a friend of yours is loudly singing a single note at 500 hz while racing toward you at 27.2 m/s on a day when the speed of sound is 343 m/s. what frequency do you hear?

The magnitude of impulse acting directly upwards during the slap is -0.00273 N.s.

The magnitude of the impulse that is downward on the lizard due to the gravitational force is 0.375 kg.m/s.

Given that, mass of the basilisk lizard = 90 g = 90/1000 kg = 0.09 kg

Mass of each foot = 2.53 g = 0.00253 kg

Vf = 0

Vi = 1.08 m/s

Time taken for a single step is 0.417 s

I = Δp = Mf * (Vf - Vi ) = 0.00253* (0 - 1.08) = -0.00273 N.s

On each step, the basilisk lizard must provide an upward force that counters its weight which is equal to W= mg.

We know the formula for impulse acting downwards is, I = F Δt

It can also be written as I = W Δt = Ml* g Δt

Substituting the values in the above formula,

Ml*g Δt = 0.09* 10* 0.417 = 0.375 kgm/s

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The magnitude of the constant friction force applied tangential to the rim of the grinding wheel is 217.14 N.

Friction force is a type of force that opposes the motion of two surfaces that are in contact.

This type of force tends to stop an object in motion from moving or sometimes it tends to stop an object at rest from moving.

The magnitude of the constant friction force applied tangential to the rim of the grinding wheel is calculated by applying the formula for torque as shown below.

τ = F r

where;

F = τ / r

F = ( 76 N m ) / ( 0.35 m )

F = 217.14 N

Thus, the magnitude of the friction force applied tangential to the rim of the grinding wheel is a function of the torque produced and the radius of the wheel.

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The force the seat belt exerts on a passenger in the car to bring him to a halt is  0.6 × [tex]10^{3}[/tex]

A car moving at 18 m/s crashes into a tree and stops in 1.14 s. calculate the force the seat belt exerts on a passenger in the car to bring him to a halt. the mass. of the passenger. is 69. kg

Generally the equation for the net Force  is mathematically given as

Fnet = mdv/dt

net = 69*10/1.14 = 0.6 × [tex]10^{3}[/tex]

When objects interact with one another, push and pull are generated. You can also describe force using words like stretch and squeeze.

Force is described as follows in physics:

A massed item changes its velocity in response to a push or pull.

Force is applied at a location known as the application, and the direction in which it is applied is referred to as the force's direction.

Using a spring balance, one can quantify the Force. Newton is the unit of force in the SI (N) The following list of consequences of the Force is not exhaustive.

A body at rest can move with enough force.

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(a) The amount of argon gas in the cylinder is approximately [tex]\rm \(4.09 \times 10^{-5}\)[/tex] moles.

(b) The change in temperature of the gas is approximately 976.7°C.

Part (a):

Given data:

Inner diameter of the cylinder (d) = 11.43 cm = 0.1143 m

Height of the cylinder ([tex]h_{max[/tex]) = 15.56 cm = 0.1556 m

Height of the disc above the bottom of the cylinder ([tex]h_0[/tex]) = 41.9% of h_max = 0.419 * 0.1556 m

Combined mass of the disc and weights (m) = 19.07 kg

Temperature of the gas (T) = 22.1°C = 22.1 + 273.15 = 295.25 K

We need to find the amount of argon gas in moles in the cylinder.

Step 1: Calculate the volume of the gas in the cylinder.

[tex]\[ V_{\text{gas}} = \pi \left(\frac{d}{2}\right)^2 \times h_0 \]\\\\\ V_{\text{gas}} = \pi \left(\frac{0.1143}{2}\right)^2 \times 0.419 \times 0.1556 \]\\\\\ V_{\text{gas}} \approx 0.00121 \, \text{m}^3 \][/tex]

Step 2: Calculate the number of moles of argon gas using the ideal gas law.

[tex]\[ PV = nRT \][/tex]

Where P is the pressure of the gas (we assume it is constant), V is the volume of the gas, n is the number of moles, R is the universal gas constant (8.314 J/(mol·K)), and T is the temperature in Kelvin.

