In a PA projection of the chest being used for cardiac evaluation, the heart measures 15.2 cm between its widest points. If the magnification factor is known to be 1.3, what is the actual diameter of the heart?

A. 9.7 cm
B. 11.7 cm
C. 19.7 cm
D. 20.3 cm

Answer:52.72 L

Explanation:

Given

Energy absorb by system [tex]Q=51 J[/tex]

Piston is working against a pressure of [tex]P=0.545 atm[/tex]

Final volume of system [tex]V_f=55.6 L[/tex]

change in internal Energy [tex]\Delta U=-106 J[/tex]

Work done by system [tex]W=P(V_f-V_i)[/tex]

According to first law of thermodynamics

[tex]Q=dU+PdV[/tex]

[tex]51=-106+W[/tex]

[tex]W=157 J[/tex]

[tex]157=0.545\times 10^5\times 10^{-3}(55.6-V)[/tex]

[tex]157=54.5\times (55.6-V)[/tex]

[tex]55.6-V=2.88[/tex]

[tex]V=52.72 L[/tex]

Answer:

(A) l = 0.39 m      

(B)  l =0.38 m  

(C) l = 0.14 m

Explanation:

Answer:

Explanation:

Answer:

Explanation:

from the question we are given the following values:

mass (m) = 1.6 kg

height (h) = 1.05 m

compression of spring (l) = ?

spring constant (k) = 330 N/m

acceleration due to gravity (g) = 9.8 m/s^{2}

(A) initial potential energy of the object = final potential energy of the spring

         potential energy of the object = mg(1.05 + l)  

         potential energy of the spring = 0.5 x k x l^{2}  (k= spring constant)

 therefore we now have

              mg(1.05 + l)  = 0.5 x k x l^{2}

              1.6 x 9.8 x (1.05 + l)  = 0.5 x 300 x l^{2}

               15.68 (1.05 + l) = 150 x l^{2}

                   16.5 + 15.68l = 150l^{2}

l = 0.39 m        

(B)   with constant air resistance the equation applied in part A above becomes

initial P.E of the object - air resistance = final P.E of the spring

mg(1.05 + l) - 0.750(1.05 + l) = 0.5 x k x l^{2}        

     1.6 x 9.8 x (1.05 + l) - 0.750(1.05 + l)  = 0.5 x 300 x l^{2}

         (16.5 + 15.68l) - (0.788 + 0.75l) = 150l^{2}        

          16.5 + 15.68l - 0.788 - 0.75l = 150l^{2}

            15.71 + 14.93l = 150^{2}

            l =0.38 m  

(C)   where g = 1.63 m/s^{2} and neglecting air resistance

      the equation mg(1.05 + l)  = 0.5 x k x l^{2} now becomes

        1.6 x 1.63 x (1.05 + l)  = 0.5 x 300 x l^{2}

        2.608 (1.05 +l) = 0.5 x 300 x l^{2}

        2.74 + 2.608l = 150 x l^{2}

l = 0.14 m

The compression of the spring when it is dropped from 1.05 m is 0.37 m.

The compression of the spring when air resistance is considered 0.36 m.

The compression of the spring when air resistance is neglected and gravity is 1.63 is 0.14 m.

The given parameters;

The compression of the spring is determined by applying the principle of conservation of energy;

[tex]\frac{1}{2} kx^2 = mgh\\\\\frac{1}{2} kx^2 = mg(1.05 + x)\\\\kx^2 = 2mg(1.05 + x)\\\\330x^2 = 2\times 1.6 \times 9.8(1.05 + x)\\\\330x^2 = 32.93 + 31.36x\\\\330x^2 - 31.36x - 32.93 = 0\\\\a = 330, \ b = -31.36, \ c = -32.93\\\\x = \frac{-b \ \ +/- \ \sqrt{b^2 -4ac} }{2a} \\\\x = \frac{-(-31.36) \ \ +/- \ \sqrt{(-31.36)^2 -4(330\times -32.93)} }{2(330)}\\\\x = 0.37 \ m[/tex]

Considering air resistance, the compression of the spring is calculated as follows;

[tex]\frac{1}{2} kx^2 = mg(1.05+ x) - F(1.05+ x)\\\\\frac{1}{2} \times 330 x^2 = 1.6\times 9.8(1.05 + x) - 0.75(1.05 + x)\\\\165x^2 = 16.46 + 15.68x - 0.79 - 0.75x \\\\165x^2 -14.93x - 15.67 = 0\\\\a = 165, \ \ b = -14.93 \ \ c = -15.67 = 0\\\\x = \frac{-b \ \ +/- \ \sqrt{b^2 - 4ac} }{2a} \\\\x = \frac{-(-14.93) \ \ +/- \ \sqrt{(-14.93)^2 - 4(15.67)} }{2(165)}\\\\x = 0.36 \ m[/tex]

