How is heat energy transferred from the ground to the atmosphere

The speed with which the squid ejects the water is 42 m/s. If we ignore lift and drag forces, the maximum horizontal range that the squid can achieve before splashing down is about 1.25 m

From the conservation of momentum,

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(36 kg) (3.5 m/s) - (3 kg) v = 0

(3 kg)v= (36 kg) (3.5 m/s)

v = 42 m/s

Hence, required speed is 42 m/s.

The maximum horizontal range is,

R = u ² sin 20/g

Rmax = u²sin 2 (45°)/g

= (3.5 m/s) ² sin 90° / 9.8

=1.25 m

The maximum horizontal range is 1.25 m.

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Dark matter comprises of 90% of the mass of the universe which is based on galaxy rotation curves and cluster dynamics.

The Galaxy rotation curves remains the same from the center of the galaxy to the periphery which means that there must be non-luminous matter that is spread in our galaxy. This matter is called dark matter.

The rotation of stars and gas in distant galaxies cannot be measure directly. Only the component along the line of sight is known from precise spectroscopic measurements.

If the galaxy is viewed almost face on, then it would mostly rotate within the plane of the sky. This could mislead observers into thinking that the galaxy is actually rotating very slowly, which would require them to overestimate the inclination between disk and sky planes. This inclination was estimated from how elliptical the galaxy appears.

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The speed of rocket travelling past the earth is 0.34c.

Consider the earth frame of reference as S and rocket's frame S moving along the direction.

Lorentz transformation equations for addition of relativistic velocities is

u' = u - v / 1 - (uv/c²)

Here,

u = observed speed of the projectile on earth frame of reference

u' = speed of the projectile leaving the rocket

c = speed of the light

If we rearrange the equation for v, we get

u'[1- (uv/c²)] = u - v

u' - (uu'v/c²) = u - v

v - (uu'v/c²) = u - u'

v = u-u' / [1-(uu'/c²)]

The rocket speed is given as

v = u-u' / [1-(uu'/c²)]

Substituting the values,

v = 0.90-0.95u' / [1-(0.90*o.95/c²)]

v = - 0.34c

Our assumptions contradicts that the rocket moving along the same direction of the projectile.

Therefore, the speed of rocket travelling past the earth is 0.34c.

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The speed of rocket travelling past the earth is 0.34c.

Consider the earth frame of reference as S and rocket's frame S moving along the direction.

Lorentz transformation equations for addition of relativistic velocities is

u' = u - v / 1 - (uv/c²)

Here,

u = observed speed of the projectile on earth frame of reference

u' = speed of the projectile leaving the rocket

c = speed of the light

If we rearrange the equation for v, we ge

u'[1- (uv/c²)] = u - v

u' - (uu'v/c²) = u - v

v - (uu'v/c²) = u - u'

v = u-u' / [1-(uu'/c²)]

The rocket speed is given as

v = u-u' / [1-(uu'/c²)]

Substituting the values,

v = 0.90-0.95u' / [1-(0.90*o.95/c²)]

v = - 0.34c

Our assumptions contradicts that the rocket moving along the same direction of the projectile.

Therefore, the speed of rocket travelling past the earth is 0.34c.

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Rank of particles on the basis of energy is  2 > 3 = 4 > 1 > 5 = 6 and Rank of particles on the basis of speed is 1=2=3=4=5=6.

Speed is same for all that is speed of light. Energy can produce both positive and negative charges +q and −q.  when a charged particle and antiparticle annihilate each other. If +q and −q both are 0 that implies m equals 0.

when an electron and positron annihilate to produce two real photons of zero mass moving away with the same energy as the electron positron pair to conserve energy.

Besides opposite charges, opposite times must also cancel out. For the electron time moves in the forward direction while for the positron time moves in the backward direction. Hence time also cancels out leading to two photons which move at the speed of light for which time stands still.

Antimatter has been considered as a trigger mechanism for nuclear weapons,and as the possibility of creating thermonuclear energy.

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The speed of the truck relative to the highway is 44m/s.

