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kap26 [50]
3 years ago
10

Suppose the coefficient of kinetic friction between mA and the plane in the figure(Figure 1) is μk = 0.15, and that mA=mB=2.7kg.

Physics
1 answer:
luda_lava [24]3 years ago
4 0
A ) 
T = mB g + mB a
T + mA a - mA g sin 35° = (Mi) mA g cos 35°
------------------------------------------------------------
T = 2.7 · 9.81  + 2.7 a
T = 26.487 + 2.7 a
26.487 + 2.7 a + 2.7 a - 2.7 · 9.81 · 0.574 = 0.15 · 2.7 · 9.81 · 0.819
5.4 a + 26.487 - 15.2023 = 3.2539
5.4 a = 8.0296
a = 1.487 ≈ 1.5 m/s²
B )
T = 2,7 · 9.81 = 26.487 
26.487 - 15.2035 = (Mi) · 2.7 · 9.81 · 0.819
11.2835 = (Mi) · 21.69
(Mi) = 11.2835 : 21.69 = 0.52
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Two identical mandolin strings under 200 N of tension are sounding tones with frequencies of 590 Hz. The peg of one string slips
Slav-nsk [51]

To solve this problem it is necessary to apply the concepts related to frequency and vibration of strings. Mathematically the frequency can be expressed as

f = \frac{v}{\lambda}

Then the relation between two different frequencies with same wavelength would be

\frac{f'}{f} = \frac{v'/\wavelength}{v/\wavelength}

\frac{f'}{f} = \frac{v'}{v}

The beat frequency heard when the two strings are sounded simultaneously is

f_{beat} = f-f'

f_{beat} = f(1-\frac{f'}{f})

f_{beat} = f(1-\frac{v'}{v})

We have the velocity of the transverse waves in stretched string as

v = \sqrt{\frac{T}{\mu}}

v = \sqrt{\frac{200N}{\mu}}

And,

v' = \sqrt{\frac{196N}{\mu}}

Therefore the relation between the two is,

\frac{v'}{v} = \sqrt{\frac{192}{200}}

\frac{v'}{v} = \sqrt{0.96}

Finally substituting this value at the frequency beat equation we have

f_{beat} = 590(1-\sqrt{0-96})

f_{beat} = 11.92Hz

Therefore the beats per second are 11.92Hz

4 0
3 years ago
Can a goalkeeper at his goal kick a soccer ball into the opponent’s goal without the ball touching the ground? The distance will
zmey [24]

The goalkeeper at his goal cannot kick a soccer ball into the opponent’s goal without the ball touching the ground

Explanation:

Consider the vertical motion of ball,

We have equation of motion v = u + at

     Initial velocity, u  = u sin θ

     Final velocity, v =  0 m/s    

     Acceleration = -g

     Substituting

                      v = u + at  

                      0 = u sin θ - g t

                      t=\frac{usin\theta }{g}

This is the time of flight.

Consider the horizontal motion of ball,

        Initial velocity, u =  u cos θ

        Acceleration, a =0 m/s²  

        Time, t=\frac{usin\theta }{g}  

     Substituting

                      s = ut + 0.5 at²

                      s=ucos\theta \times \frac{usin\theta }{g}+0.5\times 0\times (\frac{usin\theta }{g})^2\\\\s=\frac{u^2sin\theta cos\theta}{g}\\\\s=\frac{u^2sin2\theta}{2g}

This is the range.

In this problem

              u = 30 m/s

              g = 9.81 m/s²

              θ = 45° - For maximum range

Substituting

               s=\frac{30^2\times sin(2\times 45)}{2\times 9.81}=45.87m

Maximum horizontal distance traveled by ball without touching ground is 45.87 m, which is less than 95 m.

So the goalkeeper at his goal cannot kick a soccer ball into the opponent’s goal without the ball touching the ground

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3 years ago
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<span>A ball moving at a constant speed around a circular track produces acceleration. </span>
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Answer:

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Explanation:

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Answer:

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