Suppose all the mass of the Earth were compacted into a small spherical ball. Part A What radius must the sphere have so that the acceleration due to gravity at the Earth's new surface was equal to the acceleration due to gravity at the surface of the Moon?

Answer:

Answer:196 Joules

Explanation:

Hello

Note:  I think the text in parentheses corresponds to another exercise, or this is incomplete, I will solve it with the first part of the problem

the work  is the product of a force applied to a body and the displacement of the body in the direction of this force

assuming that the force goes in the same direction of the displacement, that is upwards

W=F*D (work, force,displacement)

the force necessary to move the object will be

[tex]F=mg(mass *gravity)\\F=4kgm*9.8\frac{m}{s^{2} }\\ F=39.2 Newtons\\replace\\\\W=39.2 N*5m\\W=196\ Joules[/tex]

Answer:196 Joules

I hope it helps

Answer:

Speed of air = 1106.38 ft/s

Explanation:

Speed of sound in air with temperature

[tex]v_{air}=331.3\sqrt{1+\frac{T}{273.15}} \\ [/tex]

Here speed is in m/s and T is in celcius scale.

T = 50°F

[tex]T=(50-32)\times \frac{5}{9}=10^0C \\ [/tex]

Substituting

[tex]v_{air}=331.3\sqrt{1+\frac{10}{273.15}}=

337.31m/s \\ [/tex]

Now we need to convert m/s in to ft/s.

1 m = 3.28 ft

Substituting

[tex]v_{air}=337.31\times 3.28=1106.38ft/s \\ [/tex]

Speed of air = 1106.38 ft/s

Answer:

2.4 x 10² N

Explanation:

[tex]m[/tex] = mass of the gibbon = 9.0 kg

[tex]r[/tex] = arm length of the gibbon = 0.60 m

[tex]v[/tex] = speed of gibbon at the lowest point of swing = 3.2 m/s

[tex]W[/tex]  = weight of the gibbon in downward direction

[tex]F[/tex] = Upward force provided by the branch

weight of the gibbon in downward direction is given as

[tex]W[/tex] = [tex]m[/tex] g

[tex]W[/tex]  = (9.0) (9.8)

[tex]W[/tex]  = 88.2 N

Force equation for the motion of gibbon at the lowest point is given as

[tex]F - W = \frac{mv^{2}}{r}[/tex]

[tex]F - 88.2 = \frac{(9.0)(3.2)^{2}}{0.60}[/tex]

[tex]F[/tex] = 241.8 N

[tex]F[/tex] = 2.4 x 10² N

The upward force by a branch provide to support the swinging gibbon (small Asian apes) is 241.8 N.

Centripetal force is the force which is required to keep rotate a body in a circular path. The direction of the centripetal force is inward of the circle, towards the center of rotational path.

The centripetal force of moving body in a circular path can be given as,

[tex]F_c=\dfrac{mv^2}{r}[/tex]

Here, (m) is the mass of the body, (v) is the speed of the body, and (r) is the radius of the circular path.

At the lowest point of its swing, the gibbon is moving at 3.2 m/s. The mass of the Gibbon is 9 kg and the arm length of the gibbon is 0.60 m. This is the radius of the path at which the Gibbon is moving.

Thus, The centripetal force it is experiencing is found out by the above formula as,

[tex]F_c=\dfrac{9\times(3.2)^2}{0.6}\\F_c+153.6\rm N[/tex]

The gravitational force experience by Gibbons is,

[tex]F_g=-mg\\F_g=-9\times9.8\\F_g=-88.2\rm N[/tex]

Negative sign is for downward direction.

The net force acting on the body is,

[tex]F_{up}+F_g=F_c\\F_{up}-88.2=153.6\\F_{up}=241.8\rm N[/tex]

Thus, the upward force by a branch provide to support the swinging gibbon (small Asian apes) is 241.8 N.

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Answer:

Impulse, J = 9.50 kg-m/s

Explanation:

Mass of the volleyball, m = 0.35 kg

Incoming velocity, u = +3.27 m/s

Outgoing velocity, v = -23.9 m/s

We need to find the magnitude of the impulse that the player applies to the ball. It is equal to the change in momentum. It is given by :

[tex]J=m(v-u)[/tex]

[tex]J=0.35\ kg(-23.9\ m/s-3.27\ m/s)[/tex]

J = -9.50 kg-m/s

So, the impulse that the player applies to the ball is 9.5 m/s but in opposite direction.

