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Natalka [10]
3 years ago
12

A football game customarily begins with a coin toss to determine who kicks off. The referee tosses the coin up with an initial s

peed of 6.00 m/s. In the absence of air resistance, how high does the coin go above its point of release
Physics
1 answer:
Cerrena [4.2K]3 years ago
8 0

Answer:

The coin travels 1.8348 meters above its point of release.

Explanation:

Initial speed = 6 m/s (Assumed vertical speed as not stated in question)

Final Speed = 0 m/s    (at the highest point in air)

Acceleration due to gravity = -9.81 m/s^2

Equation of motion used:

(V_f)^2 - (V_i)^2 = 2 *a * s

0^2 - 6^2 = 2 * (-9.81) * s

s = 1.8348 m

Thus, the coin will travel 1.8348 meters up to its highest point in the air.

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Find the wavelength of a 26 Hz sound wave traveling through air at 20ºC. Hint: speed of sound at 20ºC = 343 m/s
Usimov [2.4K]
The correct answer for the question that is being presented above is this one: "B. 13m." 

The formula of wave velocity is this:
wave velocity = f * <span>λ
</span>λ = v / f
<span>λ</span> = 343m/s / 26Hz
λ = 13.20m .. ans (b) 

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3 0
3 years ago
"A 3 kg crate slides down a ramp. The ramp is 1 m in length and inclined at an angle of 30 degrees. The crate starts from rest a
OlgaM077 [116]

Answer:

v = 2.57 m /sec

Explanation: See Annex Free Body Diagram

From free body diagram and Newton´s second law we have

There is not movements in the y axis direction

cos 30°  =  √3/2       sin 30°  =  1/2

We have  P  = mg   =  3 Kg  *  9.8 m/sec²

P  =   29.4  Kg*m/ sec²      P  =  29.4 [N]

Py  =  P * cos 30°    Py =  29.4 [N] * √3/2   ⇒    Py = 25.43 [N]

Px  =   P * sin 30°    Px =  29.4 [N] * 1/2      ⇒     Px = 14.7  [N]

∑ F   =  m* a         ⇒    ∑ Fy   =  0       ∑ Fx  =  m *a

∑ Fy   =  Fn  -  Py   =  0         Py   = P*cos30°       Py = 25.43 [N]

Fn  =  25.43 [N]

Fr  =  μk * Fn      ⇒   Fr  =  0.19 * 25.43   ⇒ Fr  =  4.83 [N]    

Now

∑ Fx  =  m *a       mg sin30° - Fr =  m*a    ⇒   Px  - Fr  = m*a

14.7 [N]   -  4.83 [N]   =  3 [kg] * a       ⇒   9.87 /3   = a [m /sec²]

a = 3.29 [m/sec²]

From uniformly accelerated movement

distance  =  x₀  + V₀*t ± at²/2     but  x₀   and  V₀    =  0

Then

d = ( 1/2 )*a*t²     ⇒  1 [m]  * 2  =  3.29 [m/sec²] * t²

t  =  0.78 sec

And finally

v =  a*t      ⇒   v  =  3.29 *(.78)     ⇒   v = 2.57 m /sec

5 0
3 years ago
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