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3 years ago

How much does coast to coast membership cost?

1 answer:

8 0

The price of coast to coast membership in united states could lie anywhere between $2,000 to $ 5,000
Unless you're a frequent user of this type of event, i think it would be economically more efficient if you pay the resort on one-day price

You might be interested in

Select the options that best complete the statement.Positively charged particle trajectories(always, never, the same as)follow e

SpyIntel [72]

Answer:

a) always. b) electric field lines are defined by the path positive test charges travel.

Explanation:

By convention, field lines always follow the direction that it would take a positive test charge (small enough so it can´t disrupt the field created by a charge distribution), under the influence of an electric field, at the same point where the test charge is located.

So any positive charge, subject to an electric field influence, moves along the field line that passes through its current position, in the same way that a positive test charge would.

We could say also that the electric force on a positively charged particle is in the same direction as the electric field that produces that force (due to some charge distribution) , which is true, but it doesn´t explain why.

3 0

4 years ago

Suppose you have a 34.9 m length of copper wire. If the wire is wrapped into a solenoid 0.240 m long and having a radius of 0.05

Nadusha1986 [10]

Answer:

the strength of the magnetic field inside the solenoid is 6.278 x 10⁻³ T.

Explanation:

Given;

length of the wire, = 34.9 m

length of solenoid, L = 0.24 m

radius of the solenoid, r = 0.051 m

current in the solenoid, I = 11.0 A

The number of turns of the wire is calculated as follow;

$N = \frac{34.9}{2\pi \times 0.051} = 109 \ turns$

The strength of the magnetic field inside the solenoid is calculated as follows;

$B = \mu_0 (\frac{N}{L} )I\\\\B = 4\pi \times 10^{-7} \times (\frac{109}{0.24} )\times 11.0 \\\\B = 6.278 \times 10^{-3} \ T$

Therefore, the strength of the magnetic field inside the solenoid is 6.278 x 10⁻³ T.

8 0

3 years ago

Box A has a mass of 35.0kg and Box B has a mass of 20.0kg. What is the tension on

lions [1.4K]

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3 0

3 years ago

Researchers at the University of Georgia have evaluated trends in streamside forests in areas within roughly 400 feet of the sta

Colt1911 [192]

<span>B: adds aesthetic value to the landscape. Think about it, out of all your options, that's the one that doesn't really help anything.

And I took the test, so take my word for it.</span>

6 0

3 years ago

Air at 207 kPa and 200◦C enters a 2.5-cm-ID tube at 6 m/s. The tube is constructed ofcopper with a thickness of 0.8 mm and a len

Serga [27]

Answer:

Temperature of air at exit = 24.32 C, After reducing hot air the temperature of the exit air becomes = 20.11 C

Explanation:

ρ = P/R(Ti) where ρ is the density of air at the entry, P is pressure of air at entrance, R is the gas constant, Ti is the temperature at entry

ρ = (2.07 x 10⁵)/(287)(473) = 1.525 kg/m³

Calculate the mass flow rate given by

m (flow rate) = (ρ x u(i) x A(i)) where u(i) is the speed of air, A(i) is the area of the tube (πr²) of the tube

m (flow rate) = 1.525 x (π x 0.0125²) x 6 = 4.491 x 10⁻³ kg/s

The Reynold's Number for the air inside the tube is given by

R(i) = (ρ x u(i) x d)/μ where d is the inner diameter of the tube and μ is the dynamic viscosity of air (found from the table at Temp = 473 K)

R(i) = (1.525) x (6) x 0.025/2.58 x 10⁻⁵ = 8866

Calculate the convection heat transfer Coefficient as

h(i) = (k/d)(R(i)^0.8)(Pr^0.3) where k is the thermal conductivity constant known from table and Pr is the Prandtl's Number which can also be found from the table at Temperature = 473 K

h(i) = (0.0383/0.025) x (8866^0.8) x (0.681^0.3) = 1965.1 W/m². C

The fluid temperature is given by T(f) = (T(i) + T(o))/2 where T(i) is the temperature of entry and T(o) is the temperature of air at exit

T(f) = (200 + 20)/2 = 110 C = 383 K

Now calculate the Reynold's Number and the Convection heat transfer Coefficient for the outside

R(o) = (μ∞ x do)/V(f) where μ∞ is the speed of the air outside, do is the outer diameter of the tube and V(f) is the kinematic viscosity which can be known from the table at temperature = 383 K

R(o) = (12 x 0.0266)/(25.15 x 10⁻⁶) = 12692

h(o) = K(f)/d(o)(0.193 x Ro^0.618)(∛Pr) where K(f) is the Thermal conductivity of air on the outside known from the table along with the Prandtl's Number (Pr) from the table at temperature = 383 K

h(o) = (0.0324/0.0266) x (0.193 x 12692^0.618) x (0.69^1/3) = 71.36 W/m². C

Calculate the overall heat transfer coefficient given by

U = 1/{(1/h(i)) + A(i)/(A(o) x h(o))} simplifying the equation we get

U = 1/{(1/h(i) + (πd(i)L)/(πd(o)L) x h(o)} = 1/{(1/h(i) + di/(d(o) x h(o))}

U = 1/{(1/1965.1) + 0.025/(0.0266 x 71.36)} = 73.1 W/m². C

Find out the minimum capacity rate by

C(min) = m (flow rate) x C(a) where C(a) is the specific heat of air known from the table at temperature = 473 K

C(min) = (4.491 x 10⁻³) x (1030) = 4.626 W/ C

hence the Number of Units Transferred may be calculated by

NTU = U x A(i)/C(min) = (73.1 x π x 0.025 x 3)/4.626 = 3.723

Calculate the effectiveness of heat ex-changer using

∈ = 1 - е^(-NTU) = 1 - e^(-3.723) = 0.976

Use the following equation to find the exit temperature of the air

(Ti - Te) = ∈(Ti - To) where Te is the exit temperature

(200 - Te) = (0.976) x (200 - 20)

Te = 24.32 C

The effect of reducing the hot air flow by half, we need to calculate a new value of Number of Units transferred followed by the new Effectiveness of heat ex-changer and finally the exit temperature under these new conditions.

Since the new NTU is half of the previous NTU we can say that

NTU (new) = 2 x NTU = 2 x 3.723 = 7.446

∈(new) = 1 - e^(-7.446) = 0.999

(200 - Te (new)) = (0.999) x (200 - 20)

Te (new) = 20.11 C

5 0

3 years ago