[tex]\[ n = \frac{PV}{RT} \]\\\\\ n = \frac{1 \times 0.00121}{8.314 \times 295.25} \]\\\\\ n \approx 4.09 \times 10^{-5} \, \text{moles} \][/tex]

Part (b):

Given data:

Additional mass placed on the disc ([tex]m_{extra[/tex]) = 5.56 kg

New height of the disc above the bottom of the cylinder (h_new) = 87.8% of h_max = 0.878 * 0.1556 m

We need to find the change in the temperature of the gas.

Step 1: Calculate the new volume of the gas in the cylinder.

[tex]\[ V_{\text{gas\_new}} = \pi \left(\frac{d}{2}\right)^2 \times h_{\text{new}} \]\\\\\ V_{\text{gas\_new}} = \pi \left(\frac{0.1143}{2}\right)^2 \times 0.878 \times 0.1556 \]\\\\\ V_{\text{gas\_new}} \approx 0.00178 \, \text{m}^3 \][/tex]

Step 2: Use the ideal gas law again to find the new temperature of the gas.

[tex]\[ n_{\text{new}} = \frac{P \times V_{\text{gas\_new}}}{RT_{\text{new}}} \]\\\\\\ T_{\text{new}} = \frac{P \times V_{\text{gas\_new}}}{n_{\text{new}} \times R} \]\\\\\ T_{\text{new}} = \frac{1 \times 0.00178}{4.09 \times 10^{-5} \times 8.314} \]\\\\\ T_{\text{new}} \approx 1271.95 \, \text{K} \][/tex]

Change in temperature:

[tex]\[ \Delta T = T_{\text{new}} - T \]\\\\\ \Delta T = 1271.95 - 295.25 \]\\\\\ \Delta T \approx 976.7 \, \text{K} \][/tex]

Result:

(a) The amount of argon gas in the cylinder is approximately [tex]\rm \(4.09 \times 10^{-5}\)[/tex] moles.

(b) The change in temperature of the gas is approximately 976.7°C.

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The flow rate is given as 5000L/s, which means that 5000 liters of fluid pass through a cross-sectional area of 1.1m2 per second. If the size of the pipe is changed by using a handle to make it narrower in the middle, the cross-sectional area of the pipe will also change.

However, if the flow rate remains the same, it means that the volume of fluid passing through the pipe per unit of time is still constant.

To maintain the same flow rate, the velocity of the fluid must increase as the cross-sectional area decreases, and vice versa. This can be observed by moving the speed gauge around to different locations inside the pipe and measuring the speed of the fluid at those locations.

As the cross-sectional area of the pipe decreases, the velocity of the fluid will increase, and as the cross-sectional area increases, the velocity of the fluid will decrease.

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The mass of ice required at 0oc that 10 grams of 100oc steam will melt completely is 80.07 g.

The amount of energy that must be supplied to a solid substance (typically in the form of heat) in order to cause a change in its physical state and convert it to a liquid is known as latent heat of fusion (when the pressure of the environment is kept constant). The latent heat of fusion of one kilogramme of water, for example, is 333.55 kilojoules, which is the amount of heat energy that must be supplied to convert one kilogramme of ice without changing the temperature of the environment (which is kept at zero degrees Celsius).

Given,

mass of steam, [tex]m_s[/tex] = 10g = 0.01 kg

Latent heat of fusion of ice ,[tex]L_f= 334kj/kg[/tex]

Latent heat of vaporization of water, [tex]L_v=2256kj/kg[/tex]

change of temperature, [tex]\delta T=100-0=100\textdegree\ C[/tex]

The amount of heat cooling can be determined by.

