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2 years ago

6

A ____________ is a term that originally was referring to a way to reproduce a technical drawing documenting an architectural or

engineering type of drawing.

1 answer:

larisa86 [58]2 years ago

7 0

Answer:

The answer is blueprint.

Explanation:

Have a nice day or night!

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While playing a game of catch on the quadrangle, you throw a ball at an initial velocity of 17.6 m/s (approximately 39.4 mi/hr),

MAXImum [283]

Answer:

a) The y-component of velocity just before the ball hits the ground is -14.860 meters per second.

b) The ball is in the air during approximately 2.890 seconds.

c) The horizontal distance covered by the ball is 32.695 meters.

d) The magnitude of the velocity of the ball just before it hits the ground is approximately 18.676 meters per second.

e) The angle of the total velocity of the ball just before it hits the ground is approximately 52.717º below the horizontal.

Explanation:

a) The ball experiments a parabolic motion, which is a combination of horizontal motion at constant velocity and vertical motion at constant acceleration. First, we calculate the time taken by the ball to hit the ground:

$y = y_{o} + (v_{o}\cdot \sin \theta) \cdot t+\frac{1}{2}\cdot g\cdot t^{2}$ (1)

Where:

$y_{o}$ , $y$ - Initial and final vertical position, measured in meters.

$v_{o}$ - Initial speed, measured in meters per second.

$\theta$ - Launch angle, measured in sexagesimal degrees.

$g$ - Gravitational acceleration, measured in meters per square second.

$t$ - Time, measured in seconds.

If we know that $y_{o} = 2\,m$ , $y = 0\,m$ , $v_{o} = 17.6\,\frac{m}{s}$ , $\theta = 50^{\circ}$ and $g = -9.807\,\frac{m}{s^{2}}$ , then the time taken by the ball is:

$-4.904\cdot t^{2}+13.482\cdot t +2 = 0$ (2)

This second order polynomial can be solved by Quadratic Formula:

$t_{1} \approx 2.890\,s$ and $t_{2} \approx -0.141\,s$

Only the first root offers a solution that is physically reasonable. That is, $t \approx 2.890\,s$ .

The vertical velocity of the ball is calculated by this expression:

$v_{y} = v_{o}\cdot \sin \theta +g\cdot t$ (3)

Where:

$v_{o,y}$ , $v_{y}$ - Initial and final vertical velocity, measured in meters per second.

If we know that $v_{o} = 17.6\,\frac{m}{s}$ , $\theta = 50^{\circ}$ , $g = -9.807\,\frac{m}{s^{2}}$ and $t \approx 2.890\,s$ , then the final vertical velocity is:

$v_{y} = -14.860\,\frac{m}{s}$

The y-component of velocity just before the ball hits the ground is -14.860 meters per second.

b) From a) we understand that ball is in the air during approximately 2.890 seconds.

c) The horizontal distance covered by the ball ( $x$ ) is determined by the following expression:

$x = (v_{o}\cdot \cos \theta)\cdot t$ (4)

If we know that $v_{o} = 17.6\,\frac{m}{s}$ , $\theta = 50^{\circ}$ and $t \approx 2.890\,s$ , then the distance covered by the ball is:

$x = 32.695\,m$

The horizontal distance covered by the ball is 32.695 meters.

d) The magnitude of the velocity of the ball just before hitting the ground ( $v$ ), measured in meters per second, is determined by the following Pythagorean identity:

$v = \sqrt{(v_{o}\cdot \cos \theta )^{2}+v_{y}^{2}}$ (5)

If we know that $v_{o} = 17.6\,\frac{m}{s}$ , $\theta = 50^{\circ}$ and $v_{y} = -14.860\,\frac{m}{s}$ , then the magnitude of the velocity of the ball is:

$v \approx 18.676\,\frac{m}{s}$ .

The magnitude of the velocity of the ball just before it hits the ground is approximately 18.676 meters per second.

e) The angle of the total velocity of the ball just before it hits the ground is defined by the following trigonometric relationship:

$\tan \theta = \frac{v_{y}}{v_{o}\cdot \cos \theta_{o}}$

If we know that $v_{o} = 17.6\,\frac{m}{s}$ , $\theta_{o} = 50^{\circ}$ and $v_{y} = -14.860\,\frac{m}{s}$ , the angle of the total velocity of the ball just before hitting the ground is:

$\theta \approx -52.717^{\circ}$

The angle of the total velocity of the ball just before it hits the ground is approximately 52.717º below the horizontal.

3 0

3 years ago

Read 2 more answers

The power supply converts the wall outlet AC power into DC power. T or F

Alika [10]

Answer:

True

Explanation:

All computer parts require DC power to operate, and wall outlets provide AC Power.

7 0

3 years ago

A 10 wt % aqueous solution of sodium chloride is fed to an evaporative crystallizer which operates at 80o C. The product is a sl

VikaD [51]

Answer: The feed rate is

17,020kg/he and the rate is 13,520kg/h

Check the attachment for step by step explanation

3 0

3 years ago

A flat site is being considered for a new school that will have a steel frame and brick façade. The steel columns will have a ma

ehidna [41]

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem

5 0

3 years ago

Pressurized steam at 400 K flows through a long, thin-walled pipe of 0.6-m diameter. The pipe is enclosed in a concrete (stone m

iris [78.8K]

Answer:

Rate of heat loss per unit length of pipe, q' = 767.01 W/m

Explanation:

Let q' be the Rate of heat loss per unit length

Let q be the Rate of heat loss

q' = q/L

Where L is the length of the pipe

Diameter, D= 0.6m

The rate of heat loss q is given by the formula: q = Sk(T₂ - T₁)

Where k is the thermal conductivity of the concrete at 300 K

k = 1.4 Wb/m-K (at 300K)

And S is the shape factor given by the formula:

S = 2πL/ ln(1.08w/D)

S = (2π*L) / ln(1.08*1.75/0.6))

S = (2π*L) / 1.147

S = 5.48 L

q = 5.48L*1.4(400-300)

q = 767.01 L

q' = q/L

q' = 767.01L/L

q' = 767.01 W/m

4 0

3 years ago