the resistance of the arteriole is approximately 1.96 × 10¹⁵ Pa·s·m⁻⁴.

Final answer:

The average force exerted on the tennis ball by Venus Williams' racquet during her serve is calculated using the formula F = Δp/Δt, resulting in approximately 660 N, with the correct answer being option a).

Explanation:

The question relates to determining the average force exerted by Venus Williams' racquet on a tennis ball during her serve at the 2007 French Open. To find the average force, we can use the formula derived from Newton's second law of motion, F = ma. However, instead of finding the acceleration first, we use the formula F = Δp/Δt (force is the change in momentum over the change in time), because we know the velocity after impact and the time of contact.

The initial momentum (pi) is 0, as the initial velocity is negligible, so the final momentum (pf) is mv, where m is the mass of the tennis ball (0.057 kg) and v is the final velocity (58 m/s). As the time of contact with the racquet is 5.0 ms (or 0.005 seconds), we calculate the force as follows:

F = (m × v)/Δt = (0.057 kg × 58 m/s)/0.005 s = (3.306 kg·m/s)/0.005 s = 661.2 N

Therefore, the correct answer is approximately 660 N, which matches option a).

To solve this problem it is necessary to consider two concepts. The first of these is the flow rate that can be defined as the volumetric quantity that a channel travels in a given time. The flow rate can also be calculated from the Area and speed, that is,

Q = V*A

Where,

A= Cross-sectional Area

V = Velocity

The second concept related to the calculation of this problem is continuity, which is defined as the proportion that exists between the input channel and the output channel. It is understood as well as the geometric section of entry and exit, defined as,

[tex]Q_1 = Q_2[/tex]

[tex]V_1A_1=V_2A_2[/tex]

Our values are given as,

[tex]A_1=\frac{1}{2}^2*\pi=0.785 in^2[/tex]

[tex]A_2=\frac{5}{8}^2*\pi=1.227 in^2[/tex]

Re-arrange the equation to find the first ratio of rates we have:

[tex]\frac{V_1}{V_2}=\frac{A_2}{A_1}[/tex]

[tex]\frac{V_1}{V_2}=\frac{1.227}{0.785}[/tex]

[tex]\frac{V_1}{V_2}=1.56[/tex]

The second ratio of rates is

[tex]\frac{V2}{V1}=\frac{A_1}{A2}[/tex]

[tex]\frac{V2}{V1}=\frac{0.785}{1.227}[/tex]

[tex]\frac{V2}{V1}=0.640[/tex]

The ratio of the flow rate of a ½ inch hose to that of a 5/8 inch hose is 16:25. This means that under equal pressures, for every 16 gallons of water that pass through the ½ inch hose, 25 gallons could flow through the 5/8 inch hose.

To compare the flow rates within the two hoses, we'll utilize a principle of fluid dynamics which states that the flow rate of an incompressible fluid (like water) is proportional to the cross-sectional area of the pipe. The area can be calculated using the formula for the area of a circle: A=πr², where r is the radius of the pipe.

To transform the diameters into radii, we divide them by 2: the radii are therefore 1/4 inch and 5/16 inch. As we're primarily interested in the ratio of areas (and consequently flow rates), we can ignore the π in the equation, leaving us with r² as directly representing 'area' for this comparison.

(1/4)² = 1/16 and (5/16)² = 25/256. Hence, the flow rate of the ½-inch hose to the 5/8 inch hose is 1/16: 25/256.

To simplify, we can multiply both portions of the ratio by 256 to achieve 16:25. This means for every 16 gallons per minute through the ½ inch hose, 25 gallons per minute could flow through the 5/8-inch hose, assuming the pressure in both systems is equal.

