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3 years ago

For questions 1-10 fill in the blank with the letter of the description that best matches the term

Physics

2 answers:

lutik1710 [3]3 years ago

7 0

Answer:

A. The player responsible for the area between 2nd and 3rd base = 10.Shortstop

B. The player on the mound in the middle of the field = 8. Pitcher

C. The player positioned behind home plate = 1. Catcher

D. Occurs when a fielder fails to make a play = 9. Error

E. A ball hit outside the 1st or 3rd base lines = 6. Foul Ball

F. A pitch thrown outside the strike zone = 3. Ball

G. Area over home plate between the armpits and the knees = 4. Strike Zone

H. A pitcher can throw three of these to get a batter out = 2. Strike

I. The official behind home plate = 5. Umpire

J. When a pitcher throws out a base runner = 7. Pick-off

Explanation:

tigry1 [53]3 years ago

4 0

B-Pitcher C-Catcher H-Strike I-Umpire G-Strike Zone E- Foul Ball F- Ball J- Pick-off D-Error A- Shortstop. I think (Sorry for them being out of order. I had to break them down)

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Why do you think fixed boundaries ""flip"" waves and loose boundaries do not?

Vinil7 [7]

Answer:

When the obstacle is fixed, the law of action and reaction, makes the reflected wave is inverted.

When the obstacle is mobile, he mobile point, it moves in the direction of the wave, therefore there is no inversion of it.

Explanation:

Waves when they reach an obstacle behave like a shock, therefore if we use the conservation of momentum the wave must reverse its speed, this explains that the speed changes sign, the wave is reflected.

When the obstacle is fixed, the wave when it reaches the obstacle exerts a force on the point, by the law of action and reaction the point exerts on the wave a force of equal magnitude but in the opposite direction, this reaction force which makes the reflected wave is inverted.

When the obstacle is mobile, this is without friction, when the wave arrives it exerts a force on the mobile point, it moves in the direction of the wave, reaching the maximum amplitude of the incident wave, when it is reflected the point begins to go down along with the wave, therefore there is no inversion of it.

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A circuit has a current of 1.2 A. If the voltage decreases to one-third of its original amount while the resistance remains

beks73 [17]

G 3.6A I think so try this answer

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2 years ago

66. Calculate the displacement and velocity at times of (a) 0.500 s, (b) 1.00 s, (c) 1.50 s, and (d) 2.00 s for a ball thrown st

kozerog [31]

Answer:

a) t=1s

y = 10.1m

v=5.2m/s

b) t=1.5s

y =11.475 m

v=0.3m/s

c) t=2s

y =10.4 m

v=-4.6m/s (The minus sign (-) indicates that the ball is already going down)

Explanation:

Conceptual analysis

We apply the free fall formula for position (y) and speed (v) at any time (t).

As gravity opposes movement the sign in the equations is negative.:

y = vi*t - ½ g*t2 Equation 1

v=vit-g*t Equation 2

y: The vertical distance the ball moves at time t

vi: Initial speed

g= acceleration due to gravity

v= Speed the ball moves at time t

Known information

We know the following data:

Vi=15 m / s

$g =9.8 \frac{m}{s^{2} }$

t=1s ,1.5s,2s

Development of problem

We replace t in the equations (1) and (2)

a) t=1s

$y = 15*1 - ½ 9.8*1^{2}$ =15-4.9=10.1m

v=15-9.8*1 =15-9.8 =5.2m/s

b) t=1.5s

$y = 15*1.5 - ½ 9.8*1.5^{2}$ =22.5-11.025=11.475 m

v=15-9.8*1.5 =15-14.7=0.3m/s

c) t=2s

$y = 15*2 - ½ 9.8*2^{2}$ = 30-19.6=10.4 m

v=15-9.8*2 =15-19.6=-4.6m/s (The minus sign (-) indicates that the ball is already going down)

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Harman [31]

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