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fiasKO [112]
3 years ago
7

How does jet stream form

Physics
1 answer:
barxatty [35]3 years ago
7 0
Even though the wind "tries" to flow from high pressure to low pressure, the turning of the Earth causes the air flow to turn to the right (in the Northern Hemisphere), so the jet stream flows around the air masses, rather than directly from one to the other.
You might be interested in
An object of mass 5 kilograms is acted upon by the forces F and F2 as shown.
Alekssandra [29.7K]

Answer:

61 N

Explanation:

Sum of forces in the x direction:

∑F = ma

F₂ cos (-55°) − 35 N = 0

F₂ cos (-55°) = 35 N

F₂ = 61 N

3 0
3 years ago
After a mass weighing 10 pounds is attached to a 5-foot spring, the spring measures 7 feet. This mass is removed and replaced wi
hoa [83]

Answer:

a) x(t) = e^(^-^2^t^)[ cos ( 4t ) + 0.75*sin ( 4t ) ]

b) x(t) = 1.25*e^(^-^a^t^)sin(4t + 0.64350)

c) t = ( 0.7957*k - 0.160875 )s  with k = 0, 1 , 2 , 3 , ....

Explanation:

Declaring variables:-

- The mass = m

- Positive damping constant = β

- Spring constant = k

- Gravitational constant, g = 32.2 ft/s^2

Given:-

- The attached initial weight, Wi = 10 lbs

- The attached second weight, Wf = 8 lbs

- Length of un-extended spring, Li = 5 ft

- The extended length, Lf = 7 ft

Find:-

(a) Find the equation of motion if the mass is initially released from a point foot below the equilibrium position with a downward velocity of 1 ft/s.

(b) Express the equation of motion in the form given in (23)

(c) Find the times at which the mass passes through the equilibrium position heading dow

Solution:-

- We will first evaluate the constants m , k and β using conditions given:

                         m = Wf / g

                         m = 8 / 32 = 0.25 slugs.

- We will use the equilibrium condition in (vertical direction) on the spring when the weight of Wi = 10 lbs was attached onto the spring. The spring restoring force (Fs) acts up while weight attached combats it by pointing downward.

                         Fs - Wi = 0

- From Hooke's Law we have:

                         Fs = Wi = k*ΔL   .......  ΔL : Extension of spring

                         k = W / ΔL = W / ( Lf - Li )

                         k = 10 / ( 7 - 5 )

                         k = 5 lb/ft

- Since we have insufficient information about the damping constant of the medium we will assume it as unit. β = 1.

- Now, we will consider the dynamic motion of the spring attached with mass with weight Wf damped in a medium with constant β. We will use Newton's second equation of motion for the spring.

                        F_net = m*a

Where, F_net : Net force acting on the attached mass

            a : Acceleration of the block

- There are two forces acting on the spring ( Damping force - D and restoring force of spring (ks) ). We will consider a displacement of mass in vertical direction as (x).

                       - D - Fs = m*a

- Using hooke's law and damping force (D) is proportional to velocity of attached mass. We have:

                       - β*dx/dt - k*x = m*( d^2 x / dt^2)

 - Plug in the constants:

                          \frac{d^2x}{dt^2} + 4*\frac{dx}{dt} + 20*x = 0

- Now solve the derived ODE. The Auxiliary equation for the above ODE is:  

                         s^2 + 4s + 20 = 0              

- Solve the quadratic and evaluate roots.

                        s = -2 +/- 4i ....... (Complex Roots)

- The complementary solution (yc) for complex roots of the auxiliary is:

                        xc (t) = e^(^-^2^t^)[ A*cos ( 4t ) + B*sin ( 4t ) ]

- Use the given initial conditions and evaluate constants A and B :

                       x ( 0 ) = 1 ft , x ' (0) = 1 ft/s

                       xc (0) = e^(^0^)[ A*cos ( 0 ) + B*sin ( 0 ) ] = 1* [ A + 0 ] = 1\\\\A = 1\\\\xc'(t) = -2*e^(^-^2^t^)[ A*cos ( 4t ) + B*sin ( 4t ) ] + e^(^-^2^t^)[ -4A*sin ( 4t ) + 4B*cos ( 4t ) ]\\\\xc'(0) = -2*e^(^0^)[ cos ( 0 ) + B*sin ( 0 ) ] + e^(^0^)[ -4sin ( 0 ) + 4B*cos ( 0 ) ]\\\\ 1 = -2*[ 1 + 0 ] + 1*[ 0 + 4B]\\\\B = 3 / 4 = 0.75

