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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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If a skier travels at a constant velocity downhill, this means the forces on her are: ................. A. Zero B. Negligible C.
Vlad1618 [11]

Answer:

F = 0

Explanation:

The net force acting on an object is given by the product of mass and acceleration. We know that acceleration is equal to the rate of change of velocity.

Net force,

F = ma

F=\dfrac{m(v-u)}{t}

The skier is traveling at a constant velocity, it means there is no change in velocity i.e. acceleration is equal to 0. Hence, the force on her is 0.

7 0
3 years ago
(II) A 20.0-kg box rests on a table. (a) What is the weight of the box and the normal force acting on it? (b) A 10.0-kg box is p
Bogdan [553]

*Fig is Attached with answer*

Answer:

(a)   Weight = Normal Force = 196.2 N

(b)   Normal force table on 20 kg box = 294.3 N

       Normal force 20 kg box on 10 kg box = 98.1 N

Explanation:

(a)   Mass = m = 20 kg                           g = 9.81 m/s²

      Weight = w = mg

                      w = 20 × 9.81

                      w = 196.2 N

As the box rests on the table so, normal force (NF) must be equal to the weight of the box.

                    NF = w = 196.2 N

(b)

m₁ = 20 kg                      m₂ = 10 kg

total mass = M = 30 kg

Total Weight = W = Mg  

                             = 30 × 9.81

                             = 294.3 N

As both the boxes rest on the table so, normal force (NF) must be equal to the total weight of the boxes.

                          NF = W = 294.3 N

Weight of 10 kg box = 10 × 9.81 = 98.1 N

As the 10 kg box is placed on the top of 20 kg box, So 20 kg box must exert a normal force that is equal to the weight of 10 kg box.

              Normal Force = Weight of 10 kg box = 98.1 N

6 0
3 years ago
anyone know where I can find stuff (answer key, tables, etc.) for my Newton's Law of Motion lab report on edge2020? need answers
Zolol [24]
Usually the full tables are found on quizlet or quizzes
7 0
2 years ago
Light waves
irga5000 [103]

Answer:

a

Explanation:

Because I searched it up

7 0
3 years ago
Read 2 more answers
A small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. When the amplitude of t
Maslowich

Answer:

a) The time taken to travel from 0.18 m to -0.18m when the amplitude is doubled = 2.76 s

b) The time taken to travel from 0.09 m to -0.09 m when the amplitude is doubled = 0.92 s

Explanation:

a) The period of a simple harmonic motion is given as T = (1/f) = (2π/w)

It is evident that the period doesn't depend on amplitude, that is, it is independent of amplitude.

Hence, the time it would take the block to move from its amplitude point to the negative of the amplitude point (0.09 m to -0.09 m) in the first case will be the same time it will take the block to move from its amplitude point to negative of the amplitude point in the second case (0.18 m to -0.18 m).

Hence, time taken to travel from 0.18 m to -0.18m when the amplitude is doubled is 2.76 s

b) Now that the amplitude has been doubled, the time taken to move from amplitude point to the negative amplitude point in simple harmonic motion, just like with waves, is exactly half of the time period.

The time period is defined as the time taken to complete a whole cycle and a while cycle involves movement from the amplitude to point to the negative amplitude point then fully back to the amplitude point. Hence,

0.5T = 2.76 s

T = 2 × 2.76 = 5.52 s

Note that the displacement of a body undergoing simple harmonic motion from the equilibrium position is given as

y = A cos wt (provided that there's no phase difference, that is, Φ = 0)

A = amplitude = 0.18 m

w = (2π/5.52) = 1.138 rad/s

When y = 0.09 m, the time = t₁₂ = ?

0.09 = 0.18 Cos 1.138t₁ (angles in radians)

Cos 1.138t₁ = 0.5

1.138t₁ = arccos (0.5) = (π/3)

t₁ = π/(3×1.138) = 0.92 s

When y = -0.09 m, the time = t₂ = ?

-0.09 = 0.18 Cos 1.138t₂ (angles in radians)

Cos 1.138t₂ = -0.5

1.138t₂ = arccos (-0.5) = (2π/3)

t₂ = 2π/(3×1.138) = 1.84 s

Time taken to move from y = 0.09 m to y = -0.09 m is then t = t₂ - t₁ = 1.84 - 0.92 = 0.92 s

Hope this Helps!!!

3 0
3 years ago
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