find the speed se of the spacecraft when it crashes into the earth. express the speed in terms of me , re , and the universal gravitational constant g .

Answers

Answer 1

Speed of the spacecraft right before the collision:[tex]\sqrt{} \frac{2G.Me}{Re}[/tex] .

Assumption: the earth is exactly spherical with a uniform density.

What is the magnitude of the gravitational force acting on the Earth due to the sun?The magnitude of the gravitational force acting on the Sun due to the Earth is 3.52×1022 3.52 × 10 22 Newtons.

This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

the kinetic energy of this spacecraft, and

the (gravitational) potential energy of this spacecraft.

Let  denote the mass of this spacecraft. At a distance of  from the center of the earth (with mass ), the gravitational potential energy () of this spacecraft would be:

[tex]GPE=\frac{G.Me.M}{R}[/tex]

Initially,  (the denominator of this fraction) is infinitely large. Therefore, the initial value of  will be infinitely close to zero.On the other hand, the question states that the initial kinetic energy () of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.Right before the collision, the spacecraft would be very close to the surface of the earth. The distance  between the spacecraft and the center of the earth would be approximately equal to , the radius of the earth.The  of the spacecraft at that moment would be:Subtract this value from zero to find the loss in the  of this spacecraft:Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the  of this spacecraft would be equal to the size of the gain in its .Therefore, right before collision, the  of this spacecraft would be:On the other hand, let  denote the speed of this spacecraft. The following equation that relates  and  to :Rearrange this equation to find an equation for :It is already found that right before the collision, . Make use of this equation to find  at that moment:

[tex]\sqrt{} \frac{2KE}{M}[/tex]=[tex]\sqrt{\frac{2G.Me}{Re} }[/tex]

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Related Questions

A copper rod has a length of 60 cm at room temperature (220C). The coefficient of expansion of copper is 17 x 10-6 /Co, its specific heat is 386 J/(kg K), its melting point is 1356 K and its heat of fusion is 207 kJ/kg. The copper rod is heated so that its length increases to 60.3 cm a. What is the new temperature of the copper rod? T 316 b. If the mass of the rod is 0.52 kg, how much thermal energy was added to the rod? Q 5.9 x104 J c. If the mass of the rod is 0.52 kg, 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ー ? x105 J

Answers

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

What is coefficient of thermal expansion?

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 fluoroscopy unit's magnification button is used, the resulting magnification is due to a decrease in the size of the: input phosphor. output phosphor. photocathode. focusing lenses.

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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.

Two blocks connected by a rope of negligible mass are being dragged by a horizontal force (see figure below). Suppose F = 68.0 N, m1 = 12.0 kg, m2 = 18.0 kg, and the coefficient of kinetic friction between each block and the surface is 0.100.
(a) Draw a free-body diagram for each block.
(b) Determine the tension T and the magnitude of the acceleration of the system.

Answers

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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Now use a handle in the middle to make the pipe narrower in the middle. Record the readings of the flux meter: Flow rate: 5000L/s, area: 1.1m2Which of these quantities stayed the same when you change the size of the pipe and why? Please also move the speed gauge around to see how the speed varies at different locations inside the pipe.
Flow rate stayed the same but why...?

Answers

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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A cyndrical flask of cross sectional area A is fitted with an airtight piston that is free to slide up and down .contained within the flask is an ideal gas. Initially the pressure applied by the piston is 140kpa and the base of the flask is 22cm. When additional mass is added to the piston the pressure increases to 380kpa. Assuming the system is always at the temperature of 290k find the new height of the piston

Answers

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

What is pressure?

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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A grinding wheel of radius 0.350 m rotating on a frictionless axle is brought to rest by applying a constant friction force tangential to its rim. The constant torque produced by this force is 76.0 N m. Find the magnitude of the friction force.

Answers

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

What is the constant friction?

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;

τ is the torque produce by the force of frictionr is the radius of the grinding wheel

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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a lunar eclipse occurs when group of answer choices the earth lies directly between the sun and the moon the sun lies directly between the earth and the sun the moon lies directly between the sun and the earth

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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.

What is a lunar eclipse?

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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Miguel (m=72) and Fernando (89 kg) board the bumper cars at the local carnival. Miguel is moving at a velocity of 4 m/s when he rear-ends Fernando who is at rest in his path. Fernando and his 125-kg car lunge forward at 2 m/s. Determine the post-collision velocity of Miguel and his 125-kg car.

Answers

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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Light of wavelength 550 nm illuminates a double slit, and the interference pattern is observed on a screen behind the slit. The third maximum is measured to be 2.8 cm from the central maximum. The slits are then illuminated with light of wavelength 440 nm.
How far is the fourth maximum from the central maximum?

Answers

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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Two blocks (X and Y) are in contact on a horizontal frictionless surface. A 36-N constant force is applied to X as shown. The force exerted by X on Y is:

Answers

D 30 N F = ma yields 36N = 24kga, or a 1.5m/s2 acceleration. Due to the 4kg block's contact force, the 20kg block is accelerating. 20 kg at 1.5 m/s2 contact equals 30 N.