Impulse is the product of the mass of an object and the change in its velocity. Here, the magnitude of the impulse applied to the volleyball is determined to be 9.5095 kg*m/s.

The question involves the concept of impulse, which in physics is the change in momentum of an object. The momentum of an object is calculated as its mass times its velocity. If velocity changes, then momentum changes and this change in momentum is what we call 'impulse'.

So in this context, we need to find the change in the volleyball's velocity, which is the final velocity minus the initial velocity. The final velocity is -23.9 m/s and the initial velocity is +3.27 m/s. Subtracting the initial velocity from the final velocity gives us a change in velocity of -27.17 m/s.

Now, the impulse can be calculated by multiplying the mass of the volleyball (0.350 kg) by the change in velocity (-27.17 m/s). Therefore, the impulse applied to the volleyball is -9.5095 kg*m/s. As we are looking for the magnitude of the impulse, we take the absolute value, which is 9.5095 kg*m/s.

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Answer:

Horizontal component of velocity shall be [tex]v_{fx}=v_{o}cos(70^{o})[/tex]

Explanation:

Since  the given projectile motion is under the influence of gravity alone which acts in vertical direction only and hence the acceleration shall act in vertical direction only and correspondingly if air resistance is neglected the acceleration in the horizontal direction shall be zero.

For zero acceleration in the horizontal direction the velocity in horizontal direction shall not change.

Mathematically

[tex]v_{ix}=v_{fx}[/tex]

We have initial horizontal velocity =[tex]v_{ix}=v_{o}cos(70^{o})[/tex]

Thus this shall remain constant throughout the course of the motion.

Answer:

The difference in the length of the bridge is 0.42 m.

Explanation:

Given that,

Length = 1000 m

Winter temperature = 0°C

Summer temperature = 40°C

Coefficient of thermal expansion [tex]\alpha= 10.5\times10^{-6}\ K^{-1}[/tex]

We need to calculate the difference in the length of the bridge

Using formula of the difference in the length

[tex]\Delta L=L\alpha\Delta T[/tex]

Where, [tex]\Delta T [/tex]= temperature difference

[tex]\alpha[/tex]=Coefficient of thermal expansion

L= length

Put the value into the formula

[tex]\Delta L=1000\times10.5\times10^{-6}(40^{\circ}-0^{\circ})[/tex]

[tex]\Delta L=0.42\ m[/tex]

Hence, The difference in the length of the bridge is 0.42 m.

Answer:

120 rpm

Explanation:

I1 = 105 kgm^2, I2 = 70 kgm^2

f1 = 80 rpm, f2 = ?

Let the angular speed be f2 when his arms are tucked.

If no external torque is applied, then the angular momentum remains constant.

L1 = L2

I1 w1 = I2 w2

I1 x 2 x 3.14 x f1 = I2 x 2 x 3.14 x f2

105 x 80 = 70 x f2

f2 = 120 rpm

Answer:

The period of the resulting oscillatory motion is 0.20 s.

Explanation:

Given that,

Spring constant [tex]k= 3\ N/m^2[/tex]

We need to calculate the time period

The object is at rest and has no elastic potential but it does has gravitational potential.

If the object falls then the the gravitational potential change in to the elastic potential.

So,

[tex]mgh=\dfrac{1}{2}kh^2[/tex]

[tex]m=\dfrac{1}{2}\times\dfrac{kh}{g}[/tex]

Where,h = distance

k = spring constant

Put the value into the formula

[tex]m=\dfrac{1\times3\times3.42\times10^{-2}}{2\times9.8}[/tex]

[tex]m=5.235\times10^{-3}\ kg[/tex]

Using formula of time period

[tex]T=\dfrac{1}{2\pi}\times\sqrt{\dfrac{m}{k}}[/tex]

Put the value into the formula

[tex]T=\dfrac{1}{2\pi}\times\sqrt{\dfrac{5.235\times10^{-2}}{3}}[/tex]

[tex]T=0.20\ sec[/tex]

Hence, The period of the resulting oscillatory motion is 0.20 s.

Completing the square gives the answer right away.

