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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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Give a calculate answer to show that the two values (English system and metric system) for the Planck Constant are equivalent.
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Answer:

Given values of Planck Constant are equivalent in English system and metric system.

Explanation:

Value of Planck's constant is given in English system as 4.14 x 10⁻¹⁵eV s.

Converting this in to metric system .

We have 1 eV = 1.6 x 10⁻¹⁹ J

Converting

     4.14 x 10⁻¹⁵eV s = 4.14 x 10⁻¹⁵x 1.6 x 10⁻¹⁹ = 6.63 x 10⁻³⁴ Joule s

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You are at a train station, standing next to the train at the front of the first car. The train starts moving with constant acce
Aleksandr [31]

The time needed for the 7th car to pass is 13.2 s

Explanation:

The motion of the train is a uniformly accelerated motion, therefore we can use suvat equations.

We start by analzying the motion of the first car, by using the equation:

s=ut+\frac{1}{2}at^2

where

s is the distance covered by the first car in a time t, which corresponds to the length of one car

u = 0 is the initial velocity

a is the acceleration

t = 5.0 s is the time

The equation can be rewritten as

a=\frac{2s}{t^2}=\frac{2L}{(5.0)^2}=0.08L[m/s^2]

where L is the length of one car.

The same equation can be written considering the first 7 cars:

7L = ut+\frac{1}{2}at'^2

where

7L is the distance covered by the 7 cars

t' is the time needed

We still have

u = 0

And the acceleration is constant so it is

a=0.08L

Substituting into the equation, we can find t':

7L = \frac{1}{2}(0.08L)t'^2\\7=0.04t'^2\\t'=\sqrt{\frac{7}{0.04}}=13.2 s

In attachment the graph of the distance covered versus the time taken: since the motion is uniformly accelerated, the relationship between the two variables is quadratical.

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

8 0
3 years ago
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