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2 years ago

An archery contestant

Physics

2 answers:

storchak [24]2 years ago

8 0

Answer: A

Explanation:

just olya [345]2 years ago

5 0

Answer:

AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA the answer is..................... aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA

Explanation:

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Say you want to make a sling by swinging a mass M of 1.9 kg in a horizontal circle of radius 0.042 m, using a string of length 0

padilas [110]

Answer: T= 715 N

Explanation:

The only external force (neglecting gravity) acting on the swinging mass, is the centripetal force, which. in this case, is represented by the tension in the string, so we can say:

T = mv² / r

At the moment that the mass be released, it wil continue moving in a straight line at the same tangential speed that it had just an instant before, which is the same speed included in the centripetal force expression.

So the kinetic energy will be the following:

K = 1/2 m v² = 15. 0 J

Solving for v², and replacing in the expression for T:

T = 1.9 Kg (3.97)² m²/s² / 0.042 m = 715 N

3 0

2 years ago

You kick a soccer ball across a field the ball travels across the field and slowly comes to a stop. Since the ball slowly comes

marysya [2.9K]

Answer:

It slowly decreases and the friction acting on it slowing it down becomes the bigger net force, if that makes sense :)

Explanation:

6 0

2 years ago

PLEASE HELP 100 POINTS

tia_tia [17]

Answer:

okay, So Um,

Explanation:

BTW, There are only 50 points, brainly subtracts half of the points

8 0

2 years ago

A student places an object with a mass of m on a disk at a position r from the center of the disk. The student starts rotating t

koban [17]

Answer:

The coefficient of static friction between the object and the disk is 0.087.

Explanation:

According to the statement, the object on the disk experiments a centrifugal force due to static friction. From 2nd Newton's Law, we can represent the object by the following formula:

$\Sigma F_{r} = \mu_{s}\cdot N = m\cdot \frac{v^{2}}{R}$ (1)

$\Sigma F_{y} = N - m\cdot g = 0$ (2)

Where:

$N$ - Normal force from the ground on the object, measured in newtons.

$m$ - Mass of the object, measured in newtons.

$g$ - Gravitational acceleration, measured in meters per square second.

$v$ - Linear speed of rotation of the disk, measured in meters per second.

$R$ - Distance of the object from the center of the disk, measured in meters.

By applying (2) on (1), we obtain the following formula:

$\mu_{s}\cdot m\cdot g = m\cdot \frac{v^{2}}{R}$

$\mu_{s} = \frac{v^{2}}{g\cdot R}$

If we know that $v = 0.8\,\frac{m}{s}$ , $g = 9.807\,\frac{m}{s^{2}}$ and $R = 0.75\,m$ , then the coefficient of static friction between the object and the disk is: