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tatiyna
3 years ago
6

A wheel has a constant angular acceleration of 1.8 rad/s2. During a certain 4.0 s interval, it turns through an angle of 45 rad.

Assuming that the wheel started from rest, how long had it been in motion before the start of the 4.0 s interval?
Physics
1 answer:
tankabanditka [31]3 years ago
7 0

Answer:

4.25 s

Explanation:

Given:

angular acceleration'α'=  1.8 rad/s²

angle 'θ'= 45 rad

time 't'= 4s

initial angular velocity 'w_{o}'=0rad^{-1}

as we know that,

θ= w_{1}t + \frac{1}{2} \alpha t^{2}

45 = 4 w_{1} + (0.5 x 1.8 x 16)

45- 14.4 = 4 w_{1}

30.6 = 4 w_{1}

w_{1}= 7.65 rad^{-1}

Next is to find t by using the equation

w_{1} =  w_{o} + \alpha t_{1}

7.65= 0 + (1.8)t_{1}

t_{1}= 7.65/1.8

t_{1}= 4.25 s

Therefore, At the start of 4s interval the motion is at 4.25 second

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A horizontal pipe of diameter 1.03m has a smooth constriction to a section of diameter 0.618 m. The density of oil flowing in th
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Explanation:

We can solve this problem by using Bernoulli's equation:

p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2 (1)

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p_2 = 5505 N/m^2 is the pressure in the constricted section

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v_2 is the velocity of the oil in the constricted section

Also, according to the continuity equation,

A_1 v_1 = A_2 v_2

where

A_1 = \pi r_1^2 is the cross-sectional area of the pipe, with

r_1 = \frac{1.03}{2}=0.515 m is the radius

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So the equation becomes

r_1^2 v_1 = r_2^2 v_2

So we can write

v_2=\frac{r_1^2}{r_2^2}v_1

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p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho (\frac{r_1^2}{r_2^2}v_1)^2

And solving the equation for v_1:

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And the velocity in the constricted section is

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Learn more about flow rate:

brainly.com/question/9805263

#LearnwithBrainly

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