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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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What are three observations an astronaut might make while viewing Russia at night

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3 years ago

A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3. 0 m is initially at rest. A 20 kg boy approaches the

Margaret [11]

Hi there!

$\boxed{\omega = 0.38 rad/sec}$

We can use the conservation of angular momentum to solve.

$\large\boxed{L_i = L_f}$

Recall the equation for angular momentum:

$L = I\omega$

We can begin by writing out the scenario as a conservation of angular momentum:

$I_m\omega_m + I_b\omega_b = \omega_f(I_m + I_b)$

$I_m$ = moment of inertia of the merry-go-round (kgm²)

$\omega_m$ = angular velocity of merry go round (rad/sec)

$\omega_f$ = final angular velocity of COMBINED objects (rad/sec)

$I_b$ = moment of inertia of boy (kgm²)

$\omega_b$ = angular velocity of the boy (rad/sec)

The only value not explicitly given is the moment of inertia of the boy.

Since he stands along the edge of the merry go round:

$I = MR^2$

We are given that he jumps on the merry-go-round at a speed of 5 m/s. Use the following relation:

$\omega = \frac{v}{r}$

$L_b = MR^2(\frac{v}{R}) = MRv$

Plug in the given values:

$L_b = (20)(3)(5) = 300 kgm^2/s$

Now, we must solve for the boy's moment of inertia:

$I = MR^2\\I = 20(3^2) = 180 kgm^2$

Use the above equation for conservation of momentum:

$600(0) + 300 = \omega_f(180 + 600)\\\\300 = 780\omega_f\\\\\omega = \boxed{0.38 rad/sec}$

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3 years ago

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3 years ago

Read 2 more answers

A monkey running through the jungle goes 0.198 km straight to the East, then turns 15.8° deflection from straight East toward th

inn [45]

Answer:

$|\vec r|=339.82\ m$

$\theta=-6.67^o$

Explanation:

<u>Displacement </u>

It's a vector magnitude that measures the space traveled by a particle between an initial and a final position. The total displacement can be obtained by adding the vectors of each individual displacement. In the case of two displacements:

$\vec r=\vec r_1+\vec r_2$

Given a vector as its polar coordinates (r,\theta), the corresponding rectangular coordinates are computed with

$x=rcos\theta$

$y=rsin\theta$

And the vector is expressed as

$\vec z==$

The monkey first makes a displacement given by (0.198 km,0°). The angle is 0 because it goes to the East, the zero-reference for angles. Thus the first displacement is

$\vec r_1==\ km=\ m$