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quester [9]
3 years ago
5

A golfer starts with the club over her head and swings it to reach maximum speed as it contacts the ball. Halfway through her sw

ing, when the golf club is parallel to the ground, does the acceleration vector of the club head point straight down, parallel to the ground, approximately toward the golfer's shoulders, approximately toward the golfer's feet, or toward a point above the golfer's head?
Physics
1 answer:
lara [203]3 years ago
3 0

Answer:

a) parallel to the ground True

c) parallel to the ground towards man True

Explanation:

To examine the possibilities, we propose the solution of the problem.

Let's use Newton's second law

      F = m a

The force is exerted by the arm and the centripetal acceleration of the golf club, which in this case varies with height.

In our case, the stick is horizontal in the middle of the swing, for this point the centripetal acceleration is directed to the center of the circle or is parallel to the arm that is also parallel to the ground;

Ask the acceleration vector

a) parallel to the ground True

b) down. False

c) parallel to the ground towards True men

d) False feet

e) the head. False

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One end of a horizontal spring with force constant 130.0 N/m is attached to a vertical wall. A 3.00 kg block sitting on the floo
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Answer:

a) v = 0

b) The aceleration is 1.41 m/s^{2}

c) The block is accelerating away from the wall.

Explanation:

First, you need to think about the effect this constant force is causing in the spring: it causes a displacement in the equilibrium point of the system, therefore we need to know where it sits now:

At equilibrium no movement is present reducing friction to 0:

\sum{F} = 0 = F_{spring} - F_{external}

F_{spring} = F_{external}

Kx = F_{external}

x = \frac{F_{external}}{K}=\frac{88}{130}=0.68m=68cm

This means that the spring can be compressed with the single force up to 68 cm, Any further compression will cause an unbalanced system and the occilation of the mass.

The spring can't be compressed by the given force to 80 cm, therefore it must have been compressed by another force and then released.

In this case, the instantanous speed is 0, since the block has just been released.

In the same instant we can stimate the free body diagram of forces by the next two equations:

\sum_y{F}={F_N-W}=0\\\sum_x{F}={F_{spring}-F_{external}-F_{friction}}=ma

For the y axis:

F_N = W = mg = 3*9.8 = 29.4N

To calculate the force of friction:

F_{friction} = \mu_k F_N=0.4*29.4 = 11.76N

Therefore for x axis:

{Kx-F_{external}-F_{friction}}=ma

a = \frac{130*0.8-88-11.76}{3} = \frac{104-88-11.76}{3}=\frac{4.24}{3}=1.41\frac{m}{s^2}

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