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ASHA 777 [7]
2 years ago
10

Starting from rest, a solid sphere rolls without slipping down an incline plane. at the bottom of the incline, what does the ang

ular velocity of the sphere depend upon? check all that apply. check all that apply. the angular velocity depends upon the length of the incline. the angular velocity depends upon the mass of the sphere. the angular velocity depends upon the radius of the sphere. the angular velocity depends upon the height of the incline
Physics
1 answer:
vfiekz [6]2 years ago
8 0
Starting from rest, a solid sphere rolls without slipping down an incline plane. at the bottom of the incline, what does the angular velocity of the sphere depend upon? check all that apply. check all that apply. the angular velocity depends upon the length of the incline. the angular velocity depends upon the mass of the sphere. the angular velocity depends upon the radius of the sphere. the angular velocity depends upon the height of the incline
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The two hot-air balloons in the drawing are 48.2m and 61.0 m above the ground.A person in the left balloon observes that the rig
mafiozo [28]

Answer:

The horizontal distance x between the two balloons is 54.15 m

Explanation:

The diagram described as obtained online is presented in the image attached to this solution.

Let the horizontal distance between the two balloons be x

Difference in height (vertical distance) between the two balloons = 61 - 48.2 = 12.8 m

Using trigonometric relations, it is evident that

Tan 13.3° = 12.8/x

x = 12.8/tan 13.3° = 12.8/0.2364 = 54.15 m

4 0
2 years ago
You may have noticed runaway truck lanes while driving in the mountains. These gravel-filled lanes are designed to stop trucks t
kati45 [8]

Answer:

The  coefficient of kinetic friction  \mu_k =  0.724

Explanation:

From the question we are told that

   The  length of the lane is  l =  36.0 \  m

    The speed of the truck is  v  =  22.6\  m/s

     

Generally from the work-energy theorem we have that  

    \Delta KE  =   N  *  \mu_k *  l

Here N  is the normal force acting on the truck which is mathematically represented as

     \Delta KE is the change in kinetic energy which is mathematically represented as

        \Delta KE =  \frac{1}{2} *  m *  v^2

=>     \Delta KE =  0.5  *  m *  22.6^2

=>      \Delta KE =  255.38m

        255.38m =    m *  9.8  *  \mu_k *   36.0

=>     255.38  =    352.8  *  \mu_k

=>   \mu_k =  0.724

 

6 0
2 years ago
Two objects, X and Y, move toward one another and eventually collide. Object X has a mass of 2M and is moving at a speed of 2v0
Sergeu [11.5K]

Answer:

c. The force exerted by X on Y is F to the right, and the force exerted by Yon X is F to the left.

Explanation:

If we take both objects as one single system, during the collision, assuming no external forces acting, the only forces present are the one that object X exerts on object Y, and the force that object Y does on object X.

These two forces, according to Newton's 3rd Law, form an action-reaction pair, and consequently, are equal in magnitude, acting in opposite directions.

As the object X is moving to the right, the force that produces, (F), is in the same direction (on Y), while for object Y, moving to the left, the force that produces (F also in magnitude) is in the same direction (on X), so the right answer is c.

The effect of the forces is different, due to masses are different, according Newton's 2nd Law.

6 0
2 years ago
Suppose you first walk 12.0 m in a direction 20 owest of north and then 20.0 m in a direction 40.0osouth of west. How far are yo
alexgriva [62]

Answer:

R=19.5m

\theta = 4.65° S of W

Explanation:

Refer the attached fig.

displacement  of the x and y components

x-component displacement is (R_{x}) = A_{x}+B_{x}

= A \sin(20°) + B \cos(40°)

= -12.0\sin(20°) + 20.0\cos(40°)

= -19.425m

x-component displacement is (R_{y}) = A_{y}+ B_{y}

=  A \cos(20°) - B \sin(40°)

= 12.0\cos(20°) - 20.0\sin(40°)

= -1.579

resultant displacement

∴

R = \sqrt{R_{x}^{2} +R_{y}^{2} }  }

=\sqrt{(-19.425)^{2}+(-1.579)^{2}  }

=19.5m

\theta = \tan^{-1}\left | \frac{R_{x}}{R_{y}} \right |

\theta = \tan^{-1}\left | \frac{1.579}{19.425} \right |

\theta = 4.65° S of W

6 0
2 years ago
A car enters a 300-m radius horizontal curve on a rainy day when the coefficient of static friction between its tires and the ro
Vlada [557]

To solve this problem it is necessary to take into account the concepts related to Centripetal Force and Friction Force.

In the case of the centripetal force, we know that it is defined as

F_c = \frac{mv^2}{R}

Where,

m=mass

v= velocity

r= Radius

In the case of the Force of Friction we have to,

F_f = \mu m*g

Where,

\mu =Friction Constant

m= mass

g= gravity

According to the information given, the centripetal force must be less than or equal to the friction force to stay on the road, in this way

\frac{mv^2}{R} \leq \mu m*g

Re-arrange to find the velocity,

v \leq \sqrt{\mu gR}

v \leq \sqrt{(0.6)(9.8)(300)}

v \leq 42m/s

Therefore la velocidad del carro debe ser igual o menor a 42m/s para mantenerse en el camino

6 0
2 years ago
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