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slava [35]
4 years ago
15

An 19.2-cm-long bicycle crank arm, with a pedal at one end is attached to a 20.6-cm-diameter sprocket, the toothed disk around w

hich the chain moves. A cyclist riding this bike increases her pedaling rate from 57 rpm to 86 rpm in 10.7 s. What is the tangential acceleration of the pedal?
Physics
2 answers:
Anon25 [30]4 years ago
3 0

Answer:

0.0545 m/s2

Explanation:

19.2 cm = 0.192 m

We can convert rpm (revolution per minute) to angular velocity rad/s knowing that each revolution is 2π rad and each minute is 60 seconds.

57 rpm = 57 * 2π / 60 = 6 rad/s

86 rpm = 86 * 2π / 60 = 9 rad/s

The angular acceleration of the sprocket is the change in angular velocity per unit of time

\alpha = \frac{\Delta \omega}{\Delta t} = \frac{9 - 6}{10.7} = 0.284 rad/s^2

The tangential acceleration of the pedal is the product of its angular acceleration and the radius of rotation, aka the pedal arm length L = 0.192 m

a_T = \alpha*L = 0.284*0.192 = 0.0545 m/s^2

Veronika [31]4 years ago
3 0

Answer:

Tangential acceleration=a_{t}=0.029233 m/s^2

Explanation

tangential acceleration=\alpha *r

\alpha=ω/t

alpha=2\pi((86/60)-(57/60))/10.7

\alpha=0.2838 m/s^2

at=\alpha*r

radius=d/2=20.6/2=10.3 cm=0.103 m

at=0.2838*0.103=0.029233 m/s^2

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rusak2 [61]
If the velocity is constant then the acceleration of the object is zero.
a=0 (m/s^2)
Thus when we apply the equation
\Delta X=vt+(at^2/2)
It remains
\Delta X =vt
or equivalent
t=(\Delta X/v) =7500/278 =26.98 (seconds)

7 0
4 years ago
Write about Archimedes principle​
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Answer:

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8 0
2 years ago
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You're driving down the highway late one night at 20 m/s when a deer steps onto the road 38 m in front of you. Your reaction tim
Bingel [31]

Answer:

given,

speed of the car = 20 m/s

final speed of car = 0 m/s

distance between car and the deer = 38 m

reaction time, t = 0.5 s

deceleration of the car = 10 m/s².

a) distance between deer and car

  distance travel in the reaction time

   d₁ = v x t

   d₁ = 20 x 0.5 = 10 m

   distance travel after you apply brake

   using equation of motion

   v² = u² + 2 a s

   0 = 20² - 2 x 10 x s

    s =  20 m

total distance traveled by the car

D = d₁ + d₂

D = 20 + 10 = 30 m

  distance between car and the deer = 38 m - 30 m

                                                              = 8 m

b) now, maximum speed car.

   distance travel in reaction time

    d₁ = s x t

    d₁ = 0.5 V

distance left between them

   d₂ = 38 - d₁

   d₂ = 38 - 0.5 V

   distance travel after you apply brake

   using equation of motion

    v² = u² + 2 a d₂

    0 = (V)² - 2 x 10 x (38 - 0.5 V)

     V² + 10 V - 760 = 0

now, solving the quadratic equation

  x = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}

  V = \dfrac{-10\pm \sqrt{10^2-4(1)(-760)}}{2(1)}

         V = 23.01 , -33.01

rejecting the negative term.

hence, maximum speed of the car could be V = 23.01 m/s

 

7 0
3 years ago
A potential difference of 107 mV exists between the inner and outer surfaces of a cell membrane. The inner surface is negative r
sergij07 [2.7K]

Answer:

The workdone is  W = 1.712 *10^{-20 } \  J  

Explanation:

From the question we are told that

    The potential difference is  V  =  107 mV =  107 *10^{-3} \  V

Generally the charge on  Na^{+} is  Q_{Na^{+}} = 1.60 *10^{-19 } \  C

 Generally the workdone is mathematically represented as

         W =  Q_{Na^{+}}V

=>     W = 1.60 *10^{-19 } *  107 *10^{-3}    

=>     W = 1.712 *10^{-20 } \  J    

8 0
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W=gm
where g - gravitation 
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w - weight
as gravitation equals to zero, multiplying by 0 gives W=0
It is not possible to tell whether and object is heavy or light 

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