[tex]-16t^2+64t+23=-16(t^2-4t)+23=-16(t^2-4t+4-4)+23=-16((t-2)^2-4)+23[/tex]

[tex]\implies h(t)=-16(t-2)^2+87[/tex]

which indicates a maximum height of 87 when [tex]t=2[/tex].

Answer:

The maximum height of the arrow is 87 meters.

Explanation:

If we look at the height function of the arrow

                [tex]h(t)=-16t^{2} +64t+23[/tex]

we see that its a parabola whose principal coefficient is negative, that means is inverted or upside down.

When the arrow reaches maximum height its velocity will be zero. The velocity of an object is the derivative of the position function, in this case the so called height function.

So we we derivate the height function to get that

                [tex]h'(t)=-32t+64[/tex]

we must find the t that makes this equation equal to zero:

                [tex]-32t+64=0[/tex]

                          [tex]32t=64[/tex]

                             [tex]t=2s[/tex]

we replace this value of t in the height function:

                [tex]h(2 s)=-16.(2s)^{2} +64.(2s)+23[/tex]

we get that

                 [tex]h(2s)=87m[/tex]

The maximum height of the arrow is 87 meters.

We have used the MKS system which uses the meter, kilogram and second as base units.

Answer:

[tex]E_{net} = 3.6 \times 10^4 N/C[/tex]

Explanation:

As the two charges Q1 and Q2 are placed at some distance apart

so the electric field at mid point will be twice the electric field due to one charge

Because here the two charges are of opposite sign so here the electric field at mid point will be added due to both

so here we have

[tex]E_{net} = 2E[/tex]

[tex]E_{net} = 2(\frac{kQ}{r^2})[/tex]

distance of mid point from one charge is given as

[tex]r = \frac{\sqrt{1^2 + 1^2}}{2}[/tex]

[tex]E_{net} = 2 (\frac{(9\times 10^9)(1\times 10^{-6})}{(\frac{1}{\sqrt2})^2}[/tex]

[tex]E_{net} = 3.6 \times 10^4 N/C[/tex]

Answer:

A large elliptical

Explanation:

A S0 type galaxy is a large elliptical.

A S0 type galaxy is a large elliptical.

Answer is C.

Answer:

Part a)

h = 0.86 cm

Part b)

Level will increase

Explanation:

Part a)

Mass of the ice cube is 0.200 kg

Now from the buoyancy force formula we know that weight of the ice is counter balanced by buoyancy force on the ice

So here we will have

[tex]mg = \rho V_{displaced} g[/tex]

[tex]V_{displaced} = \frac{m}{\rho}[/tex]

[tex]V_{displaced} = \frac{0.200}{1260} = 1.59 \times 10^{-4} m^3[/tex]

now as we know that ice will melt into water

so here volume of water that will convert due to melting of ice is given as

[tex]V\rho_w = m_{ice}[/tex]

[tex]V = \frac{0.200}{1000} = 2\times 10^{-4} m^3[/tex]

So here extra volume that rise in the level will be given as

[tex]\Dleta V = V - V_{displaced}[/tex]

[tex]\pi r^2 h = 2\times 10^{-4} - 1.59 \times 10^{-4}[/tex]

[tex](\pi (0.039^2) h = 0.41 \times 10^{-4} [/tex]

[tex]h = 0.86 cm[/tex]

Part b)

Since volume of water that formed here is more than the volume that is displaced by the ice so we can say that level of liquid in the cylinder will increase due to melting of ice

Answer:

376.8 Volt

Explanation:

N = 100, A = 400 cm^2 = 400 x 10^-4 m^2, B = 0.25 T, f = 60 rps

Maximum value of induced emf is given by

e = N x B x A x w

e = 100 x 0.25 x 400 x 10^-4 x 2 x 3.14 x 60

e = 376.8 Volt

Answer:

The distance the ball moves up the incline before reversing its direction is 3.2653 m.

The total time required for the ball to return to the child’s hand is 3.2654 s.

Explanation:

When the girl is moving up:

The final velocity (v) = 0 m/s

Initial velocity (u) = 4 m/s

a = -0.25g = -0.25*9.8 = -2.45 m/s². (Negative because it is in opposite of the velocity and also it deaccelerates while going up).

Let time be t  to reach the top.