[tex]m_{ice}L_f=m_sL_v+m_sS \delta T[/tex]

Here, S= specific heat constant i.e [tex]4.184 kj/kg[/tex]

[tex]m_{ice}L_f=(0.01*2256)+(0.01*4.184*100)\\\\m_{ice}*334=22.56+4.184\\\\m_{ice}*334=26.744\\\\m_{ice}=\frac{26.744}{334}=0.0807kg=80.07 g[/tex]

Thus, the mass of quantity of ice is 80.07 g.

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The following are effects of refraction of light: bending of light, change in wavelength of light, splitting of light rays.

Refraction is the bending of light rays as it passes from one medium into another. Light rays refracts whenever it travels at an angle into a substance with a different refractive index. This change of direction is caused by a change in speed of the light rays. For example, when light travels from air into water, it slows down, causing it to continue to travel at a different angle or direction.

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The tension T is 27.2N and the magnitude of the acceleration of the system is 1.29m/s^2 when two blocks connected by a rope of negligible mass are being dragged by a horizontal force.

Given force exerted on block F = 68.0 N

Mass of first block  m1 = 12.0 kg

Mass of second block m2 = 18.0 kg

The coefficient of kinetic friction is 0.100.

(a) Block 1's tension force on block 2 is equal to object 2's tension force on object 1, in terms of size. A light string's tension, which remains constant along its length, indicates how firmly the string is pulling on the objects at both ends.

(b) The second law of Newton is applied using the free-body diagrams.

On m1 we have net force acting = T-f = ma

Fnet = N1-m1g = 0 then N= m1g

the coefficient of friction gives f1 = μN1

f1 = 0.100x12x9.8 = 11.8N

On m2 we have net force acting = F1 - T- f = ma

We have from the y component N2 = m2g

then f2 = 0.100x18x9.8 = 17.6N

F−f1 −f2 =m2a+m1a

68-11.8-17.6 = a(12+18)

a = 1.29m/s^2

The acceleration of the system is 1.29m/s^2

we have tension T = m1a+f1 = 12x1.29+11.8 = 27.2N

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A lunar eclipse occurs when the earth is directly between the sun and the moon, thus, its shadow creates a lunar eclipse. Correct answer: letter A.

A lunar eclipse occurs when the Earth's shadow is cast on the surface of the Moon. This can only happen if the Sun, Earth, and Moon are all aligned in a straight line. In this configuration, the Earth is directly between the Sun and the Moon, blocking the light from the Sun and casting a shadow onto the Moon.

Is an astronomical event that occurs when the Moon passes directly behind Earth and into its shadow. During a lunar eclipse, the Moon will appear dark as it is illuminated by only the faint, red-orange light of the sun scattered by Earth's atmosphere.

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The fourth maximum is measured to be 2.97 cm from the central maximum when slits are illuminated with light of wavelength 440 nm.

The interference or bending of waves passing through an aperture around the corner of an obstacle or into the geometric shadow area of the obstacle/aperture is defined as diffraction. Secondary sources of propagating waves are diffractive objects or apertures. It is the process by which a ray or other wave system expands as it passes through a narrow aperture or edge. It is usually the result of interference between the generated waveforms.

Given,

[tex]\lambda = 550 nm=550*10^{-9}n[/tex]

[tex]y_3 = 2.8 cm = 2.8 * 10^{-2} m[/tex]

The third maximum from the central maximum can be determined by

[tex]y_3 = \frac{3D \lambda}{d}\\\\2.8*10^{-2}=\frac{3*550*10^{-9}D}{d}\\\\\frac{2.8*10^{-2}}{3*550*10^{-9}}=\frac{D}{d}\\\\0.0169*10^6=\frac{D}{d}[/tex]

Now, the fourth maximum from the central maximum can be determined by,

Now, [tex]\lamda = 440 nm =440 *10^{-91} m[/tex]

[tex]y_4 = \frac{4D \lambda }{d}\\\\y_4=4*0.0169*10^6*440*10^{-9}\\\\y_4=29.744*10^{-3}m=2.97 cm[/tex]

Thus, the fourth maximum is measured to be 2.97 cm from the central maximum

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The principal focus of a concave mirror is a point on the axis of the mirror at which parallel rays of light are brought to focus after being reflected by the mirror. A concave mirror is a mirror with a curved surface that is shaped like the inside of a bowl, with the center of the curve facing outward. When light rays are reflected by a concave mirror, they are brought to a focus at the principal focus of the mirror. The distance from the center of the mirror to the principal focus is known as the focal length of the mirror. The principal focus of a concave mirror is an important concept in optics and is used in the design and operation of various optical instruments, such as telescopes and cameras.