The compression of the spring when air resistance is neglected and gravity is 1.63;

[tex]\frac{1}{2} kx^2 = mg(1.05 + x)\\\\\frac{1}{2} \times 330 x^2 = 1.6 \times 1.63(1.05 + x)\\\\165 x^2 = 2.74 + 2.61x \\\\165 x^2 - 2.61x- 2.74 = 0\\\\a = 165, \ \ b = -2.61, \ c = \ -2.74\\\\x = \frac{-b \ \ +/- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\x = \frac{-(-2.61) \ \ +/- \ \ \sqrt{(-2.61) ^2 - 4(165\times -2.74)} }{2(165)}\\\\x = 0.14 \ m[/tex]

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Io experiences tidal heating primarily because of the intense gravitational pull of Jupiter which creates friction inside Io and ultimately results in heating. This process is enhanced by Io's elliptical orbit.

Io, one of Jupiter's moons, experiences tidal heating primarily due to the intense gravitational pull of Jupiter. These gravitational forces create friction within Io, resulting in heat. This process is also facilitated by Io's elliptical orbit which varies the gravitational forces acting on it during its orbit around Jupiter, thereby enhancing the frictional heating. Therefore, the main contributor to Io's tidal heating is its interaction with Jupiter's gravitational field.

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Answer:

0.06683

Explanation:

m = Mass of block = 3.85 kg

g = Acceleration due to gravity = 9.81 m/s²

[tex]\mu[/tex] = Coefficient of kinetic friction

x = Compression of spring = 0.103 m

k = Spring constant = 138 N/m

Work done against friction is given by

[tex]W=m\mu gs\\\Rightarrow W=\mu 3.85\times 9.81\times 0.29\\\Rightarrow W=10.952865\mu[/tex]

The potential energy of the spring is given by

[tex]P=\frac{1}{2}kx^2\\\Rightarrow P=\frac{1}{2}\times 138\times 0.103^2\\\Rightarrow P=0.732021\ J[/tex]

The potential energy and the work done against friction will balance the system

[tex]0.732021=10.952865\mu\\\Rightarrow \mu=\frac{0.732021}{10.952865}\\\Rightarrow \mu=0.06683[/tex]

The coefficient of kinetic friction between the horizontal surface and the block is 0.06683

Answer:

the person will be in the shore at 10.73 minutes after launch the shoe.

Explanation:

For this we will use the law of the lineal momentum.

[tex]L_i = L_f[/tex]

Also,

L = MV

where M is de mass and V the velocity.

replacing,

[tex]M_i V_i = M_{fp}V_{fp} + M_{fz}V_{fz}[/tex]

wher Mi y Vi are the initial mass and velocity, Mfp y Vfp are the final mass and velocity of the person and Mfz y Vfz are the final mass and velocity of the shoe.

so, we will take the direction where be launched the shoe as negative. then:

(70)(0) = (70-0.175)([tex]V_fp[/tex]) + (0.175)(-3.2m/s)

solving for [tex]V_fp[/tex],

[tex]V_fp[/tex] = [tex]\frac{(3.2)(0.175)}{69.825}[/tex]

[tex]V_fp[/tex] = 0.008m/s

for know when the person will be in the shore we will use the rule of three as:

1 second -------------- 0.008m

t seconds-------------- 5.15m

solving for t,

t = 5.15m/0.008m

t = 643.75 seconds = 10.73 minutes

The time to reach the shore is 643.75 seconds.

The problem presented involves conservation of momentum on a frictionless surface, which is a physics concept. To solve for the time it takes the person to reach the shore, we can use the principle that the momentum before throwing the shoe is equal to the momentum after throwing the shoe, as there are no external forces acting on the system (since friction is ignored).

The initial momentum of the system is zero because the person is not moving. After throwing the shoe, the momentum of the shoe can be calculated using the formula: p = m × v, where p is momentum, m is mass, and v is velocity. The momentum of the shoe is 0.175 kg × 3.2 m/s, which must be equal and opposite to the momentum of the person. Thus, the velocity of the person, v_p, can be found through the equation m_shoe × v_shoe = m_person × v_person.

Once the velocity of the person is calculated, the time, t, taken to cover the distance to the shore can be found using the formula: t = d / v_p, where d is the distance to the shore.

Carrying out the calculations gives us:

p_shoe = 0.175 kg × 3.2 m/s = 0.56 kg m/s

v_person = p_shoe / m_person = 0.56 kg m/s / 70 kg = 0.008 m/s

t = d / v_person = 5.15 m / 0.008 m/s = 643.75 seconds

The person will take approximately 643.75 seconds to reach the shore.

Answer:

Option B

Explanation:

Maintenance-free battery uses Recom­bination Electrolyte in order to limit   the oxygen and hydrogen formation during charging.

Each plate of the battery uses a glass micro-fiber separator that absorbs the whole liquid electrolyte in its pore thus resulting in no free acid.

As the battery charges to its fullest, the oxygen produced at the positive plate moves via the pores of the separator to the negative plate.

Initially it reacts and forms [tex]PbSO_{4}[/tex], i.e., Lead Sulphate, then on further charging it gets converted to Pb, i.e., lead.

Due to this, the negative plate is not able to reach the correct potential in order to liberate hydrogen, and hence no water formation takes place.

Thus technician B is correct.

Answer:

The work done by the man is 260 J.

(E) is correct option.

Explanation:

Given that,

Force = 80 N

Distance = 5.0 m

Angle = 30°

Acceleration = 1.5 m/s²

Let F be the man's force

We need to calculate the force

Using balance equation

[tex]F-mg\sin\theta=ma[/tex]

[tex]F-80\sin30=\dfrac{80}{9.8}\times1.5[/tex]

[tex]F=\dfrac{80}{9.8}\times1.5+80\sin30[/tex]

[tex]F=52\ N[/tex]

We need to calculate the work done by the man

Using formula of work done

[tex]W=F\dotc d[/tex]

[tex]W=52\times 5.0[/tex]

[tex]W=260\ J[/tex]

Hence, The work done by the man is 260 J.

If the speed of the crate decreases at a rate of 1.5 m/s2, then the work done by the man is E) 260 J

Work is the quantity of energy transferred by displacement. The following is formula of work done

[tex]\Delta KE = F*d[/tex]

Whereas the following is formula for  balance equation:

[tex]F-mg\sin\theta=ma[/tex]

The equation above is come from Newton's Second Law.  It applies to a wide range of physical phenomena, but it is not a fundamental principle like the Conservation Laws and also it applies only if the force is the net external force.

For the Case of fictionless slope, the speed at the bottom of a frictionless incline does not depend upon the angle of the incline

Then the work done by the man is...

[tex]w = mgd* sin\theta + \Delta KE\\w = 80 N *5.0 m *sin 30 + (F*d)\\w = 80 N *5.0 m *\frac{1}{2}  + (F*d)\\w = 40 N *5.0 m+ (F*d)\\w = 40 N *5.0 m+ (m*a*d)\\w = 200 Nm+ (8*1.5*5)\\w = 200 Nm+ 60\\w = 260 J\\[/tex]

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We can calculate the kinetic energy of the ejected electrons in the photoelectric effect by applying the equation KE = hf - BE, where 'KE' is kinetic energy, 'h' is Planck's constant, 'f' is the frequency of the light, and 'BE' is the binding energy or work function of a metal (in this case, titanium).

The photoelectric effect is the process where light of sufficient energy (in this case, frequency) shone on a metal surface can release electrons (photoelectrons). Kinetic energy of these ejected electrons can be calculated according to the equation: KE = hf - BE. Here, 'KE' is the kinetic energy we're striving to find, 'h' is Planck's constant, 'f' is the frequency of the incident light, and 'BE' is the binding energy or work function of the electron.

For titanium metal, you've provided that the work function (BE) is 6.93 × 10⁻¹⁹ J. The frequency ('f') of the light used is 1.216 × 10¹⁵ s⁻¹. And, Planck's constant ('h') approximately equals 6.63 × 10⁻³⁴ Js.

To execute the calculation: KE = (6.63 × 10⁻³⁴ Js x 1.216 × 10¹⁵ s⁻¹) - 6.93 × 10⁻¹⁹ J. The solution will provide your kinetic energy in Joules

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Final answer:

The speed of an object of mass m when it hits the planet's surface after being dropped from height h is found by using the conservation of energy, yielding the formula v = √[2GM(1/R - 1/(R+h))] where G is the gravitational constant, M the planet's mass, and R its radius.

Explanation:

To find the speed v of an object of mass m when it hits the surface of a planet, we will use the principles of energy conservation in the context of gravitational fields. The total mechanical energy (potential plus kinetic) at the beginning and at the end must be equal since we are ignoring air resistance and any other non-conservative forces.

The initial potential energy when the object is at height h above the planet is given by the gravitational potential energy formula U = -G(Mm)/(R+h) and the initial kinetic energy is zero as the object is initially at rest. When the object reaches the surface of the planet, its potential energy is U = -G(Mm)/R and its kinetic energy is K = (1/2)m*v^2. Conservation of energy dictates that the initial total energy equals the final total energy:

-G(Mm)/(R+h) = -G(Mm)/R + (1/2)m*v^2

Solving for v, the speed of the object at the planetary surface, we get:

v = √[2GM(1/R - 1/(R+h))]

This expression shows that the object's speed increases with a greater initial height h, a more massive planet M, and decreases with a larger planetary radius R.