What is accelerating?

In mechanics, acceleration refers to the rate at which an object's velocity with respect to time varies. They are accelerations and vector quantities.

The direction of an object's acceleration is determined by the direction of the net force acting on it.

Due to the 4kg block's contact force, the 20kg block is accelerating. 20 kg at 1.5 m/s2 contact equals 30 N.

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Two lasers are shining on a double slit, with slit separation d. Laser 1 has a wavelength of d/20, whereas laser 2 has a wavelength of d/ 15. The lasers produce separate interference patterns on a screen a distance 4.40 m away from the slits. Part What is 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? Express your answer in meters. View Available Hint(s) Ay mas ma 5867 k m Submit Previous Answers x incorrect: Try Again; 3 attempts remaining

Answers

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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find the mass in [g] of the quantity of ice required at 0oc that 10 grams of 100oc steam will melt completely. the heat of vaporization of water at standard temperature and pressure is 2256 kj/kg. the heat of fusion of ice is 334 kj/kg. incorrect answer:

Answers

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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which of the following is not a type of electromagnetic wave with lower energy than visible light waves? microwaves| radar waves| the waves that carry AM or FM broadcasts| the sound waves coming from your transistor radio| the waves that carry television transmissions

Answers

The sound waves coming from your transistor radio is not a type of electromagnetic wave with lower energy than visible light waves.

What is electromagnetic wave explain?
Electromagnetic waves
or EM waves are waves that are created as a result of vibrations between an electric field and a magnetic field. In other words, EM waves are composed of oscillating magnetic and electric fields.

How do electromagnetic waves travel?
Electromagnetic waves
differ from mechanical waves in that they do not require a medium to propagate. This means that electromagnetic waves can travel not only through the air and solid materials but also through the vacuum of space.

What are the 7 types of electromagnetic waves?
In order from highest to lowest energy, the sections of the EM spectrum are named: gamma rays, X-rays, ultraviolet radiation, visible light, infrared radiation, and radio waves. Microwaves (like the ones used in microwave ovens) are a subsection of the radio wave segment of the EM spectrum.

Thus, sound waves coming from your transistor radio are the correct option.

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a cube of mass m sliding without frictionat speed v0. It undergoes a perfectly elastic collision with the bottom tip of a rod of length d and mass M = 2m. The rod is pivoted about a frictionless axle through its center, and initially it hangs straight down and is at rest. What is the cube's velocity - both speed and direction - after the collision?

Answers

The cube's velocity - both speed and direction - after the collision is  [tex]v=\frac{v_0}{5}[/tex].

What is kinetic energy conservation?

Kinetic energy, which may be seen in the movement of an object, particle, or group of particles, is the energy of motion. Any moving item uses kinetic energy, such as a person walking, a baseball being thrown, a piece of food falling off a table, or a charged particle in an electric field.

From kinetic energy conservation:

[tex]\begin{gathered}\frac{m v_0^2}{2}=\frac{m v^2}{2}+\frac{I \omega^2}{2} \\I=\frac{M d^2}{12}=\frac{2 m d^2}{12}=\frac{m d^2}{6} \\\frac{m v_0^2}{2}=\frac{m v^2}{2}+\frac{m d^2 \omega^2}{12} \\v_0^2=v^2+\frac{d^2 \omega^2}{6} \Longrightarrow(1)\end{gathered}[/tex]

From angular momentum conservation:

[tex]\begin{gathered}\frac{m v d}{2}=I \omega \\\frac{m v_0 d}{2}=\frac{m d^2 \omega}{6} \\v_0=\frac{d \omega}{3} \\d^2 \omega^2=9 v_0^2 \Longrightarrow(2)\end{gathered}[/tex]

Substituting (2) into (1):

[tex]\begin{gathered}v^2=v_0^2-\frac{3 v_0^2}{2} \\v^2=-\frac{v_0^2}{2}\end{gathered}[/tex]

The correct answer is: [tex]v=\frac{v_0}{5}[/tex]

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the forearm in the figure(figure 1) accelerates a 3.2 kg ball at 7.5 m/s2 by means of the triceps muscle, as shown. a) Calculate the torque needed. Ignore the mass of the arm. Answer in m*Nb) Calculate the force that must be exerted by the triceps muscle. Answer in N

Answers

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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define the principal focus of a concave mirror

Answers

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.

figure 21-5b shows the concentration profile of a band of material after it has traveled through a column for 2 min and for 26 min. why is the band broader after 26 min? (b) explain why a chromatographic separation normally has an optimum flow rate that gives the best separation.