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Answer:

Explanation:

Electron's kinetic energy = 2 eV

= 2 x 1.6 x 10⁻¹⁹ J

1/2 m v² = 3.2 x 10⁻¹⁹

1/2 x 9.1 x 10⁻³¹ x v² = 3.2 x 10⁻¹⁹

v² = .703 x 10¹²

v = .8385 x 10⁶ m/s

Electrons revolve in a circular orbit when forced to travel in a magnetic field whose radius can be expressed as follows

r = mv / Bq

where m , v and q are mass , velocity and charge of electron .

here given magnetic field B = 90 mT

= 90 x 10⁻³ T

Putting these values in the expression above

r =  mv / Bq

= [tex]\frac{9\times10^{-31}\times.8385\times10^6}{90\times10^{-3}\times1.6\times10^{-19}}[/tex]

= .052 mm.

Answer:

The difference in pressure between points B and A is [tex]-3.9\times10^{4}\ Pa[/tex]

(D) is correct option.

Explanation:

Given that,

Density of the liquid = 1200 kg/m³

Speed of fluid at point A= 7.5 m/s

Speed of fluid at point B = 11 m/s

We need to calculate the difference in pressure between points B and A

Using formula of change in pressure

[tex]\Delta P=\dfrac{1}{2}D(v_{2}^2-v_{1}^2)[/tex]

Where, [tex]v_{1}[/tex] = Speed of fluid at point A

[tex]v_{2}[/tex] = Speed of fluid at point B

D = Density of the liquid

Put the value into the formula

[tex]\Delta P=\dfrac{1}{2}\times1200\times(11^2-7.5^2)[/tex]

[tex]\Delta P=-38850\ Pa[/tex]

[tex]\Delta P=-3.9\times10^{4}\ Pa[/tex]

Hence, The difference in pressure between points B and A is [tex]-3.9\times10^{4}\ Pa[/tex]

The difference in pressure, PB – PA, between points B and A can be calculated using Bernoulli's equation. Plugging in the given values, we find that the pressure difference is approximately -39,000 Pa.

The difference in pressure between points B and A can be calculated using Bernoulli's equation. Bernoulli's equation states that the pressure difference between two points in a fluid flow system is equal to the difference in kinetic energy and potential energy between the two points. In this case, we can calculate the pressure difference using the equation:

PB - PA = (1/2) * ρ * (VB^2 - VA^2)

PB - PA = (1/2) * 1200 kg/m3 * (11 m/s)^2 - (7.5 m/s)^2

Simplifying the equation, we find that the difference in pressure, PB - PA, is approximately -39,000 Pa. Therefore, the correct answer is D. -3.9 × 104 Pa.

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Answer:

Part A:

Unknown variables:

velocity of the astronaut after throwing the tank.

maximum distance the astronaut can be away from the spacecraft to make it back before she runs out of oxygen.

Known variables:

velocity and mass of the tank.

mass of the astronaut after and before throwing the tank.

maximum time it can take the astronaut to return to the spacecraft.

Part B:

To obtain the velocity of the astronaut we use this equation:

-(momentum of the oxygen tank) = momentum of the astronaut

-mt · vt = ma · vt

Where:

mt = mass of the tank

vt = velocity of the tank

ma = mass of the astronaut

va = velocity of the astronaut

To obtain the maximum distance the astronaut can be away from the spacecraft we use this equation:

x = x0 + v · t

Where:

x = position of the astronaut at time t.

x0 = initial position.

v = velocity.

t = time.

Part C:

The maximum distance the astronaut can be away from the spacecraft is 162 m.

Explanation:

Hi there!

Due to conservation of momentum, the momentum of the oxygen tank when it is thrown away must be equal to the momentum of the astronaut but in opposite direction. In other words, the momentum of the system astronaut-oxygen tank is the same before and after throwing the tank.

The momentum of the system before throwing the tank is zero because the astronaut is at rest:

Initial momentum = m · v

Where m is the mass of the astronaut plus the equipment (100 kg) and v is its velocity (0 m/s).

