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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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2 years ago

10). Three - fourth of the volume of a block which is floating in water extends above the surface

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3 years ago

At noon, ship A is 110 km west of ship B. Ship A is sailing east at 20 km/h and ship B is sailing north at 15 km/h. How fast is

Katyanochek1 [597]

Answer:

$4.47\ \text{km/h}$

Explanation:

$\dfrac{da}{dt}$ = Rate at which the distance between A and starting point of B is changing = -20 km/h

$\dfrac{db}{dt}$ = Rate at which the distance of B is changing = 15 km/h

$\dfrac{dc}{dt}$ = Rate at which the distance between A and B is changing

Time after which the rate at which the distance between A and B is changing is 4 hours

Distance covered by A in 4 hours = $20\times 4=80\ \text{km}$

a = Distance remaining to the start point of B = $110-80=30\ \text{km}$

b = Distance covered by B in 4 hours = $15\times 4=60\ \text{km}$

Distance between A and B after 4 hours

$c=\sqrt{a^2+b^2}\\\Rightarrow c=\sqrt{30^2+60^2}\\\Rightarrow c=67.08\ \text{km}$

$c^2=a^2+b^2$

Differentiating with respect to time we get

$c\dfrac{dc}{dt}=a\dfrac{da}{dt}+b\dfrac{db}{dt}\\\Rightarrow \dfrac{dc}{dt}=\dfrac{a\dfrac{da}{dt}+b\dfrac{db}{dt}}{c}\\\Rightarrow \dfrac{dc}{dt}=\dfrac{30\times -20+60\times 15}{67.08}\\\Rightarrow \dfrac{dc}{dt}=4.47\ \text{km/h}$

The rate at which the distance between the ships is changing at 4 PM is $4.47\ \text{km/h}$ .

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