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10 months ago

11. As and_ prices continued to rise in the late 1960’s and 70's, 4 and 6 cylinder engines began to make a comeback.

Engineering

1 answer:

zvonat [6]10 months ago

5 0

Answer:

Explanation:

whats the answeres two the question do they have a choise for you

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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

viktelen [127]

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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3 years ago

Vehicles begin to arrive at a parking lot at 8:10 am at a constant rate of 6 veh/min until 8:25 am. There is no arrival from 8:2

tresset_1 [31]

1.i am superman

2.175 175 175

3.em dilisues

4.bye classmate magbabalik pa ako

8 0

2 years ago

If a construction company is considering a new type of material to use in their construction, which factors would they

Amanda [17]

I would honestly select every one of the given options. Gor a company evaluating this new material it would be very valuable to hit each of these factors.

4 0

2 years ago

A piston-cylinder assembly contains 0.5 lb of water. The water expands from an initial state where p1 = 40 lbf/in.2 and T1 = 300

VARVARA [1.3K]

Answer:

W = 31.393 Btu

Explanation:

given data

piston-cylinder assembly contains water = 0.5 lb

p1 = 40 lbf/in² = 40 × 144 lbf/ft²

T1 = 300° F

p2 = 14.7 lbf/in² = 14.7 × 144 lbf/ft²

pv 1.2 = constant.

solution

we use here superheated table A 4E to get value for p1 = 40 and t = 300

v1 = 11.04 ft³/lbm

so we know

$P_2 \times V_2^{1.2} = P_1 \times V_1^{1.2}$ ,...............1

$14.7 \times V_2^{1.2} = 40 \times (11.04)^{1.2}$

solve it we get

v2 = 25.425 ft³/lbm

and here

W = $(\frac{P_1V_1 - P_2V_2}{n-1}) m$ ...............2

put here value we get

W = $(\frac{40\times 144\times (11.04) -14.7 \times 144\times (25.425) }{1.2-1})0.5$

solve it we get

W = 31.393 Btu

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3 years ago

Why water parameters of Buriganga river vary between wet and dry seasons?<br> Explain.

Sedaia [141]

I honestly don’t know because I honestly don’t know what I mean

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3 years ago

11. As __and___ prices continued to rise in the late 1960’s and 70's, 4 and 6 cylinder engines began to make a comeback.

11. As and_ prices continued to rise in the late 1960’s and 70's, 4 and 6 cylinder engines began to make a comeback.