First-order analysis of a seismogram record enables seismologists to do all except to develop a Modified Mercalli Intensity map.

First-order analysis of a seismogram record identifies the different kinds of seismic waves reaching to the earth's surface by the fault movement. This analysis records P waves.

The P wave will be the first wave to reach the surface. P waves are the fastest seismic waves. So, they will usually be the first ones that are recorded by a seismograph. The next seismic waves to record on a seismogram are the S waves.

A seismograph is the main instrument used to measure earthquakes. The seismograph produces a digital graphic recording of the ground motion caused by the reaching seismic waves on the earth surface. The digital recording is known as a seismogram.

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Answer:  735 N (actual), 960 N (accelerating up)

Explanation:

Actual weight = W = mg = (75 kg)(9.8 m/s²) = 735 N

Scale reading = N (normal force acting on the person by the elevator) = mg + ma = 735 N + (75 kg)(3.0 m/s²) = 735 N + 225 N = 960 N

That's why you feel heavier when you are accelerating up in an elevator

The calculated answer for the torque is 7.812 Nm and the force is 25.2 N

In the given question, we have to find the torque needed and the force exerted by the triceps muscle.

We know that, when there is a net torque on the system, it causes angular acceleration and the angular momentum is no more conserved.

Here let be the arm is massless.

The torque can be written as

              τ=Iα

Where I is the moment of inertia.

α is the angular acceleration.

So,

  τ=mR^2 α

    τ=mRa

Here we have used the relationship between angular and tangential acceleration, which is: α=a/R.

The position vector is:  R = 0.31

The mass is:   m =3.6

The tangential acceleration is:   a =7 m/s2

So, the torque will be,

τ=3.6*0.31*7

  =7.812

So, the required torque is 7.812 Nm.

And we know that torque is:  τ=F.R

So,   force is F =  τ/r

 = 7.812/0.31

 =25.2 N

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Rightward and downwards.  option B is the correct answer

Internal friction between the wedge and the block means that it won't affect the center of mass' motion. The external friction force exerted on the wedge by the ground causes the center of mass to shift to the right. The center of mass of a block moves downward due to the gravitational force (mg) acting on it.

Frictional Force:

When surfaces of two objects come into contact, the force called friction is what stops motion. In other words, friction weakens a machine's mechanical advantage and decreases the output to input ratio. In a car, reducing friction requires one-fourth of the energy required. Friction in the clutch and tires, however, also has an impact on how well the car can maintain its place on the road. One of the most significant occurrences in the physical universe is friction, a phenomenon that affects everything from machines to molecular structures to matches.

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the daily need for activated carbon in kg is 4319.327 kg of carbon. It would fill a 55-gallon tank, which is 18.

groundwater containing 10 mg/L of benzene

treatment goal of 99%

Flow rate of ground water =100 Gpm

                                            =378.541 Liter/min

now weight of benzene in water

  =378.541 Liter/min*10 mg/lit

  = 3.78541 gram/min

now per day weight of benzene

=3.78541 g/min*(24*60)min=5450.9904 gram

now frewndlich equation derives

q(mg/g)=2 c[tex]^{0.2}[/tex]

c=0.01*10=0.1 mg/L

q=2*[tex](0.1)^{0.2}[/tex] = 1.262

q=1.262 (mg/g)

