Titan, the largest satellite of Saturn, has a thick nitrogen atmosphere. At its surface, the pressure is 1.5 Earth-atmospheres and the temperature is 94 K. Part APart complete Part B Calculate the surface density in Titan's atmosphere in molecules per cubic meter. Express your answer as a number of molecules per meter cubed.

Answer:

404.3 J

Explanation:

Given that

Weight of the merry go round = 790 N

Radius if the merry go round = 1.6 m

Horizontal force applied = 45 N

Time taken = 4 s

To find the mass of the merry go round, we divide the weight by acceleration due to gravity. Thus,

m = F/g

m = 790 / 9.8

m = 80.6 kg

We know that the moment of inertia is given as

I = ½mr², on substitution, we have

I = ½ * 80.6 * 1.6²

I = 103.17 kgm²

Torque = Force applied * radius, so

τ = 45 * 1.6

τ = 72 Nm

To get the angular acceleration, we have,

α = τ / I

α = 72 / 103.17

α = 0.70 rad/s²

Then, the angular velocity is

ω = α * t

ω = 0.7 * 4

ω = 2.8 rad/s

Finally, to get the Kinetic Energy, we have

K.E = ½ * Iω², on substituting, we get

K.E = ½ * 103.17 * 2.8²

K.E = 404.3 J

Therefore, the kinetic energy is 404.3 J

Answer: C. Petal. Length

Explanation: Petal are unit of Corolla which are usually brightly colored. This part of a plant or flower, helps attracts insects to the plant for pollination. And also provide protection to the reproductive parts of the plant or flower.

Examples of flowers with petals is the Sun Flower, which coincidentally is the flower plant with most petals.

Answer:

a) 5.19 W/m²

b) 1.79 J

Explanation:

For the calculation of intensity, I. We have

I = E(rms)² / (cμ), where

c = speed of light

μ = permeability of free space

I = (62.5 / √2)² / [(2.99 x 10^8) (1.26 x 10^-6)]

I = 1954 / 376.74

I = 5.19 W/m²

Therefore, the intensity, I = 5.19 W/m²

t = 14.9 s

A = 0.0231 m²

Amount if energy flowing, U = IAt

U = (5.19) (0.0231) (14.9) J

U = 1.79 J

Answer:

The difference is height is [tex]\Delta h =6.92 \ m[/tex]

Explanation:

From the question we are told that

     The distance of ball  from the goal is [tex]d = 36.0 \ m[/tex]

    The height of the crossbar is  [tex]h = 3.05 \ m[/tex]

       The speed of the ball is [tex]v = 21.6 \ m/s[/tex]

       The angle at which the ball was kicked is [tex]\theta = 50 ^o[/tex]

The height attained by the ball is mathematically represented as

      [tex]H = v_v * t - \frac{1}{2} gt^2[/tex]

Where [tex]v_v[/tex] is the vertical component of  velocity which is mathematically represented as

     [tex]v_v = v * sin (\theta )[/tex]

substituting values

     [tex]v_v = 21.6 * (sin (50 ))[/tex]

     [tex]v_v = 16.55 \ m/s[/tex]

Now the time taken is  evaluated as

       [tex]t = \frac{d}{v * cos(\theta )}[/tex]

substituting value

     [tex]t = \frac{36}{21.6 * cos(50 )}[/tex]

    [tex]t = 2.593 \ s[/tex]

So

     [tex]H = 16.55 * 2.593 - \frac{1}{2} * 9.8 * (2.593)^3[/tex]

     [tex]H = 9.97 \ m[/tex]

The difference  in height is mathematically evaluated as

      [tex]\Delta h = H - h[/tex]

substituting value

    [tex]\Delta h = 9.97 - 3.05[/tex]

    [tex]\Delta h =6.92 \ m[/tex]

Answer: evaporation

Explanation:

Refrigerators work by causing the refrigerant circulating inside them to change from a liquid into a gas. This process, called evaporation, cools the surrounding area and produces the desired effect.