Given that the radius of curvature is [tex]$R=150 \mathrm{~cm}$[/tex]

Then the focal length is [tex]$\mathrm{f}=\mathrm{R} / 2=75 \mathrm{~cm}$[/tex]

Object distance is [tex]$\mathrm{s}=2.0 \mathrm{~m}$[/tex]

Speed of imge is [tex]$v^{\prime}=1.5 \mathrm{~m} / \mathrm{s}$[/tex]

From thin lens equation

[tex]$$\begin{aligned}1 / \mathrm{f} & =1 / \mathrm{s}+1 / \mathrm{s}^{\prime} \\\mathrm{s}^{\prime} & =\mathrm{f}^* \mathrm{~s} /(\mathrm{s}-\mathrm{f}) \\\mathrm{s}^{\prime} / \mathrm{s} & =\mathrm{f} /(\mathrm{s}-\mathrm{f})\end{aligned}$$[/tex]

Velocity of the image is [tex]$v^{\prime}=d s^{\prime} / \mathrm{dt}$[/tex]

[tex]$\mathrm{v}^{\prime} & =\left(d \mathrm{~s}^{\prime} / \mathrm{ds}\right)(\mathrm{ds} / \mathrm{dt}) $[/tex]

[tex]$ =[\mathrm{f} /(\mathrm{s}-\mathrm{f})]^2\left(\mathrm{v}^{\prime \prime}\right) $[/tex]

[tex]$ =\mathrm{f}^2 \mathrm{v} /(\mathrm{s}-\mathrm{f})^2$[/tex]

[tex]$\mathrm{v}^{\prime \prime} & =\mathrm{v}^{\prime}(\mathrm{s}-\mathrm{f})^2 / \mathrm{f}^2 $[/tex]

[tex]$= & \mathrm{v}^{\prime}(\mathrm{s}-\mathrm{f})^2 / \mathrm{f}^2$[/tex]

[tex]& =(1.5 \mathrm{~m} / \mathrm{s})(2.0 \mathrm{~m}-(-0.75))^2 /(-0.75 \mathrm{~m})^2 $[/tex]

[tex]$=20.17 \mathrm{~m} / \mathrm{s}$[/tex]

[tex]$$\text { Then we get } \quad v^{\prime \prime}=v^{\prime}(s-f)^2 / f^2$$[/tex]

Therefore the speed of the truck relative to the highway [tex]v=24+20.17=44.17 \mathrm{~m} / \mathrm{s}$ or $44 \mathrm{~m} / \mathrm{s}$[/tex].

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We would weigh 9.68x1013 N on the surface of a neutron star with the same mass as our Sun and a diameter of 20.0 km if we were to weigh 715 N on Earth.

According to the information provided, a neutron star has the same mass as the Sun, making its mass (m2) = 1.99x10^30 kg.

We must first determine our mass using the formula Fg = mg.

Given force (F) = 715N

mass (m) = 715/9.8 =  72.95kg

The distance from a planet's centre to its surface, as determined by the force of gravity equation, is equal to r.

Since the issue gives us the diameter, we can utilise it to find the radius, which is given by r = d/2.

Given distance (d) = 20km

r = 20/2 = 10km = 10000m

Now, using the formula for the force of gravity, we can determine our weight at the star: F = Gm1m2/r^2

G = gravitational constant = 6.67×10-11 N* m^2/kg^2

F = (6.67×10-11)x72.95x 1.99×10^30/10^8 = 9.68x10^13N

Hence the weight on surface of neutron star = 9.68x10^13N

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The substances that are important to life on Earth  on are in the forms of solid, liquid and gas and they are combined with one another by different physical and chemical reactions.

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c>a>f=e>b>d is the rank from largest to smallest, which scenarios on basis of the angle of interference.

single-slit slide with width 0.01 mm> diffraction grating with 1000 lines/cm> double-slit slide with spacing 0.02 mm=diffraction grating with 500 lines/cm> single-slit slide with width 0.04 mm> double-slit slide with spacing 0.08 mm

Monochromatic lights are called single-wavelength light, where mono refers to a single, and chroma means color which is essentially electromagnetic radiation derived from photon emissions from atoms.

A sodium lamp is one example of monochromatic light. There are some applications like:

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The acceleration of the block is 4.16 m/s^2 when the block slides down on inclined plane at an angle 45 degrees.

Acceleration is a dynamic characteristic of an object. Acceleration (a) can be obtained via inertial force (F) on a mass (m) subjected to acceleration (a) of the moving object:

F = - m a

The inertial force can be measured either through strain (if deformation is min) or through the deformation of elastic element. Equal increments of speed gained in equal time intervals,Distance increases in each time interval.The unit of acceleration in the System International: m/s2.

Acceleration shows how fast an object changes its velocity. If an object is moving with constant velocity, acceleration is equal zero: Positive and Negative Acceleration.Positive acceleration: velocity increases with the time and Negative acceleration: velocity decreases with the time.