Then:

initial momentum = 0

After throwing the tank, the momentum of the system is the sum of the momentums of the astronaut plus the momentum of the tank.

final momentum = mt · vt + ma · va

Where:

mt = mass of the tank

vt = velocity of the tank

ma = mass of the astronaut

va = velocity of the astronaut

Since the initial momentum is equal to final momentum:

initial momentum = final momentum

0 = mt · vt + ma · va

- mt · vt = ma · va

Now, we have proved that the momentum of the tank must be equal to the momentum of the astronaut but in opposite direction.

Solving that equation for the velocity of the astronaut (va):

- (mt · vt)/ma = va

mt = 15 kg

vt = 10 m/s

ma = 100 kg - 15 kg = 85 kg

-(15 kg · 10 m/s)/ 85 kg = -1.8 m/s

The velocity of the astronaut is 1.8 m/s in direction to the spacecraft.

Let´s place the origin of the frame of reference at the spacecraft. The equation of position for an object moving in a straight line at constant velocity is the following:

x = x0 + v · t

where:

x = position of the object at time t.

x0 = initial position.

v = velocity.

t = time.

Initially, the astronaut is at a distance x away from the spacecraft so that

the initial position of the astronaut, x0, is equal to x.

Since the origin of the frame of reference is located at the spacecraft, the position of the spacecraft will be 0 m.

The velocity of the astronaut is directed towards the spacecraft (the origin of the frame of reference), then, v = -1.8 m/s

The maximum time it can take the astronaut to reach the position of the spacecraft is 1.5 min = 90 s.

Then:

x = x0 + v · t

0 m = x - 1.8 m/s · 90 s

Solving for x:

1.8 m/s · 90 s = x

x = 162 m

The maximum distance the astronaut can be away from the spacecraft is 162 m.

To calculate the quantity of heat that must be removed from 6.1 g of 100°C steam, we need to consider both the change in temperature and the phase change from steam to liquid. The specific heat of water is used to calculate the heat required to lower the temperature, while the heat of vaporization is used to calculate the heat required to condense the steam. Adding these two heat values together gives us the total amount of heat that must be removed from the steam, which is approximately 3.61164 kcal.

When steam at 100°C condenses and its temperature is lowered to 46°C, heat must be removed from the steam. To calculate the amount of heat, we can use the specific heat of steam and the latent heat of vaporization. First, we calculate the heat required to lower the temperature of the steam from 100°C to 46°C using the specific heat of water. We then calculate the heat required to condense the steam using the latent heat of vaporization. Finally, we add these two heat values together to obtain the total amount of heat that must be removed from the steam.

Given:

Calculations:

Calculation:

Therefore, the quantity of heat that must be removed from 6.1 g of 100°C steam to condense it and lower its temperature to 46°C is approximately 3.61164 kcal.

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To condense and cool 6.1 g of 100°C steam to 46°C, 3.2879 kcal must be removed for condensation, and 0.3304 kcal for cooling, for a total of 3.6183 kcal.

Calculating the Quantity of Heat for Condensation and Cooling

To calculate the quantity of heat that must be removed from 6.1 g of 100°C steam to condense it and lower its temperature to 46°C, we need to consider two processes: condensation and cooling. For condensation, we use the heat of vaporization, and for cooling, we use the specific heat of water.

Total heat removed is the sum of the heat from both steps: 3.2879 kcal + 0.3304 kcal = 3.6183 kcal.

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Answer:

Option b

Solution:

As per the question:

Speed of the mouse, v = 1.3 m/s

Speed of the cat, v' = 2.5 m/s

Angle, [tex]\theta = 38^{\circ}[/tex]

Now,

To calculate the distance between the mouse and the cat:

The distance that the cat moved is given by:

[tex]x = v'cos\theta t[/tex]

[tex]x = 2.5cos38^{\circ}\times t = 1.97t[/tex]

The position of the cat and the mouse can be given by:

[tex]x = x' + vt[/tex]

[tex]1.97t = x' + 1.3t[/tex]

x' = 0.67 t           (1)

The initial speed of the cat ahead of the mouse:

u = [tex]v'sin\theta = 2.5sin38^{\circ} = 1.539\ m/s[/tex]

When the time is 0.5t, the speed of the cat is 0, thus:

[tex]0 = u - 0.5tg[/tex]

[tex]t = \frac{1.539}{0.5\times 9.8} = 0.314\ s[/tex]

Substituting the value of t in eqn (1):

x' = 0.67(0.314) = 0.210 m

Thus the distance comes out to be 0.210 m

The concept required to develop this problem is Hook's Law and potential elastic energy.

