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Assume that the force exerted on each crutch by the ground is directed along the crutch, as the force vectors in the drawing ind

olya-2409 [2.1K]

Answer: 43 degrees

Explanation:

The force of the crutch can be broken into components. The horizontal component of F is the static friction force keeping the crutch from sliding. The vertical component is opposing the weight and is the Normal force. Using the orientation of the angle q, we have the following

fs = Fx = F sin (angle (tita))

N = Fy = F cos (angle(tita))

Maximum angle implies maximum static friction

Therefore,

fsmax = UsN = Us x F cos(angle tita)

Where U = miyu

F sin(angle tita) = Us x Fcos (angle tita)

Sin (angle tita) / cos (angle tita) = Us

Therefore, tan (angle tita) = Us

Angle tita = tan^-1(Us) = tan^-1 (0.931) = 42.95 degrees = 43 degrees

There for Angle tita Max = 43 degrees

7 0

3 years ago

A uniform string of length 10.0 m and weight 0.32 N is attached to the ceiling. A weight of 1.00 kN hangs from its lower end. Th

Juliette [100K]

Answer: 0.0180701 s

Explanation:

Given the following :

Length of string (L) = 10 m

Weight of string (W) = 0.32 N

Weight attached to lower end = 1kN = 1×10^3

Using the relation:

Time (t) = √ (weight of string * Length) / weight attached to lower end * acceleration due to gravity

g = acceleration due to gravity = 9.8m/s^2

Weight of string = 0.32N

Time(t) = √ (0.32 * 10) / [(1*10^3) * (9.8)]

Time = √3.2 / 9800

= √0.0003265

= 0.0180701s

5 0

3 years ago

In Example 2.12, two circus performers rehearse a trick in which a ball and a dart collide. Horatio stands on a platform 6.4 m a

pickupchik [31]

Answer:

time of collision is

t = 0.395 s

$h = 5.63 m$

so they will collide at height of 5.63 m from ground

Explanation:

initial speed of the ball when it is dropped down is

$v_1 = 0$

similarly initial speed of the object which is projected by spring is given as

$v_2 = 16.2 m/s$

now relative velocity of object with respect to ball

$v_r = 16.2 m/s$

now since we know that both are moving under gravity so their relative acceleration is ZERO and the relative distance between them is 6.4 m

$d = v_r t$

$6.4 = 16.2 t$

$t = 0.395 m$

Now the height attained by the object in the same time is given as

$h = v_2 t - \frac{1}{2}gt^2$

$h = 16.2(0.395) - \frac{1}{2}(9.81).395^2$

$h = 5.63 m$

so they will collide at height of 5.63 m from ground

4 0

3 years ago

Suppose you observed the equation for a traveling wave to be y(x, t) = A cos(kx − ????t), where its amplitude of oscillations wa

OLga [1]

Answer:

Explanation:

The equation for a traveling wave as expressed as y(x, t) = A cos(kx − $\omega$ t) where An is the amplitude f oscillation, $\omega$ is the angular velocity and x is the horizontal displacement and y is the vertical displacement.

From the formula; $k =\frac{2\pi x}{\lambda} \ and \ \omega = 2 \pi f$ where;

$\lambda \ is\ the \ wavelength \ and\ f \ is\ the\ frequency$

Before we can get the transverse speed, we need to get the frequency and the wavelength.

frequency = 1/period

Given period = 2/15 s

Frequency = $\frac{1}{(2/15)}$

frequency = 1 * 15/2

frequency f = 15/2 Hertz

Given wavelength $\lambda$ = 2m

Transverse speed $v = f \lambda$

$v = 15/2 * 2\\\\v = 30/2\\\\v = 15m/s$

Hence, the transverse speed at that point is 15m/s

8 0

3 years ago

A hailstone traveling with a velocity of 43 meters/second comes to a virtual stop 0.28 seconds after hitting water. What is the

lisabon 2012 [21]

Magnitude of acceleration = (change in speed) / (time for the change) .

Change in speed = (ending speed) - (starting speed)

= zero    - (43 m/s)

=    -43 m/s .

Magnitude of acceleration = (-43 m/sec) / (0.28 sec)

      = (-43 / 0.28) (m/sec) / sec

      =    153.57... m/s²

      = 1.5... x 10² m/s² .

5 0

3 years ago

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