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iragen [17]
3 years ago
8

A 30-kg child starts at the center of a playground merry-go-round that has a radius of 2.1 m and rotational inertia of 500 kg⋅m2

and walks out to the edge. The merry-go-round has a rotational speed of 0.20 s−1 when she is at the center?
Physics
1 answer:
Sever21 [200]3 years ago
8 0

Answer:

\omega = 0.16 s^{-1}

Explanation:

As we know that there is no torque on the system of merry go round

so the angular momentum will remain conserved

so here we have

I_1\omega_0 = (I_1 + I_2)\omega

so we will have

I_1 = 500 kg m^2

I_2 = 30(2.1^2)

I_2 = 132.3 kg m^2

\omega_0 = 0.20 s^{-1}

now from above formula

500(0.20) = (500 + 132.3)\omega

\omega = 0.16 s^{-1}

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Answer:

C

Explanation:

Because everything on Earth falls at the same speed, the masses of the balls do not matter. Since the acceleration due to gravity is constant, their speeds will both be increasing at the same rate, and therefore the difference in speeds would remain constant until they hit the ground. Hope this helps!

5 0
3 years ago
How many grams of CO-60 result in 1 Millicuire of activity? How many years until the activity decays to 1 microcuire tl/2 =5.3 Y
vredina [299]

Answer:

m =8.81*10^{-6}grams

time t = 52.8 year

Explanation:

GIVEN DATA:

the half life of the CO-60 is, T_1/2 = 5.27 years = 1.663 e+8 s

activity dN/dt = 1 mCi = 3.7 X 10^7 decay/s

activity ,  dN/dt = N* \lambda

              \frac{dN}{dt} = N* \frac{ln2}{T_1/2}

                   N = \frac{(dN/dt )(T1/2)}{ln2}

                      = ( 3.7 X 10^7 )(1.663*10^8 ) / ln2

                      = 8.877*10^{16}

Number of moles:

     n = N/NA = 8.877*10^{16} / 6.022X10^23 = 1.474*10^{-7} mol

mass of the CO-60 is,

   m = n*M = [1.474*10^{-7} mol]*[59.93 grams /mol] = 8.81*10^{-6}grams

-----------------------------------------------------------------------------------------

    time t = -[T1/2 / ln2]*ln[N/N0]

             = - [5.3 years / ln2]*ln[1x10-6/1x10-3]

             = 52.8 year

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The cryosphere is composed of?
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A student jobs around a square park two times. Starting and ending at the gate to the park. The square jogging track is 40 meter
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A machine is designed to fill jars with 16 ounces of coffee. A consumer suspects that the machine is not filling the jars comple
babunello [35]

Answer:

a) Null hypothesis:  \mu \geq 16  

Alternative hypothesis :\mu  

b) t=\frac{15.6-16}{\frac{0.3}{\sqrt{8}}}=-3.77  

p_v =P(t_{(7)}  

If we compare the p value and the significance level given \alpha=0.1 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis.

c) We can conclude that the mean is significantly less than 16 ounces at 10% of significance.  so then the consumer suspect is correct.

Explanation:

Data given and notation  

\bar X=15.6 represent the mean for the sample

s=0.3 represent the sample standard deviation  

n=8 sample size  

\mu_o =16 represent the value that we want to test  

\alpha=0.1 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

Is a one tailed left test.  

What are H0 and Ha for this study?  

Null hypothesis:  \mu \geq 16  

Alternative hypothesis :\mu  

The degrees of freedom on this case are:

df=n-1=8-1=7

Compute the test statistic

The statistic for this case is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

t=\frac{15.6-16}{\frac{0.3}{\sqrt{8}}}=-3.77  

Give the appropriate conclusion for the test

Since is a one side left tailed test the p value would be:  

p_v =P(t_{(7)}  

Conclusion  

If we compare the p value and the significance level given \alpha=0.1 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis.

We can conclude that the mean is significantly less than 16 ounces at 10% of significance.  so then the consumer suspect is correct.

3 0
3 years ago
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