Given,

m=3kg

angle = 45 degrees

Uk= 0.2

by the equation ,

mgsinФ-f-T=ma

3* 9.81*sin45-UkN-T=3a----(1)

T=a----(2)

Now,N= mgcos45

          = 3*9.81/sqrt2

solve eqn (1) and put value in (2)

a=4.16 m /s^2

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a)The stone fall so that the pulley has 4.50 j of kinetic energy is  0.673 m

b) Percent of the total kinetic energy does the pulley have 45.5%

Given, the mass of the pulley disk mₚ = 2.50 kg

Its radius r = 20.0 cm = 0.2 m

Its moment of inertia I = (1/2)mₚr²

                               I = (1/2)(2.50 kg)(0.2 m)²

                               I = 0.05 kgm²

Given, the kinetic energy of the pulley disk = 4.50 J

If w the angular velocity of the pulley,

                              (1/2)Iw² = 4.50 J

                              (1/2)(0.05 kgm²)w²= 4.50 J

                              w² = 9/0.05 rad²/s²

                              w² =180 rad²/s²

                               v² = 180(0.2)^2 m²/s²

                              v² = 7.2 m²/s²

Given, the mass of the stone mₛ = 1.50 kg

Its kinetic energy = (1/2)(1.50 kg)(7.2 m²/s²) = 5.4 J

Potential energy loss of the stone = (1.50 kg)(9.80 m/s²)h

(a) Therefore, from equation (1), w get

     (1.50 kg)(9.80 m/s²)h = 4.5 J + 5.4 J

                                    h = (9.9 J)/(1.50 kg)(9.80 m/s²)

                                    h = 0.673 m

So, the. stone. mus.t fall by a distan.ce of  0.673 m.

(b) Total kinetic. energy of the. system. = 9.9 J

The kinetic energy. of the pulley = 4.50 J

The pulley’s fraction. of kinetic energy. = 4.50 J/9.9 J

                                                          = 0.455 = 45.5%

Therefore, the pulley has 45.5% of the total kinetic energy.

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The amount of work done to push the protons is 7.67 X 10⁻¹⁴ J. If the protons are both released from rest at the closer distance, then the speed at which they reach their original separation is 6.77 X 10⁶ m/s.

Initial distance = d = 2 X 10⁻¹⁰ m

Final distance = d' = 3 X 10⁻¹⁵ m

Mass of both protons = 1.6 X 10⁻²⁷ Kg each

Charge of both protons = q = 1.6 X 10⁻¹⁹ C each

The constant for proton energy = 1/(4πε) = 9 X 10⁹ Nm²/C²

The potential change changes when their is a change in the work done, thus, potential energy initially = U =

= U = 1/(4πε) X (q₁q₂ / d)

= U = ( 9 X 10⁹ X 1.6 X 10⁻¹⁹  X 1.6 X 10⁻¹⁹ ) / 2 X 10⁻¹⁰

= U = 11.52 X 10⁻¹⁹ J

Now, finial potential energy = U' =

= U' = 1/(4πε) X (q₁q₂ / d')

= U' = ( 9 X 10⁹ X 1.6 X 10⁻¹⁹  X 1.6 X 10⁻¹⁹ ) / 3 X 10⁻¹⁵

= U' = 7.68 X 10⁻¹⁴ J

Thus, the work done to push the two protons is the different between final potential energy and the initial potential energy = W =

= W = U' - U

= W =11.52 X 10⁻¹⁹ - 7.68 X 10⁻¹⁴

= W = 7.67 X 10⁻¹⁴ J

Now, the speed of protons while coming back to the original positions = V;

Work done is also equal to the sum of kinetic energy's, hence,

= W = KE + KE'

= W = (1/2)mv² + (1/2)mv'²

= W = 2 X (1/2)mV²

= V = √ (W / m)

= V = √ (7.67 X 10⁻¹⁴ / 1.6 X 10⁻²⁷)

= V = 6.77 X 10⁶ m/s

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If indeed the magnetism is removed more quickly, the magnitude of the induced electromagnet force increases. incorrect If the magnetic field is removed more quickly, the amplitude of the generated electromotive force is reduced.

huge or highly significant: An average man would have given up due of the task's size. earth science The brilliance of a sun as seen from Earth is also measured by magnitude.

The strength of a force is expressed numerically as its magnitude. Now let us say the force equals 10 N and it is coming from the east. "towards east" indicates the force's direction, while "10" indicates its strength. In essence, magnitude is the "value" or "quantity" of any numerical value.