it means 1.262 mg of benzene adsorps on one gram of activated carbon

a)now per day mass=5450.9904*10³ mg

1.262 mg needs 1g of carbon

1mg needs 1/1.262 g of carbon

5450.9904*10³  mg needs

((1/1.262)*5450.9904*10³ ) grams

=4319.327 Kg of carbon /day

b) packing density of activated carbon = 1.2 g/L

tank size =55 gallon

now 1.2g fills in 1 lit

1g fills in (1/1.2)litre

4319.327*10³ g fills in(1/1.2)*4319.327*10³ liter

                                   =3600 liter volume needed

1 tank volume =55gallon

                        =208.2 lit

number of tanks needed =3600 liter/208.2

                                          =17.29≅18 tanks

∴18 such tanks are needed

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Assuming the system is always at the temperature of 290k, the new height of the piston be 16.21 cm.

The physical force applied to an object is referred to as pressure. Per unit area, a perpendicular force is delivered to the surface of the objects. F/A is the fundamental formula for pressure (Force per unit area). Pascals are a unit of pressure (Pa).

Absolute, atmospheric, differential, and gauge pressures are different types of pressure.

As height of the piston is proportional to volume of ideal gas. From Boyle's law of gases   at constant temperature  we can write:

PV = pv

⇒PH = ph

⇒ 140 kpa × 22 cm = 290 kpa × h

⇒ h = 22 × 140/190 cm = 16.21 cm.

Hence, the new height of the piston be 16.21 cm.

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The broader band because of mobility of ions and because of peaks developed it has optimum flow rate.

There are two characteristics of the concentration profile that are important in determining the efficiency of a column and its ability to separate or separate subsequent solute zones. The first peak maximum refers to the position of maximum concentration in the peak. To obtain sufficient resolution, the maxima of two adjacent peaks must be separated from each other. Such separations depend on the nature of the solutes and the selectivity of the stationary and mobile phases.

A second characteristic that is important for efficiency and resolution is peak width. Peaks with distant maxima are still broad and may not fully resolve the solutes. For this reason, peak width is very important in chromatography.

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the distance Aymamin between the second maximum of laser 1 and the third minimum of laser 2 on the same side of the central maximum is 0.4  m

For part be we can use the equation

ym = m*λ*L/d

so for the first max for each one m =1

so for laser 1

y = (d/20)*(6.0m)/d

y = 6.0m /20 = 0.3

for laser 2 we do the same thing

y = 6.0m/15 = 0.4

y  (  laser1  )   <   y  (  laser2 )

(1)  hence   first maxima  of   laser1  is closest  to  central maxima

Δy = 0.4m   -0 .3 m = 0.1m

For part C)

we use the same equation to find the y for laser 1, except m =2

y= 2*6.0 m /  20   =  0.6 m

Now for laser 2 we use:

ym=(m+1/2)*λ*L/d

since there is no central minimum the first minimum is at m = 0.that means that the third minimum is at m = 2

simplifying  the equation we get

y = (2.5)*6.0 /  15 =   1m  

now we solve

Δy=     1m   -   0.6m   =  0.4  m

Double Slit Experiment:

The double-slit experiment shows that light and matter can exhibit properties of both classically defined waves and particles. Furthermore, it demonstrates the fundamental stochastic nature of quantum mechanical phenomena.

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Oldest: Red Giant in globular cluster M13 AND red main-sequence star in globular cluster M13

Middle: the sun

Youngest: Hot, blue main-sequence star in disk

What is globular cluster?

Globular clusters, which include tens of thousands to millions of stars, are stable, firmly bound clusters. They are connected to all different kinds of galaxies. Globular clusters are strongly gravitationally bound and often larger than open clusters.

As a result, the Milky Way's globular clusters (which house extremely old stars) show a spherical component, whilst the open clusters, other young stars, and star-forming areas show a disk-shaped component.

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The significance or understanding of a word or sentence is referred to as semantic. The way a statement in a multi-page report is understood; the semantic significance of the sentence—is an example of semantics.

What is semantics?

Semantics is the study of the significance of language. It could as well be used on entire texts or just a single word.

For instance, despite the fact that "objective" and "final stop" have identical meanings, semantics novices analyze their somewhat different shades of importance.

