when we see x-rays from an accretion disk in a binary system, we can't immediately tell whether the accretion disk surrounds a neutron star or a black hole. suppose we then observe each of the following phenomena in this system. which one would rule out the possibility of a black hole?

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 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 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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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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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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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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The translational & rotational kinetic energy is :

a) The translational kinetic energy of the disk across the level surface is 17.58J.

b) The rotational kinetic energy of the disk is 8.79J.

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The energy in joules of an x-ray photon with wavelength 1.00x10-10 m is 1.99* 10^-15J.

What is energy ?

Scientists define energy as the ability to do work. Modern civilization is possible because people have learned how to change energy from one form to another and then use it to do work.

The energy to electron volts is 1.24*10^4 eV.

Wavelength [tex]$\lambda=1.00 \times 10^{-10} \mathrm{~m}$[/tex]

Speed of light [tex]$c=3.0 \times 10^8 \mathrm{~m} / \mathrm{s}$[/tex]

P1anck's constant [tex]$h=6.626 \times 10^{-34} \mathrm{Js}$[/tex]

a) The energy of an X-rays is

[tex]$$\begin{aligned}E & =\frac{h c}{\lambda} \\& =\frac{\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(3.0 \times 10^8 \mathrm{~m} / \mathrm{s}\right)}{\left(1.00 \times 10^{-10} \mathrm{~m}\right)} \\& =1.9878 \times 10^{-15} \mathrm{~J} \\& =1.99 \times 10^{-15} \mathrm{~J}\end{aligned}$$[/tex]

b)

Convert energy of an electron [tex]$\mathrm{J}$[/tex] to[tex]$\mathrm{eV}$[/tex],

[tex]$$\begin{aligned}E & =1.9878 \times 10^{-15} \mathrm{~J}\left(\frac{1 \mathrm{eV}}{1.60 \times 10^{-19} \mathrm{~J}}\right) \\& =1.24 \times 10^4 \mathrm{eV}\end{aligned}$$[/tex]

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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 change in the internal energy of the weight lifter is -5.69 x 10⁵J.

Solution:

[tex]\Delta U = Q-W[/tex]

work done is lifting a body is [tex]1.35*10^5 J[/tex]

so, [tex]W = 1.35*10^5 J[/tex]

during perspiration, heat will be lost by the body to make water evaporate

so,

[tex]Q= -L_v \times m[/tex]

where m is the mass of water

[tex]Q= -2.44 \times 10^6 \times 0.178 J=-4.34 \times 10^5 J[/tex]

Now, change in internal energy,

[tex]\Delta U = -4.34\times 10^5- 1.35 \times 10^5 =-5.69 \times 10^5 J[/tex]

A glass of water standing on a table has no visible macroscopic potential or kinetic energy. However, water molecules have internal energy because they move randomly and each molecule contributes to the total internal energy of water. The energy involved in the system is related to the random motion of the particles and the potential energy of the molecules due to their orientation.

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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 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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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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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 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

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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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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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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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 speed of the sound in air at a temperature of 29 degree celsius is 10.9m/s.

The speed of sound at any particular temperature can be calculated by using the formula,

V = √yRT/M

Where,

V is the speed of sound,

y is the adiabatic constant,

R is the gas constant,

T is the temperature and M is the molecular mass.

We can assume the molecular mass of the air to be 29 g/mol.

Also we can assume the year to be a diatomic gas so the value of the adiabatic constant will be 1.4.

Now assigning all the values,

V = √1.4 x 8.31 x 300/29

V = 10.9 m/s.

So, the speed of the sound in the air at a temperature of 27 degree Celsius is 10.9m/s.

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

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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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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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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 moment of inertia of the system about an axis through the center of the square, perpendicular to its plane is [tex]$0.0636 \mathrm{~kg}-\mathrm{m}^2$[/tex].

What is moment of inertia?

The moment of inertia, otherwise known as the mass moment of inertia, angular mass, second moment of mass, or most accurately, rotational inertia, of a rigid body is a quantity that determines the torque

Given data:

The mass of each sphere is, [tex]$m=0.200 \mathrm{~kg}$[/tex].

Length of side of square is, [tex]$L=0.400 \mathrm{~m}$[/tex].

The expression for the moment of inertia of the system about an axis through the center of the square, perpendicular to its plane is,

[tex]$I=4 m R^2$[/tex]

Here,

[tex]$\mathbf{R}$[/tex] is the distance between center of the square and the sphere. And its value is,

[tex]$$\begin{aligned}& R=\frac{1}{2} \sqrt{L^2+L^2} \\& R=\frac{1}{2} \sqrt{0.400^2+0.400^2} \\& R=0.282 \mathrm{~m}\end{aligned}$$[/tex]

Then, moment of inertia is,

[tex]$$\begin{aligned}& I=4 \mathrm{~m} R^2 \\& I=4 \times 0.200 \times 0.282^2 \\& I=0.0636 \mathrm{~kg}-\mathrm{m}^2\end{aligned}$$[/tex]

Thus, the moment of inertia of the system about an axis through the center of the square, perpendicular to its plane is [tex]$0.0636 \mathrm{~kg}-\mathrm{m}^2$[/tex].

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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.

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 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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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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Answer:

atoms

Explanation:

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