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Bumek [7]
3 years ago
6

A Ferris wheel, rotating initially at an angular speed of 0.500 rad/s, accelerates over a 7.00-s interval at a rate of 0.040 0 r

ad/s2 . What angular displacement does the Ferris wheel undergo in this 7-s interval? a. 4.48 rad b. 2.50 rad c. 3.00 rad d. 0.500 rad
Physics
1 answer:
bazaltina [42]3 years ago
6 0

Answer:

Angular displacement will be 4.48 rad

So option (a) will be correct option

Explanation:

We have given initial angular acceleration \omega _i=0.5rad/sec

Angular acceleration \alpha =0.04rad/sec^2

Time t = 7 sec

We have to find the angular displacement

We know that angular displacement is given by

\Theta =\omega _it+\frac{1}{2}\alpha t^2=0.5\times 7+\frac{1}{2}\times 0.04\times 7^2=4.48rad

So option (a) will be correct option

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5 0
3 years ago
A small sphere is at rest at the top of a frictionless semicylindrical surface. The sphere is given a slight nudge to the right
V125BC [204]

Answer:

vi = 4.77 ft/s

Explanation:

Given:

- The radius of the surface R = 1.45 ft

- The Angle at which the the sphere leaves

- Initial velocity vi

- Final velocity vf

Find:

Determine the sphere's initial speed.

Solution:

- Newton's second law of motion in centripetal direction is given as:

                         m*g*cos(θ) - N = m*v^2 / R

Where, m: mass of sphere

             g: Gravitational Acceleration

             θ: Angle with the vertical

             N: Normal contact force.

- The sphere leaves surface at θ = 34°. The Normal contact is N = 0. Then we have:

                         m*g*cos(θ) - 0 = m*vf^2 / R

                         g*cos(θ) = vf^2 / R    

                         vf^2 = R*g*cos(θ)

                         vf^2 = 1.45*32.2*cos(34)

                        vf^2 = 38.708 ft/s

- Using conservation of energy for initial release point and point where sphere leaves cylinder:

                          ΔK.E = ΔP.E

                          0.5*m* ( vf^2 - vi^2 ) = m*g*(R - R*cos(θ))

                          ( vf^2 - vi^2 ) = 2*g*R*( 1 - cos(θ))

                          vi^2 =  vf^2 - 2*g*R*( 1 - cos(θ))

                          vi^2 = 38.708 - 2*32.2*1.45*(1-cos(34))

                          vi^2 = 22.744

                           vi = 4.77 ft/s

4 0
3 years ago
Which statement best explains the difference between greenhouse gases and other atmospheric gases?(1 point)
Dmitriy789 [7]

Greenhouse gases trap thermal energy and reflect the sun’s harmful radiation back to Earth is the answer

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3 0
2 years ago
Adam drops a ball from rest from the top floor of a building at the same time Bob throws a ball horizontally from the same locat
guapka [62]

Answer:

Both balls hit the ground at the same time

Explanation:

Adam drops the ball from rest, so the ball just "<em>falls</em>" in vertical direction, being gravity its only acceleration, for cinematic movements we use that:

y(t)=y_{0}+v_{0y}t+\frac{1}{2}gt^{2}

In this case we have that gravity is negative, and as Adam drops the ball, v_{0y}=0

Bob throws the ball horizontally, so the movement will be a <em>parabola</em>, we can divide into horizontal direction, and vertical direction.

But we only need to analize the vertical movement, in wich again the only acceleration is gravity, and compare it with Adam's ball. Again we have that gravity is negative, and as the initial throw is horizontal, v_{0y}=0

Finally, we have that

y(t)=h-\frac{1}{2}gt^{2}

where

h=y_{0}

both for Adam's vertical drop, and for Bob's vertical component of the parabolic throw.

Now, if we put y(t)=0 (the origin of the vertical coordinate), we get for both cases that

h=\frac{1}{2}gt^{2}

where we can clear the value for the time t, of the fall, wich will be the same in both cases.

Hence, both balls hit the ground at the same time.

3 0
2 years ago
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