Answer:

Wavelength, [tex]\lambda=0.011\ m[/tex]

Explanation:

Given that,

Number of cycles in a spiral spring is 2.91 in every 3.67 s

The velocity of the pulse in the spring is 0.925 cm/s, v = 0.00925 m/s

To find,

Wavelength

Solution,

Number of cycles per unit time is called frequency of a wave. The frequency of the longitudinal pulse is,

[tex]f=\dfrac{2.91}{3.67}=0.79\ Hz[/tex]

The wavelength of a wave is given by :

[tex]\lambda=\dfrac{v}{f}[/tex]

[tex]\lambda=\dfrac{0.00925\ m/s}{0.79\ Hz}[/tex]

[tex]\lambda=0.011\ m[/tex]

So, the wavelength of the longitudinal pulse is 0.011 meters. Hence, this is the required solution.

To find the wavelength, the frequency is calculated by dividing the number of cycles by the total time. The velocity is then converted from cm/s to m/s. The wavelength is calculated using these values in the equation for wave speed to get approximately 0.012 meters.

In the scenario described in your question, your hand completes 2.91 back-and-forth cycles every 3.67 seconds, resulting in longitudinal pulses in a spiral spring. Here, we have frequency and wave velocity, and we need to find the wavelength. We can determine the frequency by dividing the number of cycles by the total time and the wavelength using the formula for wave speed: v = fλ, where v is the velocity, f is the frequency, and λ is the wavelength.

Frequency (f) = number of cycles / total time = 2.91 cycles / 3.67 s = 0.793 Hz

Velocity (v) = 0.925 cm/s = 0.00925 m/s (since we need the answer in meters)

To find the wavelength, we can use the formula v = fλ, to rearrange this to solve for λ, we get λ = v/f. So, λ = 0.00925 m/s / 0.793 Hz = 0.01166 meters or approximately 0.012 meters.

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Explanation:

When a electron is collided with a photon with exactly the same energy it would require to get to any of the farther orbits,electron transition takes place to an orbit depending on the energy of the photon.

When electrons emit exactly the same amount energy that is difference between the current energy level and the new level,then the electron makes a transition to the new level.

An electron in an atom is likely to move from one energy level to another when it gains or loses energy.

Explanation:

The change in an electron’s position with respect to energy levels is termed as Atomic electron transition. In spite of having similar charge and mass, the energy level of any electron in an atom, surrounding the nucleus, differs depending on its orbital position from the nucleus.

Electrons positioned nearest to the atomic nucleus carry least energy. To move an electron from its original ground state energy level to a higher level; energize the atom and it will excite the electrons thus making it move from its lower energy stable state to an unstable state with higher energy level. Releasing energy of an atom decreases the energy level of its excited electrons, de-energizing it and thus stabilizing the atom.

Answer:

0.882 m/s²

Explanation:

m = Mass of car = 643 kg

a = Acceleration of the car = 2 m/s²

[tex]a_2[/tex] = Acceleration of the truck

Force exerted on the car

[tex]F=ma\\\Rightarrow F=643\times 2\\\Rightarrow F=1286\ N[/tex]

Mass of the truck with the woman = [tex]M=1415+43=1458\ kg[/tex]

From Newton's second law

[tex]F=Ma_2\\\Rightarrow a_2=\frac{1286}{1458}\\\Rightarrow a=0.882\ m/s^2[/tex]

The acceleration of the truck is 0.882 m/s²

Final answer:

The acceleration of the truck can be found by using Newton's second and third laws. Equal and opposite forces act on the car and the truck, and with the mass of each known, calculations can provide the truck's acceleration.

Explanation:

To find the acceleration of the truck when the woman pushes the car, we use Newton's third law of motion, which states that for every action, there is an equal and opposite reaction. When the woman pushes the car with a certain force, the car pushes back on the woman (and hence the truck she is sitting on) with an equal and opposite force. This means the force causing the car to accelerate is the same force causing the truck to accelerate, but in the opposite direction.

Let's denote F as the force with which the woman pushes the car. Given the mass of the car (mcar = 643 kg) and its acceleration (acar = 2 m/s²), we can find the force using Newton's second law of motion: F = mcar × acar.

The same force is responsible for the acceleration of the truck in the opposite direction. Therefore, for the truck with mass mtruck = 1415 kg and acceleration atruck, Newton's second law gives us the relationship: F = mtruck × atruck. By substituting the force F from the previous calculation, we can solve for atruck, which is the acceleration of the truck.