Example:

1) Sam was tired [ Adj. Description]

2) My interview was yesterday [Adv. Time].

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There are 3 questions here that need to be solved.

a. The new temperature of the copper rod is 316°C (rounded)

b. If the mass of the rod is 0.52kg, the thermal energy that was added to the rod is 5.9 * 10^4J

c. If the mass of the rod is 0.52kg, the total thermal energy will need to be added to the copper if we want to completely melt the copper rod starting at room temperature by using coefficient of thermal expansion of copper at 17 * 10^-6 / C° is 3.21 * 10^5J

It refers to the rate at which material expands with increase in temperature. More specifically, this coefficient is determined at constant pressure and without a phase change, such as the material is expected to still be in its solid or fluid form.

Now, moving to the question.

a. What is the new temperature of the copper rod?

L₂ = L₁ + L₁ * α * (t₂ - t₁)

Where

L2 is length of the rod that already increased

L1 is the original length of the rod

α is coefficient of expansion of copper

t2 is the new temperature

t1 is the original temperature

60.3cm = 60cm + 60cm * 17 * 10^-6 * (t₂ - 22°C)60.3

= 60 + 0.00102 * (t₂ - 22)0.3 = 0.00102 * (t₂ - 22)t₂ - 22

= 0.3 / 0.00102 t₂ - 22 = 294.12

t₂ = 316°C (rounded)

b. If the mass of the rod is 0.52kg, how much thermal energy was added to the rod?

Q = m * c * Δt

m = mass

c = coefficient

= 0.52 * 386 * (316 - 22)

= 59011.68

= 5.9 * 10^4 J (rounded)

c. If the mass of the rod is 0.52kg, how much total thermal energy will need to be added to the copper if we want to completely melt the copper rod starting at room temperature?

Q = m * c * Δt + m * L fusion

= m * c * (t melt - t₁) + m * L fusion

= 0.52 * 386 * (1356 - (22 + 273)) + 0.52 * 207 * 10^3Q

= 320603.92

Q = 3.21 * 10^5 J

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Answer:

atoms

Explanation:

atoms   the smallest unit of matter which can be identified as an element

Tensional faults—normal faults—cause each geologic feature, whereas shear faults—strike–slip faults—cause the remaining three.

The hanging wall, also known as the headwall or footwall, is the higher or overlaying block along the fault plane as rocks slide past one another during faulting. The footwall is the lower block.

A typical fault occurs when the rocks above the fault plane, also known as the hanging wall or footwall, move downward in relation to the rocks below the fault plane, or the footwall. A reverse fault occurs when the footwall slides up relative to the hanging wall.

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Miguel will be travelling at 1.2 m/s inside the opposite direction following the collision, according to this.

Define collision.
A collision in physics is any situation in which two or more bodies quickly exert forces on one another. Despite the fact that the most common usage of the word "collision" refers to situations in which two or more objects clash violently, the scientific usage of the word makes no such assumptions.

The conservation of momentum law allows for the calculation of Miguel's post-collision speed.
Mass times velocity (p=mv), a vector quantity, equals momentum.
Before the collision, Miguel and Fernando's car had a momentum of 782 kg*m/s (m1=72 kg) (v1=4 m/s) + (m2=125 kg) (v2=0 m/s).
The momentum of the system following the collision is equivalent to (m1=72kg)(v1=?m/s) + (m2=125kg)(v2=2m/s) = 782 kg*m/s.
Miguel's comment velocity is -1.2 m/s, as determined by solving for v1. Miguel will be travelling at 1.2 m/s inside the opposite direction following the collision, according to this.

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An elevator is traveling steadily upward while being hung by a cable. When this happens, the tension force's strength is equal to the gravitational force's strength.

The elevator, which is hung by a cable and traveling upward at a steady pace, is principally being acted upon by two forces. One of the forces is gravitational force, which is the weight of gravity acting on the elevator. Another force is tension in the cable, which is holding the elevator in place. Acceleration is the same as the velocity's time rate of change.

