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3 years ago

12

A 9 -slug mass hangs by a rope from the ceiling. Using the standard value of gravitational acceleration g = 32.2 fts 2, what is

the tension in the rope?

1 answer:

kap26 [50]3 years ago

3 0

Answer:

For the Mass to hang without falling

It simply means that the Tension that supports the weight force has to be Equal to or greater than it.

T=W

T= mg

m= 9 slugs

g=32.2ft/s².

T= 9 x 32.2

T= 289.8 slug.ft/s².

Or in normal units

1 Slug = 14.59Kg

1ft = 0.3048m

m= 9 slugs = 131.345kg

g =32.2ft/s² = 9.8m/s²

T= 9.8 x 131.345

T= 1,287.18N.

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1 year ago

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 10-lb block is pressed against the spring so as to compress it 2 ft when it is at a point A. If the plane is smooth, determi

leva [86]

Answer:

The distance measure from the wall = 36ft

Explanation:

Given Data:

w = 10

g =32.2ft/s²

x = 2

Using the principle of work and energy,

T₁ +∑U₁-₂ = T₂

0 + 1/2kx² -wh = 1/2 w/g V²

Substituting, we have

0 + 1/2 * 100 * 2² - (10 * 3) = 1/2 * (10/32.2)V²

170 = 0.15528V²

V² = 170/0.15528

V² = 1094.796

V = √1094.796

V = 33.09 ft/s

But tan ∅ = 3/4

∅ = tan⁻¹3/4

= 36.87°

From uniform acceleration,

S = S₀ + ut + 1/2gt²

It can be written as

S = S₀ + Vsin∅*t + 1/2gt²

Substituting, we have

0 = 3 + 33.09 * sin 36.87 * t -(1/2 * 32.2 *t²)

19.85t - 16.1t² + 3 = 0

16.1t² - 19.85t - 3 = 0

Solving it quadratically, we obtain t = 1.36s

The distance measure from the wall is given by the formula

d = VCos∅*t

Substituting, we have

d = 33.09 * cos 36. 87 * 1.36

d = 36ft

5 0

3 years ago

Read 2 more answers