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Colt1911 [192]
3 years ago
6

An object, initially at rest moves 250m in 17s. What is it's acceleration?

Physics
1 answer:
Gekata [30.6K]3 years ago
8 0
With the values you've given, only velocity can be found.
Acceleration is rate of change of velocity

d= 250s
t= 17s

a= d/t
 =\frac{250}{17}
 = 4.7 \frac{m}{s}
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Answer:

Option B is correct.

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What characteristic distinguishes a group from an aggregate?
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Water with a volume flow rate of 0.001 m3/s, flows inside a horizontal pipe with diameter of 1.2 m. If the pipe length is 10m an
galben [10]

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\triangle P=1.95*10^{-4}

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Generally the equation for Friction factor is mathematically given by

 F=\frac{64}{Re}

Where Re

Re=Reynolds Number

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 Re=1040

Therefore

 F=\frac{64}{Re}

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Generally the equation for Friction factor is mathematically given by

 Head loss=\frac{fLv^2}{2dg}

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H=\frac{\triangle P}{\rho g}

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6 0
3 years ago
The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 5.0 rev/s in 9.0 s. At thi
sertanlavr [38]

Answer

given,

angular speed of the tub = 5 rev/s

time = 9 s

he tub slows to rest  =  15.0 s

the angular acceleration

\omega_f - \omega_i = \alpha t

\alpha = \dfrac{5-0}{9}

\alpha = 0.556 rev/s^2

angular displacement

\theta_1 = \omega_i \ t + \dfrac{1}{2}\alpha t^2

\theta_1 = \dfrac{1}{2}\times 0.556 \times 9^2

\theta_1 = 22.52 rev

\theta_1 = 23 rev

case 2

now,

\omega_i = 5 rev/s

\omega_f = 0 rev/s

time = 15 s

the angular acceleration

\omega_f - \omega_i = \alpha t

\alpha = \dfrac{0-5}{15}

v\alpha =-0.333 rev/s^2

angular displacement

\theta_2 = \omega_i \ t + \dfrac{1}{2}\alpha t^2

\theta_2 =5\times 15 -\dfrac{1}{2}\times 0.333 \times 15^2

\theta_2 = 37.875 rev

\theta_2 =38 rev

total revolution in 24 s

= 23 + 38

= 62 revolution

8 0
3 years ago
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