By definition the force by Hooke's law is defined as

[tex]F = kx[/tex]

Where,

k = Spring Constant

x = Displacement

On the other hand, the elastic potential energy is defined as

[tex]E = \frac{1}{2} k\Delta x^2[/tex]

With the given values we can find the value of the spring constant, that is,

[tex]F = kx[/tex]

[tex]k=\frac{F}{x}[/tex]

[tex]k= \frac{83}{0.02}[/tex]

[tex]k = 4120N/m[/tex]

Applying the concepts of energy conservation then we can find the position of the block, that is,

[tex]E = \frac{1}{2} k\Delta x^2[/tex]

[tex]E = \frac{1}{2} k\Delta x^2[/tex]

[tex]4.8 = \frac{1}{2} (4120)(x^2-(-0.02)^2)[/tex]

[tex]x = \sqrt{\frac{2*4.8}{4120}+(-0.02)^2}[/tex]

[tex]x = \pm 0.05225m[/tex]

Therefore the position of the block can be then,

[tex]x_1 = 5.225cm[/tex]

[tex]x_2 = -5.225cm[/tex]

To solve the problem it is necessary to use the concepts related to the calculation of periods by means of a spring constant.

We know that by Hooke's law

[tex]F=kx[/tex]

Where,

k = Spring constant

x = Displacement

Re-arrange to find k,

[tex]k= \frac{F}{x}[/tex]

[tex]k= \frac{mg}{x}[/tex]

[tex]k= \frac{(0.35)(9.8)}{12*10^{-2}}[/tex]

[tex]k = 28.58N/m[/tex]

Perioricity in an elastic body is defined by

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

Where,

m = Mass

k = Spring constant

[tex]T = 2\pi \sqrt{\frac{0.35}{28.58}}[/tex]

[tex]T = 0.685s[/tex]

Therefore the period of the oscillations is 0.685s

The period of the oscillations can be calculated using the formula T = 2π√(m/k), where m is the mass of the package and k is the spring constant. Plugging in the values mentioned in the question, the period can be determined.

The period of the oscillations can be calculated using the formula T = 2π√(m/k), where T is the period, π is a constant (approximately 3.14), m is the mass of the package, and k is the spring constant.

In this case, the mass of the package is 0.350 kg and the spring stretches 12.0 cm (or 0.12 m). The spring constant can be determined using the equation k = mg/X, where g is the acceleration due to gravity (approximately 9.8 m/s2) and X is the displacement (0.12 m).

Plugging the values into the formula, we get T = 2π√(0.350/[(0.350 * 9.8)/0.12]). Solving this equation will give the period of the oscillations.

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a) The spring constant is 1225 N/m

b) The mass of the fish is 6.88 kg

c) The marks are 0.4 cm apart

Explanation:

a)

When the spring is at equilibrium, the weight of the load applied to the spring is equal (in magnitude) to the restoring force of the spring, so we can write

[tex]mg = kx[/tex]

where

m is the mass of the load

[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity

k is the spring constant

x is the stretching of the spring

For the load in this problem we have

m = 10.0 kg

x = 8.00 cm = 0.08 m

Substituting, we find the spring constant

[tex]k=\frac{mg}{x}=\frac{(10)(9.8)}{0.08}=1225 N/m[/tex]

b)

As before, at equilibrium, the weight of the fish must balance the restoring force in the spring, so we have