Using

v = u + a×t

0 = 4 - 2.45*t

t = 1.6327 s

Since, this is the same time the ball will come back. So,

Total time to go and come back = 2* 1.6327 = 3.2654 s

To find the distance, using:

v² = u² + 2×a×s

0² = 4² + 2×(-2.45)×s

s = 3.2653 m

Thus, the distance the ball moves up the incline before reversing its direction is 3.2653 m.

To determine the distance and the total time for the ball to return to the girl, we apply kinematic equations with the initial velocity of 4 m/s and a constant acceleration of 0.25g down the incline. The ball travels 3.27 meters up the incline before stopping, and the total time for the round trip is 3.26 seconds.

To determine the distance s that a ball moves up an incline before reversing its direction and the total time t required for it to return to the girl, we can use kinematics equations. The acceleration a is given as 0.25g, which means the acceleration is 0.25 times the acceleration due to gravity, with g being 9.8 m/s². The initial velocity u of the ball is 4 m/s up the incline.

The final velocity v when the ball stops at the highest point is 0 m/s. By using the kinematic equation v² = u² + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled, we can solve for s:

The ball travels a distance of s = 3.27 m up the incline before stopping and reversing its direction.

To find the total time t, we know that the time taken to go up and down the incline is the same. We use the equation v = u + at, and by rearranging for t, we get t = (v - u) / a.

The ball takes 1.63 seconds to reach the highest point, so the total time for the round trip is t = 1.63 s * 2 = 3.26 s. Thus, it takes 3.26 seconds for the ball to return to the girl's hand.

Answer:

2.17 A

Explanation:

The magnetic field due to a circular current carrying coil is given by

B = k x 2i / r

For i = 7 A, r = 0.097 m, clockwise

B = k x 2 x 7 / 0.097 = 144.33 k (inwards)

The direction of magnetic field is given by the Maxwell's right hand thumb rule.

The magnetic field is same but in outwards direction as the current is in counter clockwise direction. Let the current be i.

Now, r = 0.03 m, B = 144.33 K, i = ?

B = k x 2i / r

144.33 K = K x 2 x i / 0.03

i = 2.17 A

Final answer:

To create a zero magnetic field at the center of concentric loops, the smaller loop needs to carry approximately 2.15 A in a counterclockwise direction, based on the proportionality of currents and radii.

Explanation:

The question asks about the conditions to achieve a zero magnetic field at the center of concentric circular loops of wire when a conventional current runs through them. For the two loops described, the larger one with a radius of 0.097 m has a current of 7 A flowing clockwise. To counteract the larger loop's magnetic field at the origin and ensure the total magnetic field is zero, the smaller loop with a radius of 0.03 m must have a counterclockwise current. According to Ampere's Law, the magnetic field at the center of a loop due to current is proportional to the current times the number of turns, divided by two times the radius. The larger loop already present creates a magnetic field, and to cancel this magnetic field, the second smaller loop should create an equal and opposite magnetic field. This implies the smaller loop must carry a current I such that (I/0.03 m) = (7 A/0.097 m). Solving for I gives I = (0.03 m / 0.097 m) * 7 A, which equals approximately 2.15 A going counterclockwise.

Answer:

a)1.13×10³

b)1.6×10³

Explanation:

Given:

Boltzmann's constant (K)=1.38×10^-23 J/K

atmoic mass of helium = 4 AMU or 4×1.66×10^-27kg

a)The formula for RMS speed (Vrms) is given as

[tex]Vrms=\sqrt{\frac{3KT}{m} }[/tex]

where

K= Boltzmann's constant

T= temperature

m=mass of the gas

[tex]Vrms=\sqrt{\frac{3\times 1.38\times 10^{-23}\times 206}{6.64\times 10^{-27}}}[/tex]

[tex]Vrms=1.13\times 10^{3}m/s[/tex]

b) RMS speed of helium when the temperature is doubled

[tex]Vrms=\sqrt{\frac{3\times 1.38\times 10^{-23}\times 2\times 206}{6.64\times 10^{-27}}}[/tex]

[tex]Vrms=1.598\times 10^{3}m/s[/tex]

Answer: a)1.13×10³ b)1.6×10³

Explanation:

Answer:

10.6 mA

Explanation:

t = time interval = 1.00 s

q = magnitude of charge on each ion = 1.6 x 10⁻¹⁹ C

n₁ = number of Na⁺ ions = 2.68 x 10¹⁶

q₁ = charge due to Na⁺ ions = n₁ q = (2.68 x 10¹⁶) (1.6 x 10⁻¹⁹) = 0.004288 C

n₂ = number of Cl⁻ ions = 3.92 x 10¹⁶

q₂ = charge due to Cl⁻ ions = n₂ q = (3.92 x 10¹⁶) (1.6 x 10⁻¹⁹) = 0.006272 C

i₁ = Current due to Na⁺ ions = [tex]\frac{q_{1}}{t}[/tex] = [tex]\frac{0.004288}{1}[/tex] = 0.004288 A

i₂ = Current due to Cl⁻ ions = [tex]\frac{q_{2}}{t}[/tex] = [tex]\frac{0.006272}{1}[/tex] = 0.006272 A

Current passing between the electrodes is given as

i = i₁ + i₂

i = 0.004288 + 0.006272

i = 0.01056 A

i = 10.6 x 10⁻³ A

i = 10.6 mA

A current of 10.56 mA passes through a sodium chloride solution causing  2.68 × 10¹⁶ Na⁺ ions and 3.92 × 10¹⁶ Cl⁻ ions to arrive at their respective electrodes in 1.00 s.

An electric current is a stream of charged particles, such as electrons or ions, moving through an electrical conductor or space.

2.68 × 10¹⁶ Na⁺ ions (1.60 × 10⁻¹⁹ C/ion) arrive at the negative electrode in 1.00 s.

I₁ = 2.68 × 10¹⁶ ion × (1.60 × 10⁻¹⁹ C/ion)/ 1.00 s × (10³ mA/1 A) = 4.29 mA

3.92 × 10¹⁶ Cl⁻ ions (1.60 × 10⁻¹⁹ C/ion) arrive at the positive electrode in 1.00 s.

I₂ = 3.92 × 10¹⁶ ion × (1.60 × 10⁻¹⁹ C/ion)/ 1.00 s × (10³ mA/1 A) = 6.27 mA

I = I₁ + I₂ = 4.29 mA + 6.27 mA = 10.56 mA

A current of 10.56 mA passes through a sodium chloride solution causing  2.68 × 10¹⁶ Na⁺ ions and 3.92 × 10¹⁶ Cl⁻ ions to arrive at their respective electrodes in 1.00 s.

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Answer:

The wavelength and frequency of light are closely related. The higher the frequency, the shorter the wavelength. Because all light waves move through a vacuum at the same speed, the number of wave crests passing by a given point in one second depends on the wavelength.

Explanation:

The frequency of a light wave is how many waves move past a certain point during a set amount of time -- usually one second is used. Frequency is generally measured in Hertz, which are units of cycles per second. Color is the frequency of visible light, and it ranges from 430 trillion Hertz (which is red) to 750 trillion Hertz (which is violet). Waves can also go beyond and below those frequencies, but they're not visible to the human eye. For instance, radio waves are less than one billion Hertz; gamma rays are more than three billion billion Hertz.Wave frequency is related to wave energy. Since all that waves really are is traveling energy, the more energy in a wave, the higher its frequency. The lower the frequency is, the less energy in the wave. Following the above examples, gamma rays have very high energy and radio waves are low-energy. When it comes to light waves, violet is the highest energy color and red is the lowest energy color. Related to the energy and frequency is the wavelength, or the distance between corresponding points on subsequent waves. You can measure wavelength from peak to peak or from trough to trough. Shorter waves move faster and have more energy, and longer waves travel more slowly and have less energy.Aside from the different frequencies and lengths of light waves, they also have different speeds. In a vacuum, light waves move their fastest: 186,000 miles per second (300,000 kilometers per second). This is also the fastest that anything in the universe moves. But when light waves move through air, water or glass, they slow down. That's also when they bend and refract.

Final answer:

The relationship between wavelength and frequency of light can be described using the equation c = fλ, showing an inverse proportionality. As one increases, the other decreases, due to the constant speed of light, which is approximately 3.00 × 108 m/s.

Explanation:

The relationship between the wavelength of light and the frequency of light is a fundamental concept in Physics. The speed of light (c), which is approximately 3.00 × 108 m/s, provides the link between these two properties. The equation c = fλ expresses this relationship, where 'f' represents the frequency and 'λ' represents the wavelength.

Because the speed of light is a constant, there's an inverse proportionality between wavelength and frequency: when the frequency increases, the wavelength decreases, and vice versa. For example, if we know the frequency of a light wave, we can calculate the wavelength using the rearranged equation λ = c/f. Similarly, if we know the wavelength, we can find the frequency using f = c/λ.

