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2 years ago

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A wheel rotates with a constant angular acceleration of 6.0 rad/s2. During a certain time interval its angular displacement is 8

.0 rad. At the end of the interval its angular velocity is 14.0 rad/s. What is its angular velocity at the beginning of the interval

1 answer:

kari74 [83]2 years ago

3 0

The angular velocity at the beginning of the interval of the wheel rotating at a constant angular acceleration is determined as 10 rad/s.

<h3>Initial angular velocity of the wheel</h3>

The initial angular velocity of the wheel is determined by applying the kinematic equation as shown below;

ωf² = ωi² + 2αθ

where;

ωf is the final angular velocity
ωi is the initial angular velocity
α is angular acceleration
θ is angular displacement

Substitute the given parameters and solve for the initial angular velocity.

14² = ωi² + 2(6)(8)

196 = ωi² + 96

ωi² = 196 - 96

ωi² = 100

ωi = √100

ωi = 10 rad/s

Thus, the initial angular velocity of the wheel is 10 rad/s.

Learn more about angular velocity here: brainly.com/question/6860269

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A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.

galina1969 [7]

Answer:

44.64 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 4.2\times 1180+80.6^2}\\\Rightarrow v=128.01\ m/s$

$v=u+at\\\Rightarrow 128.01=80.6+4.2t\\\Rightarrow t=\frac{128.01-80.6}{4.2}=11.29\ s$

<u>Time taken to reach 1180 m is 11.29 seconds</u>

$v=u+at\\\Rightarrow 0=128.01-9.8t\\\Rightarrow t=\frac{128.01}{9.8}=13.06\ s$

<u>Time the rocket will keep going up after the engines shut off is 13.06 seconds.</u>

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-128.01^2}{2\times -9.8}\\\Rightarrow s=836.05\ m$

The distance the rocket will keep going up after the engines shut off is 836.05 m

Total distance traveled by the rocket in the upward direction is 1180+836.05 = 2016.05 m

The rocket will fall from this height

$s=ut+\frac{1}{2}at^2\\\Rightarrow 2016.05=0t+\frac{1}{2}\times 9.8\times t^2\\\Rightarrow t=\sqrt{\frac{2016.05\times 2}{9.8}}\\\Rightarrow t=20.29\ s$

<u>Time taken by the rocket to fall from maximum height is 20.29 seconds</u>

Time the rocket will stay in the air is 11.29+13.06+20.29 = 44.64 seconds

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3 years ago

A ball is thrown upward with a speed of 40 m/s. Approximately how much time does it take the ball to travel from the release loc

zvonat [6]

I'm going to assume that this gripping drama takes place on planet Earth, where the acceleration of gravity is 9.8 m/s². The solutions would be completely different if the same scenario were to play out in other places.

A ball is thrown upward with a speed of 40 m/s. Gravity decreases its upward speed (increases its downward speed) by 9.8 m/s every second.

So, the ball reaches its highest point after (40 m/s)/(9.8 m/s²) = <em>4.08 seconds</em>. At that point, it runs out of upward gas, and begins falling.

Just like so many other aspects of life, the downward fall is an exact "mirror image" of the upward trip. After another 4.08 seconds, the ball has returned to the height of the hand which flung it. In total, the ball is in the air for <em>8.16 seconds</em> up and down.

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3 years ago

Steam enters a well-insulated nozzle at 200 lbf/in.2 , 500F, with a velocity of 200 ft/s and exits at 60 lbf/in.2 with a velocit

Ede4ka [16]

Answer:

$386.2^{\circ}F$

Explanation:

We are given that

$P_1=200lbf/in^2$

$P_2=60lbf/in^2$

$v_1=200ft/s$

$v_2=1700ft/s$

$T_1=500^{\circ}F$

$Q=0$

$C_p=1BTU/lb^{\circ}F$

We have to find the exit temperature.

By steady energy flow equation

$h_1+v^2_1+Q=h_2+v^2_2$

$C_pT_1+\frac{P^2_1}{25037}+Q=C_pT_2+\frac{P^2_2}{25037}$

$1BTU/lb=25037ft^2/s^2$

Substitute the values

$1\times 500+\frac{(200)^2}{25037}+0=1\times T_2+\frac{(1700)^2}{25037}$

$500+1.598=T_2+115.4$

$T_2=500+1.598-115.4$

$T_2=386.2^{\circ}F$

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