[tex]mg=kx[/tex]

where this time we have:

m = mass of the fish

[tex]g=9.8 m/s^2[/tex]

k = 1225 N/m is the spring constant

x = 5.50 cm = 0.055 m is the stretching of the spring

Substituting,

[tex]m=\frac{kx}{g}=\frac{(1225)(0.055)}{9.8}=6.88 kg[/tex]

c)

To solve this part, we just need to find the change in stretching of the spring when a load of half-kilogram is hanging on the spring. Using again the same equation,

[tex]mg=kx[/tex]

where this time we have:

m = 0.5 kg

[tex]g=9.8 m/s^2[/tex]

k = 1225 N/m

x = ? is the distance between the half-kilogram marks on the scale

Substituting,

[tex]x=\frac{mg}{k}=\frac{(0.5)(9.8)}{1225}=0.004 m = 0.4 cm[/tex]

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To solve this problem it is necessary to apply the concepts presented in Kepler's third law in which the period is described, as well as the theorems developed for acceleration based on gravity.

Acceleration in gravitational terms can be expressed as

[tex]a = \frac{GM}{r^2}[/tex]

Where,

G = Gravitational Universal Constant

M = Mass of Earth

r = Distance

At the same time the Period by Kepler's law the Period is described as

[tex]T^2 = \frac{4\pi^2r^3}{GM}[/tex]

That is equal to

[tex]T^2 = \frac{4\pi^2r}{\frac{GM}{r^2}}[/tex]

Using the equation of acceleration,

[tex]T^2 = \frac{4\pi^2r}{a}[/tex]

Re-arrange to find a,

[tex]a = \frac{4\pi^2r}{T^2}[/tex]

Our values are given as

[tex]r = 1.5*10^{11}m[/tex]

[tex]T = 365days(\frac{86400s}{1days}) = 31536000s[/tex]

Replacing we have,

[tex]a = \frac{4\pi^2r}{T^2}[/tex]

[tex]a = \frac{4\pi^2(1.5*10^{11})}{(31536000)^2}[/tex]

[tex]a = 0.005954m/s^2[/tex]

Therefore the magnitude of the acceleration of the earth is [tex]0.005954m/s^2[/tex]

The magnitude of the acceleration of the Earth in its orbit around the sun, assuming a circular orbit, is approximately 0.0059 m/s².

The concept at play here is centripetal acceleration, which is the rate of change of tangential velocity and points toward the center of the circle around which Earth, or any object, orbits. For Earth orbiting the sun, we can use the formula ac = v²/r (where v is velocity, r is radius), but given that v isn't directly stated, one must compute it first by v = 2πr/T (where T is the period of the orbit in seconds).

So, the velocity of the Earth in its orbit is approximated as 2π(1.50x10¹¹ m)/(365.25x24x60x60 s) = 29,785.6 m/s. To get the acceleration, plug this computed v value into the ac equation: ac = (29,785.6 m/s)²/(1.50 x 10¹¹ m) = 0.0059 m/s².

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Answer:

Explanation:

Given

mass of baseball [tex]m=0.15 kg[/tex]

radius of ball [tex]r=3.7 cm[/tex]

angular speed of ball [tex]\omega =45 rad/s[/tex]

linear speed of ball [tex]v=42 m/s[/tex]

Transnational Kinetic Energy is given by

[tex]K.E.=\frac{mv^2}{2}[/tex]

[tex]K.E.=\frac{1}{2}\times 0.15\times 42^2[/tex]

[tex]k.E.=\frac{1}{2}\times 0.15\times 1764[/tex]

[tex]  k.E.=132.3 J[/tex]

Considering the ball as solid sphere its moment of inertia is given by  

[tex]I=\frac{2}{5}mr^2=\frac{2}{5}\times 0.15\times (0.037)^2[/tex]

[tex]I=8.21\times 10^{-5} kg-m^2[/tex]

Rotational Kinetic Energy

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

[tex] =\frac{1}{2}\times 8.21\times 10^{-5}\times 45^2[/tex]

[tex] =\frac{1}{2}\times 8.21\times 10^{-5}\times 2025[/tex]

[tex] =0.0831 J[/tex]

Answer:

C. increase because of conservation of angular momentum.

Explanation:

We know that if there is no any external torque on the system then the angular momentum of the system will be conserved.

