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3 years ago

An airplane flies with a velocity of 55.0 m/s [ 35° N of W] with respect to the air (this is

Physics

1 answer:

rodikova [14]3 years ago

6 0

Answer:

21 m/s.

Explanation:

The computation of the wind velocity is shown below:

But before that, we need to find out the angles between the vectors

53° - 35° = 18°

Now we have to sqaure it i.e given below

v^2 = 55^2 + 40^2 - 2 · 55 · 40 · cos 18°

v^2 = 3025 + 1600 - 2 · 55 · 40 · 0.951

v^2 = 440.6

v = √440.6

v = 20.99

≈ 21 m/s

Hence, The wind velocity is 21 m/s.

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Pani-rosa [81]

Answer:

a ) option 2 is correct

b) -ve acceleration for upward motion ,0 acceleration at top point ,+ve acceleration on downward motion ...

Explanation:

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2 years ago

Read 2 more answers

The pressure drop needed to force water through a horizontal 1-in diameter pipe if 0.60 psi for every 12-ft length of pipe. Dete

oksian1 [2.3K]

Answer:

The shear stress at a distance 0.3-in away from the pipe wall is 0.06012lb/ft²

The shear stress at a distance 0.5-in away from the pipe wall is 0

Explanation:

Given;

pressure drop per unit length of pipe = 0.6 psi/ft

length of the pipe = 12 feet

diameter of the pipe = 1 -in

Pressure drop per unit length in a circular pipe is given as;

$\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\$

make shear stress (τ) the subject of the formula

$\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\\tau = \frac{\delta P *r}{2L}$

Where;

τ is the shear stress on the pipe wall.

ΔP is the pressure drop

L is the length of the pipe

r is the distance from the pipe wall

Part (a) shear stress at a distance of 0.3-in away from the pipe wall

Radius of the pipe = 0.5 -in

r = 0.5 - 0.3 = 0.2-in = 0.0167 ft

ΔP = 0.6 psi/ft

ΔP, in lb/ft² = 0.6 x 144 = 86.4 lb/ft²

$\tau = \frac{\delta P *r}{2L} = \frac{86.4 *0.0167}{2*12} =0.06012 \ lb/ft^2$

Part (b) shear stress at a distance of 0.5-in away from the pipe wall

r = 0.5 - 0.5 = 0

$\tau = \frac{\delta P *r}{2L} = \frac{86.4 *0}{2*12} =0$

3 0

3 years ago

Consider a sealed 20 cm high electronic box whose base dimensions are 40cm x 40cm placed in a vacuum chamber. The emissivity of

Genrish500 [490]

Answer:

$T_{surr}=296.289\ K$

In Celsius:

$T_{surr}=296.289-273\\T_{surr}=23.289^oC$

Explanation:

The formula we are going to use is:

$\dot Q_{rad}=\epsilon\sigma A_s(T_s^4-T_{surr}^4)$

Where:

ε is the emissivity

σ is the Stefan constant

$T_s$ is the final temperature of surrounding surfaces

$T_{surr}$ is the required temperature

$A_s$ is the are of surrounding surface

Calculating The area:

$A_s=(0.4)(0.4)+4(0.4)(0.2)\\A_s=0.48\ m^2$

σ= $5.67*10^{-8}\ W/m^2.K^4$

ε =0.95

$T_s$ =55+273

$T_s$ =328 K

$\dot Q_{rad$ =100 W

$100=0.95(5.67*10^{-8})(0.48)(328^4-T_{surr}^4)\\3867693926=(328^4-T_{surr}^4)\\T_{surr}^4=7706623130\\T_{surr}=296.289\ K$

In Celsius:

$T_{surr}=296.289-273\\T_{surr}=23.289^oC$

8 0

2 years ago

What formula is used to find an object's acceleration?

nataly862011 [7]

Acceleration (a) is the change in velocity (Δv) over the change in time (Δt), represented by the equation a = Δv/Δt. This allows you to measure how fast velocity changes in meters per second squared (m/s^2).

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3 years ago

As the radius of the Gaussian surface increases, the magnitude of the electric field at the surface does which of the followingI

marshall27 [118]

Answer:

Decreases

Explanation:

Gauss's law states that the flux of the electric field through a closed surface is equal to the quotient between the charge inside that surface divided by the permittivity of free space:

$\int\limits {E\cdot dS} \,=\frac{Q_{enc}}{\epsilon_0}\\E=\frac{Q_{enc}}{A \epsilon_0}$

Therefore, the magnitude of the electric field is inversely proportional to the area, the area is directly proportional to the radius. So, as the radius of the Gaussian surface increases, the magnitude of the electric field at the surface decreases.

5 0

2 years ago