Where v is the end velocity and u is the beginning velocity, an is equal to v-u/t. However, the speed here is constant. Thus, there will be no acceleration. As a result, the net force acting on this elevator will also become zero utilizing Newton's Second Law of Motion.

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The current will be flowing in an anti-clockwise direction.

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The fighter jet notices a frequency change of 718 Hz.

from Doppler's effect, f' = f(1 - [tex]\frac{u}{c}[/tex])

Here, c stands for the speed of light, u for the speed of a fighter jet, and f for the antenna's broadcast frequency.

so, f' = f(1 - [tex]\frac{u}{c}[/tex])

⇒f'/f =( 1 - [tex]\frac{u}{c}[/tex])

⇒(f - f')/f = [tex]\frac{u}{c}[/tex]

⇒∆f = f([tex]\frac{u}{c}[/tex])

so change in frequency, ∆f = f([tex]\frac{u}{c}[/tex])

now f = 410 × 10⁶ Hz , u = 525 m/s and c = 3 × 10 m/s

so, ∆f = (410 × 10⁶ × 525)/(3 × 10⁸)

= 71750 × 10⁻²

= 717.75 Hz ≈ 718 Hz

As a result, the fighter jet recorded a 718 Hz change in frequency.

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When a digital type of fluoroscopy unit's gets magnification button and used it, then the resulting magnification decrease in size due to input phosphor. Option (a) is correct.

The front phosphor (or enter phosphor) is a element of the photo intensifier in fluoroscopic structures that converts the strength from x-rays into mild photons. It consists of a fluorescent cloth including cesium iodide activated with sodium (CsI:Na) and coats the doorway floor of the photo intensifier.

Output phosphor consists of a fluorescent compound (referred to as P20) fabricated from silver-activated zinc cadmium sulphide (ZnCdS:Ag) particles. In addition, caesium iodide has a crystalline structure, and the enter phosphor may be synthetic in order that the slim needle-like crystals (about five μm in diameter) are laid down perpendicular to the screen.

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Correct Question:

when a digital fluoroscopy unit's magnification button is used, the resulting magnification is due to a decrease in the size of the:

a). input phosphor.

b). output phosphor.

c). photocathode.

d). focusing lenses.

The magnitude and direction of net angular momentum of particles is [tex]52.7kg.m^2/s[/tex] out of the page.

Given,

particle 1:-

Mass ,m1 = 6.5 kg

speed , v1 = 2.2 m/s

distance from point O, d1 = 1.5 m

Particle 2:-

Mass ,m2 = 3.1 kg

speed, v2 = 3.6 m/s

distance from point O, d2 = 2.8 m

Angular momentum can be determined by formula,

p=mvd

angular momentum of particle-1, [tex]\rho_1=6.5*2.2*1.5=21.45 kg.m^2/s[/tex]

angular momentum of particle-2, [tex]\rho_2=3.1*3.6*2.8=31.248 kg.m^2/s[/tex]

Then, the net angular momentum is

[tex]\rho=\rho_1+\rho_2=21.45+31.248=52.69kg.m^2/s[/tex]

The direction of angular momentum of particle is always perpendicular to the motion of particle, as the particles are moving on xy-plane then the their direction will be perpendicular to xy-plane i.e. z-plane, it means out of the page.

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Your question is incomplete, here is the complete question.

In the instant shown in the diagram, two particles move in an xy-plane. Particle P1 has mass 6.5 kg and speed v1 = 2.2 m/s, and it is at distance d1 = 1.5 m from point O. Particle P2 has mass 3.1 kg and speed v2 = 3.6 m/s, and it is at distance d2 = 2.8 m from point O. What is the magnitude and direction of net angular momentum of the two particles about O?

A) 52.7 kg · m2/s out of the page B) 52.7 kg · m2/s into the page C) 21.5 kg · m2/s into the page D) 9.8 kg · m2/s into the page E) 9.8 kg · m2/s out of the page