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3 years ago

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A 10-kg block is suspended in air by a single string that passes through a pulley and is attached on the other side to a 35-kg b

lock that is on the verge of sliding on the ground. What is the coefficient of static friction between the larger block and the ground? (Note: The magnitude of the vertical force the small block exerts on the large block is 87 N.)

1 answer:

s2008m [1.1K]3 years ago

4 0

Answer:

0.286

Explanation:

Let g = 10m/s2, and assume the pulley is frictionless, the weight by the 10kg block would exert a force on the 35kg block on the ground

Weight of the 10 kg-block = mg = 10 * 10 = 100 N

This force is balanced by static friction force from the ground due to the normal force of the 35kg block. So the friction force would also be 100N

The weight and normal force of the 35kg block is N = Mg = 35*10 = 350 N

The coefficient of the friction force if friction force divided by normal force:

$\mu = \frac{F_f}{N} = \frac{100}{350} = 0.286$

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A wall that is 12 feet wide and 10 feet high is to be painted. A blackboard that is 5 feet wide and 3 feet high is affixed to th

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Answer:

The area of the portion of the wall that will be painted is 105 square feet

Explanation:

Total area of wall = 12 × 10 = 120 ft²

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So, the area of the wall to be painted = Total area of wall - Area of the wall that will not be painted = 120 - 15 = 105 ft²

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4 years ago

A major league baseball pitcher throws a pitch that follows these parametric equations: x(t) = 142t y(t) = –16t2 + 5t + 5. The t

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Answer:

(a) $x'(t)= 142$

(b) 142

(c) $y'(t)= -32t+5$

(d) 96.8 mph

(e) 0.426 s

(f) 0.061 rad

Explanation:

Velocity is a time-derivative of position.

(a) $x(t) = 142t$

$x'(t)= 142$

(b) Since $x'(t)= 142$ is independent of $t$ , it follows it was constant throughout. Hence, at any point or time, the horizontal velocity is 142.

(c) $y(t) = - 16t^2+5t+5$

$y'(t)= -32t+5$

(d) When it passes the home plate, the ball has travelled 60.5 ft (from the question). This is horizontal, so it is equivalent to $x(t)$ .

$x(t)= 142t = 60.5$

$t=\dfrac{60.5}{142}= 0.426$ .

In this time, the vertical velocity, $y'(t)$ is

$y(t)= -32\times0.426+5 = -8.632$

The speed of the ball at thus point is $s=\sqrt{142^2+(-8.632)^2}=142$ ft/s

To convert this to mph, we multiply the factor 3600/5280

$s=142\times\dfrac{3600}{5280}=96.8 \text{ mph}$

(e) The time has been determined from (d) above.

$t= 0.426$

(f) This angle is given by

$\theta=\tan^{-1}\dfrac{y'(t)}{x'(t)}$

$\theta=\tan^{-1}\dfrac{-8.632}{142}=\tan^{-1}-0.0607=3.47$ (Note here we are considering the acute angle so we ignore the negative sign)

In radians, this is

$\theta=3.47\times\dfrac{\pi}{180}=0.061 \text{ rad}$

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4 years ago

A stone is dropped at t = 0. A second stone, with 6 times the mass of the first, is dropped from the same point at t = 59 ms. (a

xxTIMURxx [149]

Answer: $y_{com}=0.707 m$

$v_{com}=3.713 m/s$

Explanation:

Given

first stone mass is m

second stone mass is 6 m

distance traveled by first stone in 430 ms

$y_1=ut+\frac{at^2}{2}$

$y_1=0+\frac{g(0.43)^2}{2}$

$y_1=0.9069 m$

Distance traveled by stone 2 in t=430-59=371 ms

$y_2=ut+\frac{at^2}{2}$

$y_2=0+\frac{g(0.0.371)^2}{2}$

$y_2=0.674 m$

velocity of first stone after t=0.43 s

$v_1=u+at$

$v_1=0+9.8\times 0.43=4.214 m/s$

velocity of second stone after t=0.371 s

$v_2=u+at$

$v_2=0+9.8\times 0.371=3.63 m/s$

Position of Center of mass of system

$y=\frac{y_1m_1+y_2m_2}{m_1+m_2}$

$y=\frac{0.9069\times m+0.674\times 6m}{m+6m}$

$y=\frac{4.95m}{7m}=0.707 m$

Velocity of COM

$v_{com}=\frac{v_1m_1+v_2m_2}{m_1+m_2}$

$v_{com}=\frac{4.214\times m+3.63\times 6m}{m+6m}$

$v_{com}=3.713 m/s$

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4 years ago

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LUCKY_DIMON [66]

The Mechanical Energy is given by the sum of the all energies of the system.

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3 years ago