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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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Two identical pebbles are dropped. The first is dropped from a height of 256 feet and the second is dropped from a height of 400
ehidna [41]

Answer:

4.022 seconds and 4.99 seconds

Explanation:

Hello!

The free fall of the stone corresponds to a uniformly varied rectilinear movement

d=V_0*t+1/2*g*t^2

Being a free fall the initial speed is zero.

The distance is positive when considered in the same direction and direction as acceleration and speed.

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79.25 m=0 m⁄s*t+1/2*9,8 m⁄s^2 *t^2

t = 4.022 seconds

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success with your homework!

Download pdf
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A bob of mass m = 0.250 kg is suspended from a fixed point with a massless string of length L = 22.0 cm. You will investigate th
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To solve the problem, it is necessary to use the concepts of gravitational force, centripetal force and trigonometric components that can be extrapolated from the statement.

By definition we know that the Force of Gravity is given by

F_g=mg

Where,

m= Mass

g = Gravitational Acceleration

The centripetal force is given by,

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

Where,

m = Mass

v = Velocity

R = Radius

For the case described in the problem, the Force of gravity the net component would be given by sin?, While for the centripetal force the net component is in the horizontal direction, therefore it corresponds to the cos\theta

Then,

F_g = mg sin\theta

F_c = \frac{mv^2}{r}cos\theta

From the radius we have its length but not the net height, which would be given by

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So equating the equations we have to

F_g = F_c

mg sin\theta=\frac{mv^2}{r}cos\theta

mg sin\theta=\frac{mv^2}{Lsin\theta}cos\theta

Re-arrange to find v,

v = \sqrt{\frac{gLsin^2\theta}{cos\theta}}

Replacing with our values

v = \sqrt{\frac{(9.8)(22*10^{-2})(sin^2 24)}{cos24}}

v = 0.624

Therefore the tangential velocity of the mass is 0.624m/s

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4 years ago
A 40-kg worker climbs a ladder upwards for 15m. What work was done during their climb upwards?
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Answer:

Explanation:

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2 years ago
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