Angular momentum L

L = I ω

I=Mass moment of inertia ,  ω=Angular speed

If angular momentum is conserved

I₁ω₁ = I₂ω₂

If people migrates towards  Earth's equator then mass moment of inertia will increases.

I₂ > I₁

Then we can say that ω₁  > ω₂ ( angular momentum is conserve)

We know that time period T given as

[tex]T=\dfrac{2\pi}{\omega}[/tex]

[tex]T_1=\dfrac{2\pi}{\omega_1}[/tex]

[tex]T_2=\dfrac{2\pi}{\omega_2}[/tex]

If ω₂ decrease then T₂ will increase .It means that length of the day will increase.

Therefore answer is C.

If a large number of people migrated towards the Earth's equator, the day length would remain largely unaffected. This is due to the law of conservation of angular momentum, considering the insignificant change to the Earth's total mass.

The scenario mentioned in the question refers to a law in physics called the conservation of angular momentum. The length of a day on Earth is predominantly determined by how fast our planet spins on its axis. If there were a large migration of people towards the Earth's equator, hypothetically, the day length would remain largely unaffected. The mass of the people in this context is insignificant when compared to the total mass of the Earth. As such, their migration would not have a noticeable impact on Earth's rotational speed due to conservation of angular momentum. Therefore, the correct answer is that the length of the day would remain unaffected (option b).

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Answer:

Power, P = 924.15 watts

Explanation:

Given that,

Length of the ramp, l = 12 m

Mass of the person, m = 55.8 kg

Angle between the inclined plane and the horizontal, [tex]\theta=25^{\circ}[/tex]

Time, t = 3 s

Let h is the height of the hill from the horizontal,

[tex]h=l\ sin\theta[/tex]

[tex]h=12\times \ sin(25)[/tex]

h = 5.07 m

Let P is the power output necessary for a person to run up long hill side as :

[tex]P=\dfrac{E}{t}[/tex]

[tex]P=\dfrac{mgh}{t}[/tex]

[tex]P=\dfrac{55.8\times 9.8\times 5.07}{3}[/tex]

P = 924.15 watts

So, the minimum average power output necessary for a person to run up is 924.15 watts. Hence, this is the required solution.

We find the height of the 25° incline and then calculate the potential energy per kg, which is then turned into power by dividing by time. The result is 16.6 W or 0.022 hp. Therefore, the minimum average power output necessary for a person to run up a 12.0 m, 25° long hillside in 3.00 s is approximately 2.2% of a horsepower.

This problem is solved in several steps. First, we need to determine the potential energy gain of the person running up the hillside. The formula for potential energy (PE) is PE = m*g*h, where m is mass, g is gravity (approximately 9.8 m/s²), and h is the height of the hill. Since we're not given the person's mass, we'll imagine the person has a mass of 1 kg just to calculate the potential energy gain per kg, but it isn't necessary to know the exact weight for calculating the minimum average power.

The height of the incline is given by 12.0m*sin(25°) = 5.09 m. So, the potential energy gain is PE = 1 kg * 9.8 m/s² * 5.09 m = 49.88 J. Converting this to power (P) by dividing energy by time, P = 49.88 J / 3.0 s = 16.6 W.

Since one horsepower (hp) is approximately 746 watts (W), the power in horsepower is 16.6 W / 746 W/hp = 0.022 hp, or 2.2% of a horsepower. This is the minimum average power output required for a person to climb this hill in 3 seconds.