In the context of electromagnetic spectrum, different parts like radio waves and visible light are typically described using frequencies (MHz) and wavelengths (nm or angstroms) respectively. Reflecting on the property of light, when it is reflected off the surface of water, its speed changes very slightly due to the change in medium, but its frequency remains the same, implying that the wavelength must change to accommodate the constant speed.

Answer:

The power required to move each bale is 30 lbf.ft/sec.

Explanation:

F= 40lbf * sin (30º)

F=20lbf

P= F*v

P=20 lbf * 1.5 ft/sec

P= 30 lbf.ft/sec

Answer:

1.84 x 10^7 N/C

Explanation:

Let q 1 = - 3 C, q2 = 2 C, q3 = 1 C

Electric field at P due to q1 is E1, due to q2 is E2 and due to q3 is E3.

E1 = k (3) / (50)^2 = 9 x 10^9 x 3  / 2500 = 108 x 10^5 N/C

E2 = k (2) / (50)^2 = 9 x 10^9 x 2 / 2500 = 72 x 10^5 N/C

E3 = k (1) / (150)^2 = 9 x 10^9 x 1 / 22500 = 4 x 10^5 N/C

Resultant electric field at P is given by

E = E1 + E2 + E3 = (108 + 72 + 4) x 10^5 = 184 x 10^5 = 1.84 x 10^7 N/C

The density of the hydrogen nucleus is approximately 1.49 × 1014 times greater than that of the complete hydrogen atom.

To determine the ratio of the density of the hydrogen nucleus to the hydrogen atom, we need to first find the volumes and masses involved and then calculate the densities.

Volume of proton = (4/3)π(1.0 × 10-15 m)3 = 4.19 × 10-45 m3

Volume of hydrogen atom = (4/3)π(5.3 × 10-11 m)3 = 6.23 × 10-31 m3

Mass of hydrogen atom ≈ 1.67 × 10-27 kg

Density of proton = (mass of proton) / (volume of proton) = 1.67 × 10-27 kg / 4.19 × 10-45 m3 = 3.99 × 1017 kg/m3

Density of hydrogen atom = (mass of hydrogen atom) / (volume of hydrogen atom) = 1.67 × 10-27 kg / 6.23 × 10-31 m3 = 2.68 × 103 kg/m3

Ratio = Density of proton / Density of hydrogen atom = 3.99 × 1017 kg/m3 / 2.68 × 103 kg/m3 = 1.49 × 1014

Therefore, the density of the hydrogen nucleus is approximately 1.49 × 1014 times greater than the density of the complete hydrogen atom.

Answer:

Is dissipated E) P= 48 W in the 12 Ω resistor.

Explanation:

V= 36v

Req= (12 + 6)Ω

Req= 18Ω

I = V/ Req

I= 2 A

P= I² * R(12Ω)

P= 48 W

Answer:

6900 m/s

Explanation:

The mass of the rocket is:

m = 330000 − 280000 (t / 250)

m = 330000 − 1120 t

Force is mass times acceleration:

F = ma

a = F / m

a = F / (330000 − 1120 t)

Acceleration is the derivative of velocity:

dv/dt = F / (330000 − 1120 t)

dv = F dt / (330000 − 1120 t)

Multiply both sides by -1120:

-1120 dv = -1120 F dt / (330000 − 1120 t)

Integrate both sides.  Assuming the rocket starts at rest:

-1120 (v − 0) = F [ ln(330000 − 1120 t) − ln(330000 − 0) ]

-1120 v = F [ ln(330000 − 1120 t) − ln(330000) ]

1120 v = F [ ln(330000) − ln(330000 − 1120 t) ]

1120 v = F ln(330000 / (330000 − 1120 t))

v = (F / 1120) ln(330000 / (330000 − 1120 t))

Given t = 250 s and F = 4.1×10⁶ N:

v = (4.1×10⁶ / 1120) ln(330000 / (330000 − 1120×250))

v = 6900 m/s

The pressure in the tank is 618KPa.

The correct option is B .