- The complementary solution becomes:

                         xc (t) = e^(^-^2^t^)[ cos ( 4t ) + 0.75*sin ( 4t ) ]

- Since there is no excitation force acting on the system( Homogenous ). The particular solution does not exist and the general solution to equation of motion is:

                         x(t) = e^(^-^2^t^)[ cos ( 4t ) + 0.75*sin ( 4t ) ]

- An alternative form of the displacement (x) - Time (t) motion is:

                        x(t) = Pe^(^-^a^t^)sin(w_dt + theta)

Where,

                        P = √ ( A^2 + B^2)   .... Amplitude

                        theta (θ) = arctan ( B / A )

                        Auxiliary roots = ( a +/- bi) ..... a = -2 , b = 4  

                        wd = w*√ 1 - ρ^2 = w = b = 4 .......... (β = 1 => ρ = 0)

- Evaluate P and theta (θ):

                        P = √ ( 1^2 + 0.75^2) = 1.25 m

                        theta (θ) = arctan ( 3 / 4 ) = 0.64350 rads

- The alternative form of Equation (23) is:

                        x(t) = 1.25*e^(^-^a^t^)sin(4t + 0.64350)

c)

- To determine the range of times (t) when the mass passes the the equilibrium position can be evaluated by setting x(t) form (23) equal to 0.

                       x(t) = 1.25*e^(^-^a^t^)sin(4t + 0.64350) = 0\\\\e^(^-^a^t^) \neq 0 , sin(4t + 0.64350) = 0\\\\4t + 0.64350 = 0 , \pi , 2\pi , 3\pi = k\pi\ \ .......\ \ (k = 0, 1 , 2 , 3 , ...)\\\\t = \frac{k\pi - 0.64350}{4} = k*0.78571429 - 0.160875

- The domain for time (t) is as follows:

                       t = ( 0.7957*k - 0.160875 )s  with k = 0, 1 , 2 , 3 , ....

                         

7 0
3 years ago
The water hose at your home outputs 24 gallons of water a minute. how many cubic centimeters per second does the hose output? th
adoni [48]

When we convert 24 gallon/min to cubic centimeters per second (cm³/s), the result obtained is 1514.164 cm³/s

<h3>How to convert 24 gallon/min to cm³/min</h3>

We'll begin by converting 24 gallon/min to cm³/min. This can be obtained as follow:

1 gallon/min = 3785.41 cm³/min

Therefore,

24 gallon/min = (24 gallon/min × 3785.41 cm³/min) / 1 gallon/min

24 gallon/min = 90849.84 cm³/min

<h3>How to convert 90849.84 cm³/min to cm³/s</h3>

We can convert 90849.84 cm³/min to cm³/s as follow:

60 cm³/min = 1 cm³/s

Therefore,

90849.84 cm³/min = (90849.84 cm³/min × 1 cm³/s) / 60 cm³/min

90849.84 cm³/min = 1514.164 cm³/s

Thus,

24 gallon/min = 1514.164 cm³/s

Learn more about conversion:

brainly.com/question/2139943

#SPJ1

7 0
1 year ago
A 68-kg skydiver has a speed of 52 m/s at an altitude of 670 m above the
Arlecino [84]

Answer:

91936J

Explanation:

We know that kinetic energy= 1/2 mv^2

M= mass = 68 Kg

v= velocity= 52 m/s

KE=1/2 × 68 × (52)^2

KE= 1/2 × 68 × 2704

KE= 91936J

4 0
3 years ago
Electric current is produced when ____ move through a _____.
joja [24]
<span>Electric current is produced when a changing magnetic field is applied to a conductor, an electromotive force (EMF) is produced, thus causing a suitable path.</span>
5 0
3 years ago
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