All categories

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)

You might be interested in

Spider-Man and Ned were testing the distance he could shoot his web depending on the angle at which he points his web shooter. H

Serga [27]

Answer:

3. Madison, WI

4. Quito, Equador

6. Main Sequence stars are becoming red giants

Explanation:

3 0

3 years ago

Read 2 more answers

A. List at least 5 examples of events or factors that could cause extinction. 

baherus [9]

Disease
Lack of natural reourses
war
asteroid
Polution.

8 0

3 years ago

Read 2 more answers

He throws a second ball (B2) upward with the same initial velocity at the instant that the first ball is at the ceiling. c. How

Gwar [14]

Answer:

hello your question has some missing parts

A juggler performs in a room whose ceiling is 3 m above the level of his hands. He throws a ball vertically upward so that it just reaches the ceiling.

answer : c) 0.39 sec

d) 2.25 m

e) 1.92 m/sec

Explanation:

The initial velocity of the first ball = 7.67 m/sec ( calculated )

Time required for first ball to reach ceiling = 0.78 secs ( calculated )

Determine how long after the second ball is thrown do the two balls pass each other

Distance travelled by first ball downwards when it meets second ball can be expressed as : d = 1/2 gt^2 = 9.8t^2 / 2

hence d = 4.9t^2 ----- ( 1 )

Initial speed of second ball = first ball initial speed = 7.67 m/sec

3 - d = 7.67t - 4.9t ---- ( 2 )

equating equation 1 and 2

3 = 7.67t therefore t = 0.39 sec

Determine how far the balls are above the Juggler's hands ( when the balls pass each other )

form equation 1 ;

d = 4.9 t^2 = 4.9 *(0.39)^2 = 0.75 m

therefore the height the balls are above the Juggler's hands is

3 - d = 3 - 0.75 = 2.25 m

determine their velocities when the pass each other

velocity = displacement / time

velocity = d / t = 0.75 / 0.39 sec = 1.92 m/sec

7 0

3 years ago

Paying off your debts will most likely

g100num [7]

I think the answer is d

4 0

2 years ago

At what speed, as a fraction of c , is a particle's total energy twice its rest energy

WINSTONCH [101]

The rest energy of a particle is
$E_0=m_0 c^2$
where $m_0$ is the rest mass of the particle and c is the speed of light.

The total energy of a relativistic particle is
$E=mc^2 = \frac{m_0 c^2}{ \sqrt{1- \frac{v^2}{c^2} } }$
where v is the speed of the particle.

We want the total energy of the particle to be twice its rest energy, so that
$E=2E_0$
which means:
$\frac{m_0c^2}{ \sqrt{1- \frac{v^2}{c^2} } }=2m_0 c^2$
$\frac{1}{ \sqrt{1- \frac{v^2}{c^2} } }=2$
From which we find the ratio between the speed of the particle v and the speed of light c:
$\frac{v}{c}= \sqrt{1- (\frac{1}{2})^2 } =0.87$
So, the particle should travel at 0.87c in order to have its total energy equal to twice its rest energy.

3 0

3 years ago