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Dmitrij [34]
2 years ago
11

Junior slides across home plate during a baseball game. If he has a mass of 115 kg, and the coefficient of kinetic friction betw

een him and the ground is 0.35, what is the force of friction acting on him?
Physics
2 answers:
uysha [10]2 years ago
6 0

Answer:

like the other person said, it is 395

Explanation:

tatyana61 [14]2 years ago
4 0

Weight Force of Junior = m g = 115kg x 9.81 m/s^2 = 1128.15N then compute for the friction force


Friction Force= WF x (coefficient of kinetic friction) = 1128.15N x 0.35 =  394.8525N or 395N

 

But you can compute in a straightway:

Solution:

= 115 x 9.81 x 0.35

= 394.85

= 395 N

 

It will still give the same results.

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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
The maximum tension that a 0.80 m string can tolerate is 15 N. A 0.35-kg ball attached to this string is being whirled in a vert
zimovet [89]

Answer:

v=5.86 m/s

Explanation:

Given that,

Length of the string, l = 0.8 m

Maximum tension tolerated by the string, F = 15 N

Mass of the ball, m = 0.35 kg

We need to find the maximum speed the ball can have at the top of the circle. The ball is moving under the action of the centripetal force. The length of the string will be the radius of the circular path. The centripetal force is given by the relation as follows :

F=\dfrac{mv^2}{r}

v is the maximum speed

v=\sqrt{\dfrac{Fr}{m}} \\\\v=\sqrt{\dfrac{15\times 0.8}{0.35}} \\\\v=5.86\ m/s

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2 years ago
A passenger compartment of a rotating amusement park ride contains a bench on which a book of mass
Basile [38]

a) 120 s

b) v = 0.052R [m/s]

Explanation:

a)

The period of a revolution in a simple harmonic motion is the time taken for the object in motion to complete one cycle (in this case, the time taken to complete one revolution).

The graph of the problem is missing, find it in attachment.

To find the period of revolution of the book, we have to find the time between two consecutive points of the graph that have exactly the same shape, which correspond to two points in which the book is located at the same position.

The first point we take is t = 0, when the position of the book is x = 0.

Then, the next point with same shape is at t = 120 s, where the book returns at x = 0 m.

Therefore, the period is

T = 120 s - 0 s = 120 s

b)

The tangential speed of the book is given by the ratio between the distance covered during one revolution, which is the perimeter of the wheel, and the time taken, which is the period.

The perimeter of the wheel is:

L=2\pi R

where R is the radius of the wheel.

The period of revolution is:

T=120 s

Therefore, the tangential speed of the book is:

v=\frac{L}{T}=\frac{2\pi R}{120}=0.052R

8 0
2 years ago
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SpyIntel [72]

Answer:

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Plugging in the relevant values, we have;

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v = √(GM/r)

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E = (GmM/r) - ½mv²

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