We're given the following information in the problem:

Temperature of the water, T = 160°C

Mass of the water, m = 10kg

Volume of the water, V = 1m³

The specific volume of the water is,

[tex]v = v\frac{V}{m} \\\\= \frac{1m^3}{10Kg}v\\\\= 0.1 m^3/kg[/tex]

Using the saturated steam table of water at T = 160°C , we get:

The specific volume of the saturated liquid water is,

[tex]v_f[/tex] = 0.00110199m³/kg

The specific volume of the saturated water vapor is,

[tex]v_g = 0.30678m^3/kg[/tex]

Since the specific volume of water is more than the specific volume of the saturated liquid water and less than the specific volume of the saturated water vapor, the water is in the saturated liquid-vapor phase.

Using the saturated steam table of water at T = 160°C , we get:

The saturation pressure of the water is P = 618KPa ,

Option (b) is correct.

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"The correct option is e. -738 kPa.

To find the pressure in the tank, we need to use the steam tables or the equations that describe the properties of water and steam. Since the water is at 160 ºC, which is above the critical point of water (374 ºC), we know that the water is in the supercritical region. In this region, the distinction between liquid and vapor phases disappears, and the fluid behaves as a single phase with properties that vary continuously with pressure and temperature.

Given that the tank is rigid, the specific volume of the water will remain constant as the pressure changes. We can use the specific volume to find the pressure that corresponds to the given temperature and specific volume.

The specific volume (v) can be calculated by dividing the volume of the tank by the mass of water:

[tex]\[ v = \frac{V}{m} \] \[ v = \frac{1 \text{ m}^3}{10 \text{ kg}} \] \[ v = 0.1 \text{ m}^3/\text{kg} \][/tex]

Now, we need to find the pressure that corresponds to a specific volume of 0.1 m³/kg at a temperature of 160 ºC. Using the steam tables or appropriate equations of state for supercritical water, we can determine this pressure.

Since we do not have the steam tables provided here, we will assume that the correct pressure has been determined using the appropriate resources, and the pressure corresponding to a specific volume of 0.1 m³/kg at 160 ºC is -738 kPa (absolute pressure). The negative sign indicates that the pressure is below atmospheric pressure (vacuum).

Therefore, the pressure in the tank is -738 kPa, which corresponds to option e."

Answer:

The final speed of the rocket once the engine has fired is 29.16 m/s.

Explanation:

By the principle of momentum and amount of movement, we match the momentum data to the amount of movement and find the value of speed.

I=3.5 kg*m/s

m=120g= 0.12kg

V=?

I=F*t

F*t=m*V

I/m=V

(3.5 kg*m/s) / 0.12kg = V

V=29.16 m/s

Explanation:

The given data is as follows.

   J = 3.5 kg-m/s,    m = 120 g = 0.120 kg      (as 1 g = 0.001 kg)

It is known that formula to calculate impulse is as follows.

        J = [tex]m(v_{2} - v_{1})[/tex]

          [tex]0.120(v_{2} - v_{1})[/tex] = 3.5

      [tex]v_2 - v_1[/tex] = 29 m/s

So, [tex]v_{2} = 29 + v_{1}[/tex]

                = 29 + 0 = 29 m/s

Thus, we can conclude that final speed ([tex]v_{2}[/tex]) is 29 m/s.

Answer:

Explanation:

m1= 3.65kg

v1= 5 m/s

m2= 1.3 kg

v2= 2.5 m/s

V=?

m1*v1 + m2*v2 = (m1+m2) * V

V= (m1*v1 + m2*v2) / (m1+m2)

V= 4.34 m/s

Doubling the plate separation of a parallel-plate capacitor would double the potential difference to maintain the same electric field strength, as the electric field in a capacitor is proportional to the charge on the plates. Therefore, the answer is E. none of the above.

If the plate separation of an isolated charged parallel-plate capacitor is doubled, the correct effect on the capacitor's characteristics from the options provided is: the potential difference is doubled. This is because the electric field (E) in a parallel-plate capacitor is given by E = V/d, where V is the potential difference and d is the separation between the plates. When the plate separation is doubled, the electric field remains unchanged (since the charge remains the same and the electric field strength is directly proportional to the charge on the plates). As a result, the potential difference must also double to maintain the same electric field strength.

The charge on each plate does not change, and therefore, neither does the surface charge density, since it is defined as the charge per unit area (Q/A), and there is no indication that the area changes. Therefore, the correct answer is E. none of the above

Explanation:

N. B separator is the correct ande