Answers

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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An elevator, suspended by a cable, is moving upward at constant speed. The correct relationship between the magnitude of force of gravity on the elevator and tension force in the cable is: O A) the magnitude of force of gravity equals the magnitude of tension force B) the magnitude of force of gravity is greater than the magnitude of tension force C) the magnitude of force of gravity is less than the magnitude of tension force D) the magnitude of force of gravity is less than or equal to the magnitude of tension force E) the magnitude of force of gravity is equal to or greater than the magnitude of tension force

Answers

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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When block is placed on a wedge as shown in figure, the block starts sliding down and the wedge also starts sliding on ground. All surfaces are rough. The centre of mass of (wedge + block) system will move
A. leftward and downward
B. rightward and downward
C. leftward and upwards
D. only downward

Answers

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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Each item below belongs either with the population of disk stars or the population of halostars of the Milky Way Galaxy. Match each item to the appropriate population.Disk stars:-high-mass stars-youngest stars-stars that all orbit in nearly the same plane-the SunHalo stars:-globular clusters-oldest stars-stars whose orbits can be inclined at any angle-stars with the smallest abundance of heavy elements

Answers

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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T/F what is the semantic (meaning) of the following sentence? [0.5 pt] show detailed steps to decompose sentence into cnf format [0.5 pt]

Answers

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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determine whether each geologic feature is being caused by tensional, compressional, or shear stresses by analyzing the directions of the forces being applied. drag the appropriate items to their respective bins. each item may be used only once.

Answers

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

What names do rocks above and below a fault go by?

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.

What error occurs when the footwall shifts uphill while the hanging wall shifts downward?

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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Ultraviolet radiation has frequencies from 3.0×1015 to 3.0×1016 Hz, whereas the frequency region for gamma ray radiation is 3.0×1019 to 3.0×1024 Hz.
We can say that:
1. The speed of ultraviolet radiation is _________ higher than/ lower than/the same as gamma ray radiation.
2. The wavelength of ultraviolet radiation is ________ longer than /shorter than/ the same as gamma ray radiation.
What is the maximum number of orbitals that can be identified with the following quantum numbers: n = 3, ℓ = 0
a. 7
b. 8
c. 18
d. 1
e. 2

Answers

1) The same as gamma rays. As we know all electromagnetic radiations travel with speed of light.

2) longer than gamma rays

As we know, higher the frequency lower is the wavelength.as ultraviolet has lower frequency than gamma rays.so it has longer wavelength.

3) n=3 ,l=0 ,hence it is 3s orbital so it has 1 orbital.

(d) is correct.

The frequency of a repeated event is its number of instances per unit of time. It differs from angular frequency and is sometimes referred to as temporal frequency for clarification. One event occurs per second when measuring frequency in hertz.

By comparing the values of the fields you provide, a straightforward frequency analysis generates a report that shows each value for those fields along with the frequency at which it appears.

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Problem 2: Students of College Physics use an experimental set-up like the one shown, except, of course, the cut-away section to aid visualization is only in the drawing. A cylinder with a d=11.43cm inner diameter and an hmax=15.56cm height is placed on a heating plate. A weighted circular disc fits snugly into the cylinder, creating a gas-tight seal; the combined mass of the disc and the weights it supports is 19.07kg.50% Part (a) The disc initially rests on the bottom of the cylinder, but argon gas is introduced through a valve until the bottom of the disc is at a height h0 which is 41.9% of hmax. How much argon gas, in moles, is in the cylinder if it has a temperature of 22.1∘C? (Assume that the gas in the cylinder approximates an ideal gas.)50% Part (b) Suppose an additional mass of 5.56kg is placed on the disc, but at the same time, the heating plate is used to heat the gas in the cylinder. If the height of the disc above the cylinder bottom is now 87.8% of hmax, then what is the change in the temperature, in degrees Celsius, of the gas?

Answers

(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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A car moving at 18 m/s crashes into a tree and stops in 1.14 s. The mass of the passenger inside is 69 kg.Calculate the magnitude of the average force, in newtons, that the seat belt exerts on the passenger in the car to bring him to a halt.

Answers

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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in the instant of fig. 11-41, 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 dis- tance 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 are the (a) magnitude and (b) direction of the

Answers

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

8
Figure 2.24 shows a model of a skyscraper which Omar made
for the school exhibition. It is mounted on a firm plastic base
and has a total weight, W, of 4.0 N. Omar applies a force Fas
shown to test the stability of his model.
Calculate the minimum value of F required to cause the model
to rotate about point P.

Answers

Answer:

24-997

Explanation:

A 75 kg person stands on a metric bathroom scale that reads in Newtons while in an elevator. The
elevator accelerates upward at 3 m/s/s. Determine the person's actual weight and the reading of the
scale under these conditions.

Answers

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

a fighter jet is traveling at 525 m/s directly away from a communication antenna that broadcasts at 410 mhz . What change in frequency does the fighter jet observe

Answers

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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❧ What are the effects of Refraction of Light?

Answers

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

What is Refraction?

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.

Effects of RefractionBending of lightChange in wavelength of lightSplitting of light rays.

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