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horsena [70]
3 years ago
6

Suppose that you can throw a projectile at a large enough v0 so that it can hit a target a distance R downrange. Given that you

know v0 and R, determine the general expressions for the two distinct launch angles θ1 and θ2 that will allow the projectile to hit D. For v0 = 37 m/s and R = 70 m, determine numerical values for θ1 and θ2.
Engineering
1 answer:
viktelen [127]3 years ago
5 0

Answer:

\theta_1=15^o\\\theta_2=75^o

Explanation:

<u>Projectile Motion</u>

In projectile motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration (assuming no friction), and the acceleration in the vertical direction is always the acceleration of gravity. The basic formulas are shown below:

V_x=V_{ox}=V_ocos\theta

Where \theta is the angle of launch respect to the positive horizontal direction and Vo is the initial speed.

V_y=V_{oy}-gt=V_osin\theta-gt

The  horizontal and vertical distances are, respectively:

x=V_{o}cos\theta t

\displaystyle y=y_o+V_{o}sin\theta t-\frac{gt^2}{2}

The total flight time can be found as that when y = 0, i.e. when the object comes back to ground (or launch) level. From the above equation we find

\displaystyle t_f=\frac{2V_osin\theta}{g}

Using this time in the horizontal distance, we find the Range or maximum horizontal distance:

\displaystyle R=\frac{V_o^2sin2\theta}{g}

Let's solve for \theta

\displaystyle sin2\theta=\frac{R.g}{V_o^2}

This is the general expression to determine the angles at which the projectile can be launched to hit the target. Recall the angle can have to values for fixed positive values of its sine:

\displaystyle \theta_1=\frac{asin\left(\frac{R.g}{V_o^2}\right)}{2}

\displaystyle \theta_2=\frac{180^o-asin\left(\frac{R.g}{V_o^2}\right)}{2}

Or equivalently:

\theta_2=90^o-\theta_1

Given Vo=37 m/s and R=70 m

\displaystyle \theta_1=\frac{asin\left(\frac{70\times 9.8}{37^2}\right)}{2}

\theta_1=15^o

And

\theta_2=90^o-15^o=75^o

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REY [17]

Answer:

0.5°c

Explanation:

Humidity ratio by mass can be expressed as

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Humidity ratio is normally expressed in kilograms (or pounds) of water vapor per kilogram (or pound) of dry air.

Humidity ratio expressed by mass:

x = mw / ma                                  (1)

where

x = humidity ratio (kgwater/kgdry_air, lbwater/lbdry_air)

mw = mass of water vapor (kg, lb)

ma = mass of dry air (kg, lb)

It can be as:

x = 0.005 (100) / [(100 - 100)]

x = 0.005 x 100 / (100 - 100)

x = 0.005 x 100 / 0

x = 0.5°c

So the temperature to which atmospheric air must be cooled in order to have humidity ratio of 0.005 lb/lb is 0.5°c

6 0
3 years ago
An air conditioner removes heat steadily from a house at a rate of 750 kJ/min while drawing electric power at a rate of 6 kW. De
Paraphin [41]

Answer:

a. 2.08, b. 1110 kJ/min

Explanation:

The power consumption and the cooling rate of an air conditioner are given. The COP or Coefficient of Performance and the rate of heat rejection are to be determined. <u>Assume that the air conditioner operates steadily.</u>

a. The coefficient of performance of the air conditioner (refrigerator) is determined from its definition, which is

COP(r) = Q(L)/W(net in), where Q(L) is the rate of heat removed and W(net in) is the work done to remove said heat

COP(r) = (750 kJ/min/6 kW) x (1 kW/60kJ/min) = 2.08

The COP of this air conditioner is 2.08.

b. The rate of heat discharged to the outside air is determined from the energy balance.

Q(H) = Q(L) + W(net in)

Q(H) = 750 kJ/min + 6 x 60 kJ/min = 1110 kJ/min

The rate of heat transfer to the outside air is 1110 kJ for every minute.

5 0
3 years ago
A nozzle receives an ideal gas flow with a velocity of 25 m/s, and the exit at 100 kPa, 300 K velocity is 250 m/s. Determine the
Margaret [11]

Given Information:

Inlet velocity = Vin = 25 m/s

Exit velocity = Vout = 250 m/s

Exit Temperature = Tout = 300K

Exit Pressure = Pout = 100 kPa

Required Information:

Inlet Temperature of argon = ?

Inlet Temperature of helium = ?

Inlet Temperature of nitrogen = ?

Answer:

Inlet Temperature of argon = 360K

Inlet Temperature of helium = 306K

Inlet Temperature of nitrogen = 330K

Explanation:

Recall that the energy equation is given by

$ C_p(T_{in} - T_{out}) = \frac{1}{2} \times (V_{out}^2 - V_{in}^2) $

Where Cp is the specific heat constant of the gas.

Re-arranging the equation for inlet temperature

$ T_{in}  = \frac{1}{2} \times \frac{(V_{out}^2 - V_{in}^2)}{C_p}  + T_{out}$

For Argon Gas:

The specific heat constant of argon is given by (from ideal gas properties table)

C_p = 520 \:\: J/kg.K

So, the inlet temperature of argon is

$ T_{in}  = \frac{1}{2} \times \frac{(250^2 - 25^2)}{520}  + 300$

$ T_{in}  = \frac{1}{2} \times 119  + 300$

$ T_{in}  = 360K $

For Helium Gas:

The specific heat constant of helium is given by (from ideal gas properties table)

C_p = 5193 \:\: J/kg.K

So, the inlet temperature of helium is

$ T_{in}  = \frac{1}{2} \times \frac{(250^2 - 25^2)}{5193}  + 300$

$ T_{in}  = \frac{1}{2} \times 12  + 300$

$ T_{in}  = 306K $

For Nitrogen Gas:

The specific heat constant of nitrogen is given by (from ideal gas properties table)

C_p = 1039 \:\: J/kg.K

So, the inlet temperature of nitrogen is

$ T_{in}  = \frac{1}{2} \times \frac{(250^2 - 25^2)}{1039}  + 300$

$ T_{in}  = \frac{1}{2} \times 60  + 300$

$ T_{in}  = 330K $

Note: Answers are rounded to the nearest whole numbers.

5 0
3 years ago
A series AC circuit contains a resistor, an inductor of 250 mH, a capacitor of 4.40 µF, and a source with ΔVmax = 240 V operatin
slega [8]

Answer:

Explanation:

Inductance = 250 mH = 250 / 1000 = 0.25 H

capacitance = 4.40 µF = 4.4 × 10⁻⁶ F ( µ = 10⁻⁶)

ΔVmax = 240, f frequency = 50Hz and I max = 110 mA = 110 /1000 = 0.11A

a) inductive reactance = 2πfl =  2 × 3.142 × 50 × 0.25 H =78.55 ohms

b) capacitive reactance = \frac{1}{2\pi fC} = 1 / ( 2 × 3.142× 50 × 4.4 × 10⁻⁶ ) = 723.34 ohms

c) impedance = \frac{Vmax}{Imax} = 240 / 0.11 = 2181.82 ohms

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