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3 years ago

A solid ball is released from rest and slides down a hillside that slopes downward at 65.0" from the horizontal

1 answer:

3 0

Setting reference frame so that the x axis is along the incline and y is perpendicular to the incline
X: mgsin65 - F = mAx 
Y: N - mgcos65 = 0 (N is the normal force on the incline) N = mgcos65 (which we knew) 
Moment about center of mass: 
Fr = Iα 
Now Ax = rα 
and F = umgcos65 
mgsin65 - umgcos65 = mrα -------------> gsin65 - ugcos65 = rα (this is the X equation m's cancel) 
umgcos65(r) = 0.4mr^2(α) -----------> ugcos65(r) = 0.4r(rα) (This is the moment equation m's cancel) 
ugcos65(r) = 0.4r(gsin65 - ugcos65) ( moment equation subbing in X equation for rα) 
ugcos65 = 0.4(gsin65 - ugcos65) 
1.4ugcos65 = 0.4gsin65 
1.4ucos65 = 0.4sin65 
u = 0.4sin65/1.4cos65 
u = 0.613

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nika2105 [10]

Answer:

Part 1)

Boat A will win the race

Part 2)

Boat A will win the race by 48 km as the 2nd boat will reach the other end while boat A will just touches the finish line

Part 3)

average velocity must be zero

Explanation:

As we know that the distance moved by the boat is given as

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now the time taken by the boat to move to and fro is given as

$t = \frac{d}{v}$

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Time taken by Boat B to cover the distance

$t = \frac{48}{24} + \frac{48}{72}$

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Part 1)

Boat A will win the race

Part 2)

Boat A will win the race by 48 km as the 2nd boat will reach the other end while boat A will just touches the finish line

Part 3)

Since the displacement of Boat A is zero

so average velocity must be zero

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3 years ago

Shay reacts solid zinc and aqueous copper sulfate to form aqueous zinc sulfate and solid copper. If he reacts 10.1 grams of zinc

zhenek [66]

Answer:

Now since mass of reactant is equal to mass of the product after the reaction so we can say that mass conservation is applicable here

Explanation:

As we know that zinc reacts with copper sulfate

so the reaction is given as

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so here we have

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$CuSO_4 = 18.6 g$

$ZnSO_4 = 20 g$

$Cu = 8.7 g$

Now total mass of reactant is given as

$M_1 = 10.1 + 18.6 = 28.7 g$

Mass of the product is given as

$M_2 = 20 + 8.7 = 28.7 g$

Now since mass of reactant is equal to mass of the product after the reaction so we can say that mass conservation is applicable here

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Dennis_Churaev [7]

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Explanation:

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