Answer: 192 ft/s

Explanation:

The initial height of the object is:

576ft above the ground.

The position equation is:

p(t) = -16*t^2 + 576

in the position equation, we only can see the therm of the initial height and the term of the acceleration (that is equal to the gravitational acceleration g = 32 ft/s^2 over 2)

So we have no initial velocity, this means that at the beginning we only have potential energy:

U = m*g*h

where m is the mass of the object, g = 32m/s^2 and h = 576 ft.

Now, as the object starts to fall down, the potential energy is transformed into kinetic energy, and when the object is about to hit the ground, all the potential energy has become kinetic energy.

The kinetic energy equation is:

K = (m/2)*v^2

where v is the velocity of the object, then the maximum kinetic energy (when the object reaches the ground) is equal to the initial potential energy:

m*g*h = (m/2)*v^2

now we can solve this for v.

v = √(2*g*h) = √(2*32ft/s^2*576ft) = 192 ft/s

Answer:

visual basics

Explanation:

not suited for programming, slower than the other languages. hard to translate to other operating systems

Answer:

B = 0.024T positive z-direction

Explanation:

In this case you consider that the direction of the motion of the electron, and the direction of the magnetic field are perpendicular.

The magnitude of the magnetic force exerted on the electron is given by the following formula:

[tex]F=qvB[/tex]     (1)

q: charge of the electron = 1.6*10^-19 C

v: speed of the electron = 1.6*10^7 m/s

B: magnitude of the magnetic field = ?

By the Newton second law you also have that the magnetic force is equal to:

[tex]F=qvB=ma[/tex]       (2)

m: mass of the electron = 9.1*10^-31 kg

a: acceleration of the electron = 7.0*10^16 m/s^2

You solve for B from the equation (2):

[tex]B=\frac{ma}{qv}\\\\B=\frac{(9.1*10^{-31}kg)(7.0*10^{16}m/s^2)}{(1.6*10^{-19}C)(1.6*10^7m/s)}\\\\B=0.024T[/tex]

The direction of the magnetic field is found by using the right hand rule.

The electron moves upward (+^j). To obtain a magnetic forces points to the positive x-direction (+^i), the direction of the magnetic field has to be to the positive z-direction (^k). In fact, you have:

-^j X ^i = ^k

Where the minus sign of the ^j is because of the negative charge of the electron.

Then, the magnitude of the magnetic field is 0.024T and its direction is in the positive z-direction

Answer:

this is the required pressure transmitted in the hydraulic system.

Answer:

1.603 s

Explanation:

Given that

Initial mass, = 0.45 kg

Initial period, = 1.45 s

Initial radius, = 0.14 m

Final mass, = 0.55 kg

Final period, = ?

Final radios, = 0.14 m

Since we are finding the rotation period of two masses of same radius, we can assume that the outward force is the same in both cases. This means that

m₁r₁ω₁² = m₂r₂ω2²

Where, ω = 2π/T, on substituting, we have

0.45 * 0.14 * (2π / 1.45)² = 0.550 * 0.14 * (2π / T₂)²

0.45 / 1.45² = 0.550 / T₂²

T₂² = 0.550 * 1.45² / 0.45

T₂² = 2.56972

T₂ = √2.56972

T₂ = 1.603 sec

The  rotation period for a mass of 0.550 kg at the same radius is 1.603 s

Since

Initial mass = 0.45 kg

Initial period = 1.45 s

Initial radius = 0.14 m

Final mass = 0.55 kg

Final radios = 0.14 m

Now the following formulas should be used.

m₁r₁ω₁² = m₂r₂ω2²

here ω = 2π/T

So,

0.45 * 0.14 * (2π / 1.45)² = 0.550 * 0.14 * (2π / T₂)²

0.45 / 1.45² = 0.550 / T₂²

T₂² = 0.550 * 1.45² / 0.45

T₂² = 2.56972

T₂ = √2.56972

T₂ = 1.603 sec

hence, The  rotation period for a mass of 0.550 kg at the same radius is 1.603 s

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Answer:

The angular speed of the new system is [tex]3\,\frac{rad}{s}[/tex].