The given parameters;

The potential energy of the hydrogen atoms is calculated as follows;

[tex]V = \frac{kq}{r}[/tex]

where;

The potential energy when the distance = 0.74 x 10⁻¹⁸ m

[tex]V = \frac{9\times 10^{9} \times 1.602\times 10^{-19}}{0.74\times 10^{-18}} = 1.95\times 10^9 \ N.m[/tex]

The potential energy when the distance = 0.5 x 10⁻¹⁸ m

[tex]V = \frac{9\times 10^{9} \times 1.602\times 10^{-19}}{0.5\times 10^{-18}} = 2.88 \times 10^9 \ N.m[/tex]

Thus, we can conclude that the following;

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When hydrogen atoms approach each other, their potential energy decreases and they begin to form a bond. At an optimal distance of 0.74 å, a stable covalent bond is formed. The potential energy being zero or less does not signify a bond formation.

When two hydrogen atoms are far apart, the potential energy indeed approaches zero because there is no overlap of their atomic orbitals. As they approach each other, their atomic orbitals start overlapping, the energy of the system decreases, initiating the process of bond formation. When the distance between the two hydrogen atoms reaches 0.74 å (Angstroms), their energy is at its lowest point which signifies the formation of a covalent bond. This is because at this optimal distance, both repulsive and attractive forces combine to achieve the lowest possible energy configuration, leading to a stable covalent bond.

However, if the distance were to reduce to 0.50 å, the atoms would be too close and the energy of the system would increase due to stronger nuclear repulsions and electron confinement. In essence, a scenario where the potential energy is either less than or equals to zero does not correlate to the formation of a covalent bond. Instead, it's the optimal balance of forces at a specific distance that gives rise to a stable covalent bond in hydrogen atoms.

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Wind-driven currents are found NEAR THE SURFACE of the oceans

Explanation:

Surface currents are at the interphase between the hydrosphere and atmosphere. Therefore the feel the greatest effect of drag by wind currents, especially prevailing winds (that blow predominantly in one direction like westerlies and easterlies) within the lower atmosphere.  The deep currents, on the other hand, are more influenced by Coriolis effect of the earth’s rotation. It is these differences in influences of surface and deeper currents that cause Ekman transport in oceans.

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Answer:

(a) A = 0.650 m

(b) f = 1.3368 Hz

(c) E = 17.1416 J

(d)  K = 11.8835 J

     U = 5.2581 J

Explanation:

Given

m = 1.15 kg

x = 0.650 cos (8.40t)

(a) the amplitude,

A = 0.650 m

(b) the frequency,

if we know that

ω = 2πf = 8.40    ⇒   f = 8.40 / (2π)

⇒   f = 1.3368 Hz

(c) the total energy,

we use the formula

E = m*ω²*A² / 2

⇒  E = (1.15)(8.40)²(0.650)² / 2

⇒  E = 17.1416 J

(d) the kinetic energy and potential energy when x = 0.360 m.

We use the formulas

K = (1/2)*m*ω²*(A² - x²)       (the kinetic energy)

and

U = (1/2)*m*ω²*x²              (the potential energy)

then

K = (1/2)*(1.15)*(8.40)²*((0.650)² - (0.360)²)

⇒  K = 11.8835 J

U = (1/2)*(1.15)*(8.40)²*(0.360)²

⇒  U = 5.2581 J

The amplitude (A) of the oscillation is 0.650 m, and the frequency (f) is approximately 1.34 Hz. To determine the total energy and the kinetic and potential energy at the point when x = 0.360 m, the spring constant is needed.

In order to answer your question about a 1.15-kg mass oscillating according to the equation x = 0.650 cos(8.40t), we can use the facts about simple harmonic motion.

(a) The amplitude is the coefficient of the cosine function, which is 0.650 m.

(b) The frequency (f) can be determined from the coefficient of the t inside the cosine function. In simple harmonic motion, the angular frequency (ω) equals 2πf. Therefore f = ω / 2π = 8.40 / 2π = 1.34 Hz.

(c) The total energy (E) of an oscillator is given by the formula E = 1/2 kA². However, without the spring constant (k) given in the question, this cannot be determined.

(d) When x = 0.360 m, the potential energy (PE) is given by PE = 1/2 kx², and the kinetic energy (KE) can be calculated by subtracting the potential energy from the total energy. But again, without the spring constant, these cannot be determined.