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The velocity function is v(t)=1/2 t^2+4t + 3

The acceleration function is given by a(t) = t + 4 where v(0) = 3. The acceleration function a(t) is a known function of time. The velocity function is integral of the acceleration function plus a constant of integration. The time derivative of the velocity function is acceleration, hence d/dt  v(t)=a(t)

Taking indefinite integral of both sides,

∫d/dt  v(t)dt=∫a(t)dt+C1

Where C1 is the constant of integration. Since

∫d/dt  v(t)dt=v(t)

Hence,

v(t)=∫a(t)dt + C1

Inserting the acceleration function, a(t) = t + 4

v(t)=∫(t +4)dt + C1

v(t)=1/2 t^2+4t + C1

In order to find C1, let v = 3 when t = 0 as given

3=1/2 (0)^2+4(0) + C1

3=0 + C1

C1=3

Hence,

v(t)=1/2 t^2+4t + 3

Note: The question is incomplete. The complete question probably is: The acceleration function (in m/s-) and the initial velocity are given for a particle moving along a line a(t) = t + 4, v(0) = 3, 0 ≤ t ≤ 11 (a) Find the velocity at time t.

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The order of events that that describes effects of light exposure on the retina are;

Light acts directly on the retina in mammals to perform two important functions; visual function via rod and cone photoreceptor cells and non-image forming tasks such as circadian rhythm synchronization to a 24 hours solar cycle, pineal melatonin suppression, and pupil light reflexes.

This process takes place first in the rhodopsin and trans-retina isomerizes from cis-retinal. Opsin initiates a signaling cascade in rod cell outer segments where rod cell glutamate release is decreased. Changes in glutamate release are detected by bipolar cells and lastly action potentials are emitted by ganglion cells.

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A 300 kw load has a trailing power factor of 0.65. The reactive power required in order to improve the power factor to unity is 346.12kVAR

Given Power of load (P) = 300KW

power factor = 0.65

We know that kVA = KW/cosФ

kVAR = kVA sinФ

sin = √1-cos^2Ф

here kVA is apparent power(S)

KW is active power(P)

kVAR is reactive power(Q)

cosФ = power factor

S = P/cosФ = 300/0.65 = 461.5kVA

sinФ  = √1-(0.65)^2 = 0.75

Now Q = (461.5)(0.75) = 346.12kVAR lagging

the additional load is capacitive

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Sort the scenarios based on how long it takes the block n to reach equilibrium T[tex]_{c}[/tex]>T[tex]_{A}[/tex]>T[tex]_{B}[/tex]=T[tex]_{D}[/tex].

F[tex]_{net}[/tex]=kx₁=kx₂

x₁+x₂=A

[tex]\frac{F}{k}[/tex]+[tex]\frac{F}{k}[/tex]=A

A=[tex]\frac{2F}{k}[/tex]

F[tex]_{net}[/tex]=[tex]\frac{kA}{2}[/tex]

point of connecting spring displaced by A/2

Ranking ⇒[(case-B)=(case-D)>(case-A)>(case-C)]

Part-(b)

Now, time period T=2π[tex]\sqrt{\frac{m}{k_{eq} } }[/tex]

case-A [tex]k_{q}[/tex]=k,

case-B [tex]k_{eq}[/tex]=2k

case-C [tex]k_{eq}[/tex]=k/2

case-D [tex]k_{eq}[/tex]=2k

so,T[tex]_{c}[/tex]>T[tex]_{A}[/tex]>T[tex]_{B}[/tex]=T[tex]_{D}[/tex]

A process is in equilibrium when the observable properties, such as color, temperature, pressure, concentration, etc., do not change.

Equilibrium, which literally means "balance," denotes that the reactants and products involved in a chemical process are in balance with one another. The equilibrium condition can also be seen in some physical processes, such as the melting point of ice at 0 degrees Celsius, when both ice and water are present.

The equilibrium is known as a physical equilibrium in the case of physical processes like the melting of solids, the dissolution of salt in water, etc., whereas the equilibrium connected to chemical reactions is known as a chemical equilibrium.

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The gauge pressure at the heart of the Argentinosaurus is 1000torr and its feet is 1722 torr.

(a)Heart of the Argentinosaurus has gauge pressure of

pheart = pbrain + ρgh = 80 + 1.06 × [tex]10^{3}[/tex] * 9.8 (21 - 9) = 1000torr

b) The Argentinosaurus's foot have a pressure gauge that reads

Pfeet = pbrain +  ρgh = 80 + 1642 = 1722 torr

When the atmospheric pressure is above a certain level, the gauge pressure is positive; when it is below that level, it is negative. In actuality, air pressure raises the pressure in any fluid that isn't contained by a hard container. Pascal's principle is the reason why this occurs. Pg denotes gauge pressure, which is related to absolute pressure in the manner shown below: Where pa is the local atmospheric pressure, the equation is pg = p - pa. Example: Tire pressure of 32.0 psi is determined using an automobile tire gauge. There are 14.2 psi in the nearby atmosphere. Gauge pressure and atmospheric pressure are added to determine the total pressure, or absolute pressure: Absolute pressure (Pabs), gauge pressure (Pg), and atmospheric pressure (Patm) are all expressed as Pabs=Pg+Patm.

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