Explanation:

Due to the absence of external forces between both disks, the Principle of Angular Momentum Conservation is observed. Since axes of rotation of each disk coincide with each other, the principle can be simplified into its scalar form. The magnitude of the Angular Momentum is equal to the product of the moment of inertial and angular speed. When both disks begin to rotate, moment of inertia is doubled and angular speed halved. That is:

[tex]I\cdot \omega_{o} = 2\cdot I \cdot \omega_{f}[/tex]

Where:

[tex]I[/tex] - Moment of inertia of a disk, measured in kilogram-square meter.

[tex]\omega_{o}[/tex] - Initial angular speed, measured in radians per second.

[tex]\omega_{f}[/tex] - Final angular speed, measured in radians per second.

This relationship is simplified and final angular speed can be determined in terms of initial angular speed:

[tex]\omega_{f} = \frac{1}{2}\cdot \omega_{o}[/tex]

Given that [tex]\omega_{o} = 6\,\frac{rad}{s}[/tex], the angular speed of the new system is:

[tex]\omega_{f} = \frac{1}{2}\cdot \left(6\,\frac{rad}{s} \right)[/tex]

[tex]\omega_{f} = 3\,\frac{rad}{s}[/tex]

The angular speed of the new system is [tex]3\,\frac{rad}{s}[/tex].

Answer:

a. 6.41 m/s

b. 120.85 m/s^2

Explanation:

The computation is shown below:

a. Pebble speed is

As we know that according to the tangential speed,

[tex]v = r \times \omega[/tex]

[tex]= \frac{0.68}{2} \times 18.84[/tex]

= 6.41 m/s

The 18.84 come from

[tex]= 2 \times 3.14 \times 3[/tex]

= 18.84

b. The pebble acceleration is

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

[tex]= \frac{6.41^2}{0.34}[/tex]

= 120.85 m/s^2

We simply applied the above formulas so that the pebble speed and the pebble acceleration could come and the same is to be considered

Answer:

mass 20 times of an amazing and all its motion

Answer:

the mass drop by 6.5cm before coming to rest.

Explanation:

Given that:

the mass of the block M= 1.40 kg

angle of inclination θ = 30°

spring constant K = 40.0 N/m

mass of the suspended block m = 60.0 g = 0.06 kg

initial downward speed = 1.40 m/s

The objective is to determine how far does it drop before coming to rest?

Let assume it drops at y from a certain point in the vertical direction;

Then :

Workdone by gravity on the mass of the block is:

[tex]w_1g = 1.4*9.81*y*sin30[/tex] = - 6.867y

Workdone by gravity on the mass of the suspended block is:

[tex]w_2g = 0.06*9.81[/tex] = 0.5886

The workdone by the spring = -1/2ky²

= -0.5 × 40y²

= -20 y²

The net workdone is = -20 y² - 6.867y + 0.5886y

According to the work energy theorem

Net work done = Δ K.E

-20 y² - 6.867y + 0.5886 = 1/2 × 0.06 × 1.4²

-20 y²  - 6.867y + 0.5886 = 0.5 × 0.1176

-20 y²  - 6.867y + 0.5886 = 0.0588

-20 y²  - 6.867y + 0.5886 - 0.0588

-20 y²  - 6.867y + 0.5298 = 0

multiplying through by (-)

20 y²  + 6.867y - 0.5298 = 0

Using the quadratic formula:

[tex]\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex]

where;

a = 20 ; b = 6.867 c= - 0.5298

[tex]\dfrac{-(6.867) \pm \sqrt{(6.867)^2-(4*20*-0.5298)}}{2*(20)}[/tex]

[tex]= \dfrac{-(6.867) + \sqrt{(6.867)^2-(4*20*-0.5298)}}{2*(20)} \ \ \ OR \ \ \ \dfrac{-(6.867) - \sqrt{(6.867)^2-(4*20*-0.5298)}}{2*(20)}[/tex]= 0.0649 OR  −0.408

We go by the positive integer

y = 0.0649 m

y = 6.5 cm

Therefore; the mass drop by 6.5cm before coming to rest.

-- As she lands on the air mattress, her momentum is (m v)

Momentum = (60 kg) (5 m/s down) = 300 kg-m/s down

-- As she leaves it after the bounce,

Momentum = (60 kg) (1 m/s up) = 60 kg-m/s up

-- The impulse (change in momentum) is

Change = (60 kg-m/s up) - (300 kg-m/s down)

Magnitude of the change = 360 km-m/s

The direction of the change is up /\ .

The direction of a body or object's movement is defined by its velocity.In its basic form, speed is a scalar quantity.In essence, velocity is a vector quantity.It is the speed at which distance changes.It is the displacement change rate.

Solve the problem ?

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Complete question:

The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the exit velocity.

Answer:

The exit velocity is 629.41 m/s

Explanation:

Given;

initial temperature, T₁ = 1200K

initial pressure, P₁ = 150 kPa

final pressure, P₂ = 80 kPa

specific heat at 300 K, Cp = 1004 J/kgK

k = 1.4

Calculate final temperature;

[tex]T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}[/tex]

k = 1.4

[tex]T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}}\\\\T_2 = 1200(\frac{80}{150})^{\frac{1.4-1 }{1.4}}\\\\T_2 = 1002.714K[/tex]

Work done is given as;

[tex]W = \frac{1}{2} *m*(v_i^2 - v_e^2)[/tex]

inlet velocity is negligible;

[tex]v_e = \sqrt{\frac{2W}{m} } = \sqrt{2*C_p(T_1-T_2)} \\\\v_e = \sqrt{2*1004(1200-1002.714)}\\\\v_e = \sqrt{396150.288} \\\\v_e = 629.41 \ m/s[/tex]

Therefore, the exit velocity is 629.41 m/s

Answer:

A. a (-321.393, 383.022) b (-76.40, -64.278)

B. (-397.991, 318.744)

C. a. resulting speed 509.9mph  b. bearing of the plane = 51.6°

Explanation:

Answer:

we see that the lights with the most extreme wavelength are blue and red

we see that the separation between the interference lines (y) increases linearly with the wavelength for which the phenomenon is best observed in the RED response 2

Explanation:

In Young's double-slit experiment, constructive interference is written by the equation

       d sin θ = m λ

where you give the gap separation, lam the length of the donda used and m the order of interference

in many he uses trigonometry to express the synth in confusing the distances on a very distant screen

so θ = y / L

in this experiment the angles are generally very small, so

     tan θ = sin θ / cos θ = sin θ

     sint θ = y / L

let's replace

      d y / L = mλ

      y = (m L / d) λ

now let's examine the effect of changing the wavelength

1 yellow lam = 600 10⁻⁹ m

2) red lam = 750 10⁻⁹m

3) blue lam = 450 10⁻⁸ nm

4) green lam = 550 10⁻⁹ nm

we see that the lights with the most extreme wavelength are blue and red

we see that the separation between the interference lines (y) increases linearly with the wavelength for which the phenomenon is best observed in the RED response 2

Answer:

-67 m/s

Explanation:

We are given that

Mass of ball,m=0.4 kg

Initial speed,u=14 m/s

Impulse,I=-32.4 N-s

Time,t=27 ms=[tex]27\times 10^{-3} s[/tex]

We have to find the mass's velocity in the x- direction.

We know that

[tex]Impulse=mv-mu[/tex]

Substitute the values

[tex]-32.4=0.4v-0.4(14)[/tex]

[tex]-32.4+0.4(14)=0.4 v[/tex]

[tex]-26.8=0.4v[/tex]

[tex]v=\frac{-26.8}{0.4}=-67m/s[/tex]

Answer:

q1 = 7.6uC , -2.3 uC

q2 = 7.6uC , -2.3 uC

( q1 , q2 ) = ( 7.6 uC , -2.3 uC ) OR ( -2.3 uC , 7.6 uC )

Explanation:

Solution:-

- We have two stationary identical conducting spheres with initial charges ( q1 and q2 ). Such that the force of attraction between them was F = 0.6286 N.