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Answer:

e=2356.125 V

Explanation:

Given that

D= 10 cm  = 0.1 m

R= 0.05

B= 10,000

Angular speed ,ω = 1800 rev/min

Speed in rad/s given as

[tex]\omega =\dfrac{2\pi N}{60}[/tex]

[tex]\omega =\dfrac{2\pi \times 1800}{60}[/tex]

ω= 188.49 rad/s

The potential difference given as

e= B R v

v=average speed

[tex]v=\dfrac{\omega}{2}R[/tex]

[tex]e=\dfrac{B}{2}R^2 {\omega}[/tex]

By putting the values

[tex]e=\dfrac{10000}{2}0.05^2 \times {188.49}[/tex]

e=2356.125 V

Answer:

water is in the vapor state,

Explanation:

We must use calorimetry equations to find the final water temperatures. We assume that all energy is transformed into heat

        E = Q₁ + [tex]Q_{L}[/tex]

Where Q1 is the heat required to bring water from the current temperature to the boiling point

      Q₁ = m [tex]c_{e}[/tex] ( [tex]T_{f}[/tex] -T₀)

      Q₁ = 50 4180 (100 - 37)

      Q₁ = 1.317 10⁷ J

Let's calculate the energy so that all the water changes state

     [tex]Q_{L}[/tex]  = m L

     [tex]Q_{L}[/tex]  = 50 2,256 106

    [tex]Q_{L}[/tex]  = 1,128 10⁸ J

Let's look for the energy needed to convert all the water into steam is

     Qt = Q₁ + [tex]Q_{L}[/tex]  

     Qt = 1.317 107 + 11.28 107

     Qt = 12,597 10⁷ J

Let's calculate how much energy is left to heat the water vapor

     ΔE = E - Qt

     ΔE = 10¹⁰ - 12,597 10⁷

     ΔE = 1000 107 - 12,597 107

     ΔE = 987.4 10⁷ J

With this energy we heat the steam, clear the final temperature

     Q = ΔE = m [tex]c_{e}[/tex] ( [tex]T_{f}[/tex]-To)

    ( [tex]T_{f}[/tex]-T₀) = ΔE / m [tex]c_{e}[/tex]

      [tex]T_{f}[/tex] = T₀ + ΔE / m [tex]c_{e}[/tex]

      [tex]T_{f}[/tex] = 100 + 987.4 10⁷ / (50 1970)

      [tex]T_{f}[/tex] = 100 + 1,002 10⁵

      [tex]T_{f}[/tex] = 1,003 10⁵ ° C

This result indicates that the water is in the vapor state, in realizing at this temperature the water will be dissociated into its hydrogen and oxygen components

Calculating the final state and temperature of 50 kg of water at 37°C with 10¹°J of energy involves determining the energy needed to heat the water to 100°C and the energy to vaporize it. Using the given specific heat of water, heat of vaporization, and specific heat of steam, one can find whether all the water turns to steam and whether any energy is left to further heat the steam.

If a lightning flash releases about 1010J of electrical energy, and all this energy is added to 50 kg of water at 37°C, we can determine the final state and temperature of the water. Given that the specific heat of water is 4180 J/kg·°C, the heat of vaporization at the boiling temperature for water is 2.256×106J/kg, and the specific heat of steam is 1970 J/kg·°C, these values can be used to calculate the amount of water that can be heated, brought to a boil, and then completely turned to steam. We can break this problem down into different stages, each requiring a certain amount of energy.

Firstly, we would need to calculate how much energy is required to bring the water from 37°C to 100°C (its boiling point):

Afterward, we would need to calculate the energy required to vaporize the water:

The sum of Q1 and Q2 should be equal to or less than the energy released by the lightning flash for all the water to be vaporized. If there is any excess energy after vaporizing the water, it would further heat the steam. If all the energy is not used in heating and vaporizing the water, the final state would still include some liquid water. A detailed calculation using the provided information and considering the stages involved will give the final temperature and state of the water.