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Yuri [45]
3 years ago
14

Calculate the following forces for a box with a mass=20.0 kg sitting on an incline of 29.0 degrees and a coefficient of friction

which is 0.455. Calculate the parallel force, normal force, and frictional force.
Physics
1 answer:
Anna007 [38]3 years ago
3 0
Vertical force on the box=mg 
<span>the component of gravity parallel=mg*SinTheta </span>
<span>the component of gravity normal=mg*CosTheta </span>
<span>frictional force up the plane: mg*cosTheta*mu max, but if it is sitting still, it is equal and opposite to mg*cosTheta (it cannot be greater than this or it would go up the plane).</span>
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Answer:

There is no short answer.

Explanation:

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The electric forces on both beads are equal.
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3 years ago
If a rock is thrown upward on the planet mars with a velocity of 11 m/s, its height (in meters) after t seconds is given by h =
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For most of these problems, you're being asked the velocity of the rock as a function of t, while you've been given the position as a function of t. So first calculate the first derivative of the position function using the power rule. 
y = 11t - 1.86t^2 
y' = 11 - 3.72t 
Now that you have the first derivative, it will give you the velocity as a function of t. 
(a) Velocity after 2 seconds. 
y' = 11 - 3.72t 
y' = 11 - 3.72*2 = 11 - 7.44 = 3.56 
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(b) Velocity after a seconds. 
y' = 11 - 3.72t 
y' = 11 - 3.72a  
So the answer is 11 - 3.72a  
(c) Use the quadratic formula to find the zeros for the position function y = 11t-1.86t^2. Roots are t = 0 and t = 5.913978495. The t = 0 is for the moment the rock was thrown, so the answer is t = 5.9 seconds.  
(d) Plug in the value of t calculated for (c) into the velocity function, so: 
y' = 11 - 3.72a
 y' = 11 - 3.72*5.913978495
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 y' = -11 
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Answer:

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