- To model the electrostatic force ( F ) between two stationary charged objects we can apply the Coulomb's Law, which states:

                              [tex]F = k\frac{|q_1|.|q_2|}{r^2}[/tex]

Where,

                     k: The coulomb's constant = 8.99*10^9

- Coulomb's law assume the objects as point charges with separation or ( r ) from center to center.  

- We can apply the assumption and approximate the spheres as point charges under the basis that charge is uniformly distributed over and inside the sphere.

- Therefore, the force of attraction between the spheres would be:

                             [tex]\frac{F}{k}*r^2 =| q_1|.|q_2| \\\\\frac{0.6286}{8.99*10^9}*(0.5)^2 = | q_1|.|q_2| \\\\ | q_1|.|q_2| = 1.74805 * 10^-^1^1[/tex] ... Eq 1

- Once, we connect the two spheres with a conducting wire the charges redistribute themselves until the charges on both sphere are equal ( q' ). This is the point when the re-distribution is complete ( current stops in the wire).

- We will apply the principle of conservation of charges. As charge is neither destroyed nor created. Therefore,

                             [tex]q' + q' = q_1 + q_2\\\\q' = \frac{q_1 + q_2}{2}[/tex]

- Once the conducting wire is connected. The spheres at the same distance of ( r = 0.5m) repel one another. We will again apply the Coulombs Law as follows for the force of repulsion (F = 0.2525 N ) as follows:

                          [tex]\frac{F}{k}*r^2 = (\frac{q_1 + q_2}{2})^2\\\\\sqrt{\frac{0.2525}{8.99*10^9}*0.5^2} = \frac{q_1 + q_2}{2}\\\\2.64985*10^-^6 = \frac{q_1 + q_2}{2}\\\\q_1 + q_2 = 5.29969*10^-^6[/tex]  .. Eq2

- We have two equations with two unknowns. We can solve them simultaneously to solve for initial charges ( q1 and q2 ) as follows:

                         [tex]-\frac{1.74805*10^-^1^1}{q_2} + q_2 = 5.29969*10^-^6 \\\\q^2_2 - (5.29969*10^-^6)q_2 - 1.74805*10^-^1^1 = 0\\\\q_2 = 0.0000075998, -0.000002300123[/tex]

                          [tex]q_1 = -\frac{1.74805*10^-^1^1}{-0.0000075998} = -2.3001uC\\\\q_1 = \frac{1.74805*10^-^1^1}{0.000002300123} = 7.59982uC\\[/tex]

Answer:

12 m/s²

Explanation:

Maximum acceleration is a = Aω², where A is the amplitude and ω is the frequency.

a = (0.0030 m) (10 rev/s × 2π rad/rev)²

a = 12 m/s²

Answer:

Corresponding raft speed = -0.875 m/s (the minus sign indicates that the raft moves in the direction opposite to the diver)

Explanation:

Law of conservation of momentum gives that the momentum of the diver and the raft before the dive is equal to the momentum of the diver and the raft after the dive.

And since the raft and the diver are initially at rest, the momentum of the diver after the dive is equal and opposite to the momentum experienced by the raft after the dive.

(Final momentum of the diver) + (Final momentum of the raft) = 0

Final Momentum of the diver = (mass of the diver) × (diving velocity of the diver)

Mass of the diver = 73 kg

Diving velocity of the diver = 5.54 m/s

Momentum of the diver = 73 × 5.54 = 404.42 kgm/s

Momentum of the raft = (mass of the raft) × (velocity of the raft)

Mass of the raft = 462 kg

Velocity of the raft = v

Momentum of the raft = 462 × v = (462v) kgm/s

404.42 + 462v = 0

462v = -404.42

v = (-404.42/462) = -0.875 m/s (the minus sign indicates that the raft moves in the direction opposite to the diver)

Hope this Helps!!!

