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4vir4ik [10]
3 years ago
7

Determine the stopping location of the prize wheel. At this moment it is centered on the number 11. It is spinning at a rate of

124.40 rpm. It is slowing at a rate of 1.400 rad/s/s.
Predict the time the wheel will remain spinning, the angular displacement it will go through, and then use this to predict the number on which the wheel will stop. To be safe, you will also get one number the right and to the left of the number you chose.
Physics
1 answer:
Goshia [24]3 years ago
5 0

Answer:

The wheel spins for 11.43s. The number cannot be determined.

Explanation:

We know that initial velocity is 124.4 rpm = 13.027137522 rad/s, acceleration is -1.14 rad/s^2, and final velocity is assumed to be 0 rad/s. We are asked to find t and displacement. We use the equation \omega=\omega_0+\alpha t where \omega=0 rad/s, \omega_0=13.027 rad/s, \alpha=-1.14rad/s^2, and t=?.

Rearrange the equation to obtain \frac{\omega-\omega_0}{\alpha}=t. Plug in the numbers and solve to obtain t=11.43s.

The number of the wheel cannot be determined as we do not know the placement of numbers on the wheel.

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A 5kg bag falls a verticle height of 10m before hitting the ground.
g100num [7]

Answer:

u = 7m {s}^{ - 1}

Explanation:

We know that when we don't have air friction on a free fall the mechanical energy (I will symbololize it with ME) is equal everywhere. So we have:

me(1) = me(2)

where me(1) is mechanical energy while on h=10m

and me(2) is mechanical energy while on the ground

Ek(1) + DynamicE(1) = Ek(2) + DynamicE(2)

Ek(1) is equal to zero since an object that has reached its max height has a speed equal to zero.

DynamicE(2) is equal to zero since it's touching the ground

Using that info we have

m \times g \times h   =   \frac{1}{2}  \times m \times u {}^{2} \\

we divide both sides of the equation with mass to make the math easier.

9.8 \times 10 =  \frac{1}{2}  \times u {}^{2}  \\  \frac{98}{2}  = u {}^{2}  \\ u { }^{2} = 49 \\ u = 7

7 0
4 years ago
A phonograph record accelerates from rest to 28.0 rpm in 5.73 s.
Arturiano [62]

Answer:

a) \alpha=0.5117\ rad.s^{-2}

b) \theta=8.4\ rad

Explanation:

Given:

  • initial rotational speed of phonograph, \omega_i=0\ rad.s^{-1}
  • final rotational speed of phonograph, N_f=28\ rpm \Rightarrow \omega_f=2\pi\times\frac{28}{60} =2.932\ rad.s^{-1}
  • time taken for the acceleration, t=5.73\ s

a)

Now angular acceleration:

\alpha=\frac{\omega_f-\omega_i}{t}

\alpha=\frac{2.932-0}{5.73}

\alpha=0.5117\ rad.s^{-2}

b)

Using eq. of motion:

\theta=\omega_i.t+\frac{1}{2} \alpha.t^2

\theta=0+\frac{1}{2}\times 0.5117\times 5.73^2

\theta=8.4\ rad

5 0
3 years ago
A water tank is filled with water. What will be the pressure of the water when the level of the water is 6m?
Luden [163]

Answer:

pressure = density x g x height

= 1000 x 10 x 6 Pascal

=60000 Pascal

OR 60 kP

3 0
3 years ago
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A hot-air balloon is drifting in level flight due east at 2.8 m/s due to a light wind. The pilot suddenly notices that the ballo
fiasKO [112]

a. There is nothing suggesting that the balloon is accelerating horizontally, so we can assume that its horizontal speed is constant. The time before the balloon crash into the hill is simply the distance between the balloon and the hill divided by its velocity. Remember that velocity is simply the amount of distance that a object travels in a certain amount of time:

t = \frac{130m}{2.8 m/s} = 46.43 s

b. Know that you know the maximum amount of time that the balloon can take to gain 28m of altitude, the minimum acceleration can be found using the  equations constant acceleration motion:

x = \frac{1}{2}at^2 + v_ot +x_0

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d. Acceleration is the amount of change in velocity after a given amount of time. So, with the acceleration and the time we can fin the velocity:

v_y = a_y*t = 0.026m/s^2*46.43s=1.206 m/s

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The answer is B. because when u add 3 on O2 and multiplied together makes 6

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