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IrinaK [193]
3 years ago
14

if you toss a stick into the air, it appears to wobble all over the place. Specifically, about what place does it wobble?

Physics
1 answer:
12345 [234]3 years ago
3 0
Foolish, ambiguous question, with no correct answer, but at a wild guess, it MIGHT wobble about it own axis????
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A sled is on an icy (frictionless) slope that is 30° above the horizontal. When a 40-N force, parallel to the incline and direct
son4ous [18]

Answer: 5.8kg

Explanation:

F - mg sinΦ = ma

Given that

M=?

g= 9.8

Φ= 30

a= 2

F= 40 then

40 - 9.8 * m * sin 30 = 2 * m

40 = 2 * m + 9.8 * m * sin 30

40 = (2 + 9.8 * sin 30) m

m = 40 / (2 + 9.8 * 0.5)

m = 40 / (2 + 4.9)

m = 40 / 6.9

m = 5.797kg

m = 5.8kg

4 0
3 years ago
The reactivity of an atom arises primarily from ___.
kherson [118]

Answer:

E. existence of unpaired electrons in the outermost shell.

Explanation:

7 0
3 years ago
What happens when two aqueous solutions are combined in a precipitation reaction and no precipitate is formed?   A. The solution
inysia [295]

There's no reaction for me.

5 0
3 years ago
A sport car moving at constant speed travels 110m in 5.0 s. if it then brakes and comes to a stop in 4.0 s, what is the magnitud
stiks02 [169]
D = 110 m,  t = 5 s
v o = 110 cs : 5 m = 22 m/s
-------------------------------------
v = v o - a t
v = 0 m/s,  v o =  22 m/s,  t = 4 s
0 = 22 - 4 a
4 a = 22
a = 22 : 4
a = 5.5 m/s²
g = 9.80 m/s²
9.80 : 5.5 = 0.56 
Answer:
The magnitude of its acceleration is 5.5 m/s or 0.56 g. 
7 0
3 years ago
At a baseball game, the batter hit a fly ball at time t = 0 s. The outfielder caught the ball at t = 5.8 s. When was the ball at
Agata [3.3K]
We have the following equation for height:
 h (t) = (1/2) * (a) * t ^ 2 + vo * t + h0
 Where,
 a: acceleration
 vo: initial speed
 h0: initial height.
 The value of the acceleration is:
 a = -g = -9.8 m / s ^ 2
 For t = 0 we have:
 h (0) = (1/2) * (a) * 0 ^ 2 + vo * 0 + h0
 h (0) = h0
 h0 = 0 (reference system equal to zero when the ball is hit).
 For t = 5.8 we have:
 h (5.8) = (1/2) * (- 9.8) * (5.8) ^ 2 + vo * (5.8) + 0
 (1/2) * (- 9.8) * (5.8) ^ 2 + vo * (5.8) + 0 = 0
 vo = (1/2) * (9.8) * (5.8)
 vo = 28.42
 Substituting values we have:
 h (t) = (1/2) * (a) * t ^ 2 + vo * t + h0
 h (t) = (1/2) * (- 9.8) * t ^ 2 + 28.42 * t + 0
 Rewriting:
 h (t) = -4.9 * t ^ 2 + 28.42 * t
 The maximum height occurs when:
 h '(t) = -9.8 * t + 28.42
 -9.8 * t + 28.42 = 0
 t = 28.42 / 9.8
 t = 2.9 seconds.
 Answer:
 
The ball was at maximum elevation when:
 
t = 2.9 seconds.
8 0
3 years ago
Read 2 more answers
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