Answer:

The wavelength is  [tex]\lambda = 886 \ nm[/tex]

Explanation:

From the question we are told that

   The  energy band gap is  [tex]E = 1.4 eV[/tex]

Generally the energy of  a single photon of light emitted for an electron jump in a GaAS solar cell is mathematically represented as

      [tex]E = \frac{hc}{\lambda }[/tex]

Where  h is the Planck's  constant with values

     [tex]h = 4.1357 * 10^{-15} eV[/tex]

and  c is  the speed of light with values  [tex]c = 3*10^{8} \ m/s[/tex]

So  

     [tex]\lambda = \frac{hc}{E}[/tex]

substituting values

    [tex]\lambda = \frac{4.1357 *10^{-15} * 3.0 *10^{8}}{1.4}[/tex]

  [tex]\lambda = 886 \ nm[/tex]

Answer:

it changes into liquid

Answer:

It changes in to liquids

Explanation:

This is because the water vapor cools down and condenses it attaches it self to dust forming water droplets. Those water droplets are water.

Answer:

0.1 sec

Explanation:

frequency of the clicks produced = 55 kHz = 55000 Hz

depth of the bottom of ocean from the dolphin = 75 m

we know that the speed of sound  in water is generally accepted to be ≅ 1480 m/s.

the total distance traveled by the sound from the dolphin, to the bottom of the ocean, and then back to the dolphin = 2 x 75 = 150 m

time elapsed will then be

time = distance traveled ÷ speed of sound

time = 150/1480 ≅0.1 sec

Answer:

[tex]F_g=461lb_f[/tex]

Explanation:

First calculate the mass of the asteroid. To do so, you need to find the volume and know the density of iron.

If r = d/2 = 645ft, then:

[tex]V = \frac{4}{3} \pi r^3[/tex]

[tex]V = \frac{4}{3} \pi r^3\\V = 1.124\times10^{9}ft^3\delta_{iron}=m/V=491lb/ft^3m=V\times\delta=5.519\times10^{11}lb[/tex]

In order to find force, use Newton's universal law of gravitation:

[tex]F_g=G\frac{m_1m_2}{d^2}[/tex]

Where,

G= the gravitational constant:

[tex]G= 1.068846 \times10^{-9} ft^3 lb^{-1} s^{-2}[/tex]

[tex]F_g=461lb_f[/tex]

Answer:

The bottom two lines.

Explanation:

They need their own line of voltage quantity. A parallel circuit has the definition of 'two or more paths for current to flow through.' The voltage does stay the same in each line.

Answer:

[tex]solution \\ distance \: travelled = 1.8 \: km \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 1.8 \times 1000m \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 1800 \: m \\ time \: taken = 5 \: minute \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 5 \times 60 \: seconds \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 300 \: seconds \\ average \: velocity = \frac{distance \: travelled}{time \: taken} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{1800}{ 300} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 6 \: {ms}^{ - 1} [/tex]

hope this helps..

Its average speed is (1,800 m) / (300 sec) = 6 m/s .

There's not enough information in the question to calculate the velocity with.  We would need to know the straight-line distance and direction from the place he started from to the place he ended at.

Answer:

Dear user,

Answer to your query is provided below

The angle for which the rope will break θ = 41.8°

Explanation:

Explanation of the same is attached in image

A bag is gently pushed off the top of a wall at A and swings in a vertical plane at the end of a rope of length l. The angle θ for which the rope will break, is 41.81°

The tension is a kind of force which acts on linear objects when subjected to pull.

The maximum tension Tmax =2W

From the work energy principle,

T₂ = 1/2 mv²

Total energy before and after pushing off

0+mglsinθ = 1/2 mv²

v² = 2gflsinθ..............(1)

From the equilibrium of forces, we have

T= ma +mgsinθ = mv²/l +mgsinθ

2mg = mv²/l +mgsinθ

2g = v²/l +gsinθ

Substitute the value of v² ,we get the expression for θ

θ = sin⁻¹(2/3)

θ =41.81°

Hence, the angle θ for which the rope will break, is 41.81°

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