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Answer:

0.027m

Explanation:

the bolt loses contact with the piston only when acceleration due to gravity equals acceleration of piston

ω² * A = g where ω is angular velocity, A amplitude, g acceleration due to gravity

ω is given by 2πf, ω² is 4π²f²

A= g/4π²f² depending on the value of g used either 10m/s² or 9.8m/s²,

i used 10m/s² in this answer

Answer:

 W = 24.28 kN

Explanation:

given,

Mass of satellite = 5850 Kg

height , h = 4.1 x 10⁵ m

Radius of planet = 4.15 x 10⁶ m

Time period = 2 h

                    = 2 x 3600 = 7200 s

Time period of satellite

[tex]T = \dfrac{2\pi}{R}\sqrt{\dfrac{(R+h)^3}{g}}[/tex]

R is the radius of planet

h is the height of satellite

[tex]T^2 = \dfrac{4\pi^2}{R^2}\ {\dfrac{(R+h)^3}{g}}[/tex]

now calculation of acceleration due to gravity

[tex]g = \dfrac{4\pi^2}{R^2}\ {\dfrac{(R+h)^3}{T^2}}[/tex]

[tex]g = \dfrac{4\pi^2}{(4.15\times 10^6)^2}\ {\dfrac{(4.15\times 10^6+4.1\times 10^5)^3}{(7200)^2}}[/tex]

g = 4.15 m/s²

True weight of satellite

W = m g

W = 5850 x 4.15

W = 24277.5 N

 W = 24.28 kN

True weight of the satellite is   W = 24.28 kN

The true weight of the satellite, when the satellite is at rest on the surface of the planet, is 24.28 kN.

Time period of satellites is the total time taken by a satellite to complete a full orbit around a body. It can be given as,

[tex]T=\dfrac{2\pi}{R}\sqrt{\dfrac{(R+h)^3}{g}}[/tex]

Here, (R) is the radius of the body, and (g) is the gravitational acceleration force.

In a circular orbit 4.1 x10 to the 5th power m above the surface of a planet. The period of the orbit is 2 hours and the radius of the planet is 4.15 x 10 to the 6th power m.

To find the weight of the satellite, first find the value of gravitation acceleration using the time period formula as,

[tex]2=\dfrac{4\pi^2}{4.15\times10^6}\sqrt{\dfrac{(4.15\times10^6+4.1\times10^5)^3}{g}}\\g=4.15\rm m/s^2[/tex]

The weight of the body is mass time gravity. As the satellite has a mass of 5850 kg and value of g is 4.15 m/s². Thus, the weight of it is,

[tex]W=5850\times4.15\\W=24277.5\rm N\\W=24.28\rm \; kN[/tex]

Thus, the true weight of the satellite, when the satellite is at rest on the surface of the planet, is 24.28 kN.

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Answer:

Angle of reflection  at plane surface of prism = 30°

Explanation:

Angle of reflection = angle of incidence at any surface plane or curved .

Here angle of incidence at plane surface of prism = 30°

Angle of reflection  at plane surface of prism = 30°

If i be the angle of incidence and r be angle of refraction

sin i / sin r = refractive index

= 1.52

sin r = sin i / 1.52

=  sin 30 / 1.52

= .5 / 1.52

= .3289

r = 19.2 °

Answer:

Three times

Explanation:

The amount of heat added to a substance when its temperature is increased from T_1 to T_2 is given by

[tex]Q=mc(T_2-T_1)[/tex]

c= specific heat capacity

m= mass of the substance

⇒Q∝c

that is if c is increased three times the amount of heat added is also increased three times. Therefore, the amount of heat added to the copper is three times the heat added to lead.

Answer:

option C

Explanation:

The correct answer is option C

A light that transmits through n₂ travels t distance before reflection off the n₁ medium and again travels distance t before reaching the point from where it entered n₂  medium. Hence it travels 2 t distance more than the light that is reflected off n₂.

It( light entering n₂) also travels an additional distance equal to, half of the wavelength, when reflected off n₁ ( as n₁ is greater than n₂).  

Wavelength in n₂ is = [tex]\dfrac{\lambda}{n_2}[/tex]  

Hence, path length difference = [tex]2t +\dfrac{\lambda}{2 n_2}[/tex]

The effective path length difference between the light that reflected off of the n2 medium and the light that reflected off the n1 medium is

Path length difference = 2t+λ/(2n2)

Generally, Wavelength is simply defined as the distance between points in the adjacent cycles of a waveform signal channeled through a space.

In conclusion, A light that travels through n2, travels in a time t, a distance before reflection from the n1 channel,  travels a distance d before reaching the point from where it entered n2  channel.

Path length difference = 2t+λ/(2n2)

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Answer:

Air found in the soil is trapped in it and is not exposed to the air currents, poor  in oxygen and rich in moisture.

Explanation:

Air found in the soil is trapped in it and is not exposed to the air currents.

It contains more moisture than the atmospheric air.

It is rich in carbon dioxide and poor in oxygen.

The air found in the soil forms a part of the lithosphere and not the atmosphere.

Nitrogen trapped in the soil is used by the plants to make protein, carbon dioxide is used for the photosynthesis and oxygen is used for the respiration of roots and microorganisms.

Answer:

One of the most important difference between the two galaxies" is that the sbb galaxy has a lot of material and bar structures passing thourgh the center and the sb galaxy does not have this".

