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klemol [59]
3 years ago
10

A projectile is launched horizontally 1m above the ground. If it lands 300m away from the initial launch position, find: a)-the

initial launch velocity, u, and b)-the angle 0, at which the projectile contacts the ground.
Engineering
1 answer:
PIT_PIT [208]3 years ago
4 0

Answer:

(a): The launch velocity is Vx= 666.66 m/s.

(b): The angle wich the projectile contacts the ground is α= 0.38°

Explanation:

h= 1m

g= 9.8 m/s²

h= g*t²/2

t= 0.45 s

Vy= g*t

Vy= 4.42 m/s

d=Vx* t

Vx= 666.66 m/s (a)

α= tg⁻¹ ( Vy/Vx)

α= 0.38° (b)

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The pressure distribution over a section of a two-dimensional wing at 4 degrees of incidence may be approximated as follows: Upp
Aliun [14]

Answer:

The lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

Explanation:

The Upper Surface Cp is given as

Cp_u=-0.8 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.8*0.6+0.4*0.1

The Lower Surface Cp is given as

Cp_l=-0.4 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.4*0.6+0.4*0.1

The difference of the Cp over the airfoil is given as

\Delta Cp=Cp_l-Cp_u\\\Delta Cp=-0.4*0.6+0.4*0.1-(-0.8*0.6-0.4*0.1)\\\Delta Cp=-0.4*0.6+0.4*0.1+0.8*0.6+0.4*0.1\\\Delta Cp=0.4*0.6+0.4*0.2\\\Delta Cp=0.32

Now the Lift Coefficient is given as

C_L=\Delta C_p cos(\alpha_i)\\C_L=0.32\times cos(4*\frac{\pi}{180})\\C_L=0.3192

Now the coefficient of moment about the leading edge is given as

C_M=-0.3*0.4*0.6-(0.6+\dfrac{0.4}{3})*0.2*0.4\\C_M=-0.1306

So the lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

5 0
3 years ago
Which statements describe how the Fed responds to high inflation? Check all that apply.
Sveta_85 [38]
Answer:
• it charges banks more interest
• it sells more securities
• it decreases the money supply

In response to high inflation, the Fed charges banks more interests and pays the banks less interests. It also sells not securities.
8 0
3 years ago
Tech A says never use a water hose to clean up dust after a repair. Tech B says never use a floor scrubber to clean up dust afte
DENIUS [597]

Answer:

Tech B

Explanation:

You ruin your mop that way. it will clump up and than you'll have a mud mess.

5 0
4 years ago
A generator operating at 50 Hz delivers 1 pu power to an infinite bus through a transmission circuit in which resistance is igno
olga55 [171]

Answer:

critical clearing angle = 70.3°

Explanation:

Generator operating at = 50 Hz

power delivered = 1 pu

power transferable when there is a fault = 0.5 pu

power transferable before there is a fault = 2.0 pu

power transferable after fault clearance = 1.5 pu

using equal area criterion to determine the critical clearing angle

Attached is the power angle curve diagram and the remaining part of the solution.

The power angle curve is given as

= Pmax sinβ

therefore :  2sinβo = Pm

                   2sinβo = 1

                   sinβo = 0.5 pu

                   βo = sin^{-1} (0.5) = 30⁰

also ;   1.5sinβ1 = 1

               sinβ1 = 1/1.5

               β1 = sin^{-1} (\frac{1}{1.5} ) = 41.81⁰

∴ βmax = 180 - 41.81  = 138.19⁰

attached is the remaining solution

The critical clearing angle = cos^{-1} 0.3372  ≈ 70.3⁰

3 0
3 years ago
One kilogram of water fills a 0.140 m^3 rigid container at an initial pressure of 1.8 MPa. The container is then cooled to 40°C.
Pavel [41]

Answer:

T1 = 299.18 °C

P2 = 0.00738443 MPa

Explanation:

From the data, we can get two properties for the initial condition. These are pressure and specific volume.

The pressure is 1.8 MPa and the specific volume, we can get it with the mass and volume of the container, since it’s filled this is also the volume of the water in it.

v=\frac{vol (m^{3})}{mass (kg)} = \frac{0.140 m^{3}}{1 kg} = 0.140 \frac{ m^{3}}{kg}

When we check in the thermodynamic tables, the conditions for saturation at 1.8 MPa we found the following:

P^{sat} = 1.8 MPa

T^{sat} = 207.12 C

v_{g} = 0.1103\frac {m^{3}}{kg} specific volume for the saturated vapor  

v_{l} = 0.001167 \frac{m^{3}}{kg} specific volume for the saturated liquid  

Since the specific volume in our condition is higher that the specific volume for the saturated vapor, we have a superheated steam.  

Looking in the thermodynamic tables for superheated steam we found that the temperature where the steam has a specific volume of 0.140 \frac{ m^{3}}{kg} at 1.8 MPa is 299.18 °C. This is the initial temperature in the container.

Since the only information that we have about the final condition is that the container was cooled. We can assume that it was cooled until a condition of saturation. So, the final pressure for the water will be the pressure of saturation for a temperature of 40°C. From thermodynamic tables we get:

P^{sat} at 40C = 0.00738443 MPa

7 0
3 years ago
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