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tester [92]
3 years ago
8

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),

at an angle of 50 degrees above horizontal. The ball flies past your friend, and lands on the ground. You let the ball go at a height of 2 m above the ground. Define upward as the positive y direction, and the horizontal direction of the ball's travel as the positive x direction. You can assume that air resistance and any effects of the ball spinning are so small that they can be ignored; this means that gravity is the only force causing an acceleration.
Required:
a. What is the y-component of velocity just before the ball hits the ground?
b. For how much time is the ball in the air?
c. How far horizontally does the ball travel before it hits the ground?
d. What is the magnitude of the velocity of the ball just before it hits the ground?
e. What is the angle of the total velocity of the ball just before it hits the ground?
Engineering
2 answers:
Reil [10]3 years ago
6 0

Answer:

A

Explanation:

MAXImum [283]3 years ago
3 0

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.

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Serggg [28]

Answer:

Explanation:

f = 50.0 Hz, L = 0.650 H, π = 3.14

C = 4.80 μF, R = 301 Ω resistor. V = 120volts

XL = wL = 2πfL

= 2×3.14×50* 0.650

= 204.1 Ohm

Xc= 1/wC

Xc = 1/2πfC

Xc = 1/2×3.14×50×4.80μF

= 1/0.0015072

= 663.48Ohms

1. Total impedance, Z = sqrt (R^2 + (Xc-XL)^2)= √ 301^2+ (663.48Ohms - 204.1 Ohm)^2

√ 90601 + (459.38)^2

√ 90601+211029.98

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= 549.209

Z = 549.21Ohms

2. I=V/Z = 120/ 549.21Ohms =0.218Ampere

3. P=V×I = 120* 0.218 = 26.16Watt

Note that

I rms = Vrms/Xc

= 120/663.48Ohms

= 0.18086A

4. I(max) = I(rms) × √2

= 0.18086A × 1.4142

= 0.2557

= 0.256A

5. V=I(max) * XL

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= 52.250volts

6. V=I(max) × Xc

= 0.256A × 663.48Ohms

= 169.85volts

7. Xc=XL

1/2πfC = 2πfL

1/2πfC = 2πf× 0.650

1/2×3.14×f×4.80μF = 2×3.14×f×0.650

1/6.28×f×4.8×10^-6 = 4.082f

1/0.000030144× f = 4.082×f

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8 0
3 years ago
Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320
lesya [120]

Answer:

The minimum allowable bolt diameter required to support an applied load of P = 450 kN is 45.7 milimeters.

Explanation:

The complete statement of this question is "Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320 MPa, and a factor of safety of 4.2 is required with respect to fracture. Determine the minimum allowable bolt diameter required to support an applied load of P = 450 kN"

Each bolt is subjected to shear forces. In this case, safety factor is the ratio of the ultimate shear strength to maximum allowable shear stress. That is to say:

n = \frac{S_{uts}}{\tau_{max}}

Where:

n - Safety factor, dimensionless.

S_{uts} - Ultimate shear strength, measured in pascals.

\tau_{max} - Maximum allowable shear stress, measured in pascals.

The maximum allowable shear stress is consequently cleared and computed: (n = 4.2, S_{uts} = 320\times 10^{6}\,Pa)

\tau_{max} = \frac{S_{uts}}{n}

\tau_{max} = \frac{320\times 10^{6}\,Pa}{4.2}

\tau_{max} = 76.190\times 10^{6}\,Pa

Since each bolt has a circular cross section area and assuming the shear stress is not distributed uniformly, shear stress is calculated by:

\tau_{max} = \frac{4}{3} \cdot \frac{V}{A}

Where:

\tau_{max} - Maximum allowable shear stress, measured in pascals.

V - Shear force, measured in kilonewtons.

A - Cross section area, measured in square meters.

As connection consist on five bolts, shear force is equal to a fifth of the applied load. That is:

V = \frac{P}{5}

V = \frac{450\,kN}{5}

V = 90\,kN

The minimum allowable cross section area is cleared in the shearing stress equation:

A = \frac{4}{3}\cdot \frac{V}{\tau_{max}}

If V = 90\,kN and \tau_{max} = 76.190\times 10^{3}\,kPa, the minimum allowable cross section area is:

A = \frac{4}{3} \cdot \frac{90\,kN}{76.190\times 10^{3}\,kPa}

A = 1.640\times 10^{-3}\,m^{2}

The minimum allowable cross section area can be determined in terms of minimum allowable bolt diameter by means of this expression:

A = \frac{\pi}{4}\cdot D^{2}

The diameter is now cleared and computed:

D = \sqrt{\frac{4}{\pi}\cdot A}

D =\sqrt{\frac{4}{\pi}\cdot (1.640\times 10^{-3}\,m^{2})

D = 0.0457\,m

D = 45.7\,mm

The minimum allowable bolt diameter required to support an applied load of P = 450 kN is 45.7 milimeters.

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Explanation:

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If the old radiator is replaced with a new one that has longer tubes made of the same material and same thickness as those in th
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Answer: hello some parts of your question is missing attached below is the missing information

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