Explanation:

Basically we have 3 types of galaxies (ellipticals, spirals, and irregulars)

Sb galaxies

On this type of galaxies the "intermediate type of spiral typically has a medium-sized nucleus. They can contain stars, star clouds, and interstellar gas and dust. Usually show wide dispersions in terms of their shape. Other members of this subclass have arms that begin tangent to a bright, while still others reveal a small, bright spiral pattern inset into the nuclear bulge. In any of these cases, the spiral arms may be set at different pitch angles. is important to mention that a pitch angle is defined as the angle between an arm and a circle centred on the nucleus and intersecting the arm".

SB galaxies

"For these types of galaxies the luminosities, dimensions, spectra, and distributions of the barred spirals tend to be indistinguishable from those of normal spirals".

One special case occuers at the SBb systems, that "have a smooth bar as well as relatively smooth and continuous arms. In some galaxies of this type, the arms start at or near the ends of the bar, with conspicuous dust lanes along the inside of the bar that can be traced right up to the nucleus"

One of the most important difference between the two galaxies" is that the sbb galaxy has a lot of material and bar structures passing thourgh the center and the sb galaxy does not have this".

The centripetal acceleration is [tex]0.63 m/s^2[/tex]

Explanation:

The centripetal acceleration for an object in circular motion is given by

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

where

v is the linear speed

r is the radius of the circle

The body in the problem cover half of revolution in

t = 10 s

And the corresponding linear distance covered is

L = 10 m

which corresponds to half of the circumference, so

[tex]L=2\pi r[/tex]

From this equation we find the radius of the circle:

[tex]r=\frac{L}{2\pi}=\frac{10}{2\pi}=1.59 m[/tex] [m]

While the linear speed is:

[tex]v=\frac{L}{t}=\frac{10}{10}=1 m/s[/tex]

Therefore, the centripetal acceleration is

[tex]a=\frac{1^2}{1.59}=0.63 m/s^2[/tex]

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Answer:

option D.

Explanation:

The correct answer is option D.

The irony is the figure of speech which represents the contradiction between what is stated and what actually the scenario is.

the statement by the poet does not have any contradictory statements.

The extended metaphor is a part of speech which is used when there is a comparison between two unlike things in the paragraph.

in the given statement of the poem, there is no comparison.

Personification is the part of speech where human quality are given to non-living things.

in the give statement wind in the wire took up the story can be taken as the human quality.

so, the statement part of speech is personification.

Answer:

A. Bohr's model focused on the nucleus

Answer:

D. 1/4 lambda

Explanation:

Given:

Initially out of phase speakers by 180°.

Then on being separated by 1.5 λ.

The initial case of speakers being out of phase by 180°

On separating the speakers by a distance of 1.5 λ.

Answer:

The z-component of the net torque is: 2.58 (N*m)

Explanation:

We need to apply the torque equation ([tex]T=RXF[/tex]), where T is the torque, R is the distance at the point where the force is applied and F is the force, and remembering the cross product ([tex]AXB=abSin(\beta)(n)[/tex]), where A and B are the vectors and ab are the magnitudes, (beta) the angle between A and B in the plane containing them, and (n) is the direction of the vector given by the right-hand rule. Knowing those things, we can get:[tex]T=RXF=4.1(i)+5.4(j)X1.8(i)+3(j)=0+4.1*3(k)+5.4*1.8(-k)+0=2.58(k)(N*m)[/tex] that is the z-component of the torque.

The torque on the object is -2.58Nm

Data;

The net torque is calculated as the product of the force and distance

[tex]t = r * F[/tex]

Let's substitute the values and solve

Since this is a dot product, we can easily multiply through

[tex]\tau = r * F\\\tau = (1.8*5.4) - (3 * 4.1) = -2.58Nm[/tex]

Using 2 by 2 matrix method, the torque on the object is -2.58Nm

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Answer:

[tex]v_r=13.68\ m.s^{-1}[/tex] is the final velocity of the racket.

[tex]F=18.86\ N[/tex]

Explanation:

Given:

By the law of conservation of momentum:

[tex]m_r.u_r+m_b.u_b=m_r.v_r+m_b.v_b[/tex]

where: [tex]v_r=[/tex] final velocity of the racket

[tex]1000\times 15+60\times 18=1000\times v_r+60\times 40[/tex]

[tex]v_r=13.68\ m.s^{-1}[/tex] is the final velocity of the racket.

By the Newton's second law of motion:

[tex]F=\frac{dp}{dt}[/tex] ............................(1)

where:

dp = change in momentum

dt = change in time

Change in momentum of ball:

[tex]\Delta p_b=m_b.v_b-m_b.u_b[/tex]

[tex]\Delta p_b=60\times 10^{-3}\times (40-18)[/tex]

[tex]\Delta p_b=1.32\ kg.m.s^{-1}[/tex]

Now, using eq.(1):

[tex]F=\frac{1.32}{0.07}[/tex]

[tex]F=18.86\ N[/tex]

The tennis racket will be moving at 11.52 m/s immediately after the impact. The average force exerted on the ball by the racket, during the brief 7 ms contact time, is approximately 497 N.

