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crimeas [40]
2 years ago
5

If an airplane were traveling eastward with a thrust force 450 N and there was a tailwind of 200 N, what would the resulting net

force be on the airplane?
*20 points*

Physics
2 answers:
Vsevolod [243]2 years ago
4 0

Answer: The resulting net force is 650N

Explanation:

I jus know stuff HAVE A GOOD DAY MERRY CHRISTMAS AND HAPPY NEW YEAR ECT.

Finger [1]2 years ago
3 0

Answer:

650N

Explanation:

450+200=650

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If my glass is filled with ice and water, will it overflow once the ice has melted?
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7 0
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Read 2 more answers
A certain lightbulb has a tungsten filament with a resistance of 26 Ω when cold and 170 Ω when hot. If the equation R = R0 [1 +
iris [78.8K]

Answer:

Explanation: The equation that relates resistance of tungsten at different temperatures is as follows

R = R₀ [1 + α ∆T]  , R₀ is resistance at lower temperature , R is resistance at higher temperature . α is temperature coefficient of resistivity and ∆T is rise in temperature .

Putting the values

170 = 26 [1 + .0045 ∆T]

∆T = 1230.75

lower temperature = 40◦C

higher temperature = 1230 + 40

= 1270◦C

5 0
3 years ago
Two small insulating spheres with radius 3.50×10−2 m are separated by a large center-to-center distance of 0.555 m. One sphere i
xz_007 [3.2K]

Answer:

E = 7.83 \times 10^5 N/C

Explanation:

Since we know that two sphere is oppositely charged so net electric field at the mid point of two balls will be sum of the electric field due to each ball at the mid point

So we know that

E = \frac{kq_1}{r^2} + \frac{kq_2}{r^2}

here we know that

q_1 = 4.20 \mu C

q_2 = 2.50 \mu C

r = \frac{0.555}{2}

so we have

E = \frac{(9\times 10^9)(4.20 + 2.50) \times 10^{-6}}{0.2775^2}

E = 7.83 \times 10^5 N/C

4 0
2 years ago
1) The equilibrium constant Kc for the reaction N 2(g) + O 2(g) 2NO(g) at 1200 C is 1.00x 10^-5. Calculate the molar concentrati
Elina [12.6K]

Explanation:

1) N₂ + O₂ → 2 NO

Kc = [NO]² / ([N₂] [O₂])

Set up an ICE table:

\left[\begin{array}{cccc}&Initial&Change&Equilibrium\\N_{2}&0.114&-x&0.114-x\\O_{2}&0.114&-x&0.114-x\\NO&0&+2x&2x\end{array}\right]

Plug into the equilibrium equation and solve for x.

1.00×10⁻⁵ = (2x)² / ((0.114 − x) (0.114 − x))

1.00×10⁻⁵ = (2x)² / (0.114 − x)²

√(1.00×10⁻⁵) = 2x / (0.114 − x)

0.00316 = 2x / (0.114 − x)

0.00361 − 0.00316x = 2x

0.00361 = 2.00316x

x = 0.00018

The volume is 1.00 L, so the concentrations at equilibrium are:

[N₂] = 0.114 − x = 0.11382

[O₂] = 0.114 − x = 0.11382

[NO] = 2x = 0.00036

2(a) Cl₂ → 2 Cl

Kc = [Cl]² / [Cl₂]

\left[\begin{array}{cccc}&Initial&Change&Equilibrium\\Cl_{2}&2.0&-x&2.0-x\\Cl&0&+2x&2x\end{array}\right]

1.2×10⁻⁷ = (2x)² / (2 − x)

1.2×10⁻⁷ (2 − x) = 4x²

2.4×10⁻⁷ − 1.2×10⁻⁷ x = 4x²

2.4×10⁻⁷ ≈ 4x²

x² ≈ 6×10⁻⁸

x ≈ 0.000245

2x ≈ 0.00049

2(b) F₂ → 2 F

Kc = [F]² / [F₂]

\left[\begin{array}{cccc}&Initial&Change&Equilibrium\\F_{2}&2.0&-x&2.0-x\\F&0&+2x&2x\end{array}\right]

1.2×10⁻⁴ = (2x)² / (2 − x)

1.2×10⁻⁴ (2 − x) = 4x²

2.4×10⁻⁴ − 1.2×10⁻⁴ x = 4x²

2.4×10⁻⁴ ≈ 4x²

x² ≈ 6×10⁻⁵

x ≈ 0.00775

2x ≈ 0.0155

F₂ dissociates more, so Cl₂ is more stable at 1000 K.

7 0
3 years ago
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