The problem can be solved using the principle of conservation of momentum, which states that the total momentum before the collision is equal to the total momentum after the collision if no external forces act on the system. The formula for momentum is p = mv, where p is momentum, m is mass, and v is velocity.

The momentum before the collision (pi) for both the racket and the ball is:

pi,racket = mracket \\times vi,racket = 1.0 kg \\times 15.0 m/s = 15 kg\\cdot m/s

pi,ball = mball \\times vi,ball = 0.060 kg \\times (-18.0 m/s) = -1.08 kg\\cdot m/s

The negative sign indicats that the ball is moving in the opposite direction to the racket. Total initial momentum Pi,total = pi,racket + pi,ball.

The momentum after the collision (pf) for the ball is:

pf,ball = mball \\times vf,ball = 0.060 kg \\times 40.0 m/s = 2.4 kg\\cdot m/s

Using the conservation of momentum, Pi,total = Pf,total, i.e.,

15 kg\\cdot m/s - 1.08 kg\\cdot m/s = pf,racket + 2.4 kg\\cdot m/s

Solving for pf,racket, the momentum of the racket after collision:

pf,racket = 15 - 1.08 - 2.4 = 11.52 kg\\cdot m/s

The velocity of the racket after the collision (vf,racket) can be found by dividing pf,racket by the mass of the racket:

vf,racket = pf,racket / mracket = 11.52 kg\\cdot m/s / 1.0 kg = 11.52 m/s

To calculate the average force exerted on the ball by the racket, we can use the change in momentum (impulse) and divide it by the time of contact. The change in momentum of the ball (\\Delta p) is the momentum after minus the momentum before:

\\Delta p = pf,ball - pi,ball = 2.4 kg\\cdot m/s - (-1.08 kg\\cdot m/s) = 3.48 kg\\cdot m/s

The average force (F) exerted on the ball is:

F = \\Delta p / \\Delta t = 3.48 kg\\cdot m/s / 0.007 s = 497.14 N

The racket will be moving at 11.52 m/s immediately after the impact, and the average force that the racket exerts on the ball is approximately 497 N.

Answer:

2 seconds

5.74165 seconds

-32 ft/s²

119.73 ft/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 32 ft/s²

[tex]s=-16t^2+64t+160[/tex]

[tex]v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-64}{-32}\\\Rightarrow t=2\ s[/tex]

The maximum height will be reached in 2 seconds

From the given equation

[tex]s=-16(2)^2+64\times 2+160\\\Rightarrow s=224\ ft[/tex]

The maximum height is 224 ft from the ground

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 224=0t+\frac{1}{2}\times 32\times t^2\\\Rightarrow t=\sqrt{\frac{224\times 2}{32}}\\\Rightarrow t=3.74165\ s[/tex]

Time taken to fall from the maximum height is 3.74165 seconds

Time taken from the point in time the ball is thrown is 2+3.74165=5.74165 seconds

[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 32\times 224+0^2}\\\Rightarrow v=119.73\ ft/s[/tex]

Velocity with which the ball will hit the ground is 119.73 ft/s

At t = 1

Acceleration of a body thrown is always i.e., a body in free fall is always 32 ft/s². Here as the ball is thrown up the it will be negative so -32 ft/s²

Answer:

Explanation:

Convergent lens is the lens which converges the rays of light falling on it. It is thicker at the middle and thinner at the edges. It is also known as convex lens.

Divergent lens is the lens which diverge the rays of light falling on it. It is thinner at the middle and thicker at the edges. It is also called as concave lens.

Convergent lenses, or convex lenses, thicken in the middle and focus light rays to a point, forming real or virtual images; divergent lenses, or concave lenses, are thinner in the middle, causing the light rays to spread out and only form virtual images. Snell’s law describes the refraction process in both lens types as being influenced by differences in the index of refraction.

Difference Between Convergent and Divergent Lenses

Lenses are often categorized as either convergent (convex) or divergent (concave) depending on how they refract light. A converging lens is thicker at the middle and causes parallel light rays entering the lens to converge, or come together, at a single point known as the focal point. This lens can create both real and virtual images, the nature of which depends on the position of the object relative to the lens. In contrast, a divergent lens is thinner at the middle and causes parallel light rays to spread apart or diverge after passing through the lens. Divergent lenses only form virtual images.

Snell's law helps explain the refraction in both types of lenses, considering the difference in the index of refraction between the lens material and the surrounding air. The lens power is determined by its ability to bend light, described by the lens's focal length, with more powerful lenses having shorter focal lengths.