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spin [16.1K]
3 years ago
7

A mass m = 4.6 kg hangs on the end of a massless rope L = 2.16 m long. The pendulum is held horizontal and released from rest.

Physics
1 answer:
serious [3.7K]3 years ago
4 0

ind the velocity my using the conservation of energy mgh=1/2mv^2. At the bottom, the mass is turning in a circle with tension pointing up towards the center of the circle and weight pointing down. Use the centripetal force requirement mv^2/r=T-W and solve for T. Hope this helps.

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Lightning bolts can carry currents up to approximately 20 kA. We can model such a current as the equivalent of a very long, stra
GalinKa [24]

Answer:

how large a magnetic field would you experience = 8.16 x 10∧-4T

Explanation:

I = 20KA = 20,000A

r = 4.9 m

how large a magnetic field would you experience =  u.I/2πr

how large a magnetic field would you experience = (4π x10∧-7) × 20000/2π × 4.9

how large a magnetic field would you experience = 8.16 x 10∧-4T

4 0
3 years ago
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Convert 4 radians to degrees A. 242.6° B. 9.6° C. 229.2° D. 6.8°
muminat
The central angle of a circle is 360° or 2π radians.

Therefore
1 radian = (360 degrees)/(2π radians) = 180/π degrees/radian.
4 radians = (4 radians)*(180/π degrees/radian) = 229.18 degrees.

Answer: C.  229.2°
8 0
2 years ago
Read 2 more answers
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
Help please! Will give brainliest
zmey [24]
The answer is C
All other statements are correct but C
6 0
3 years ago
A certain copper wire has a resistance of 13.0 Ω . At some point along its length the wire was cut so that the resistance of one
alekssr [168]

Answer with Explanation:

Let r be the resistance of short piece of copper wire.

Resistance of copper wire=R=13\Omega

Resistance is directly proportional to length.

If a wire has greater resistance then,the wire will be greater in length.

Therefore,resistance of long piece of wire=7r

Total resistance of copper  wire=Sum of resistance of two piece of wires

r+7r=13

8r=13

r=\frac{13}{8}ohm

Resistance of long piece of wire=7\times\frac{13}{8}=\frac{91}{8}\Omega

Resistance of short piece of wire =\frac{13}{8}\Omega

Resistivity of wire and cross section area of wire remains same .

Let L be the total  length  of wire and L' be the length of short  piece of wire.

We know that

R=\frac{\rho L}{A}=\frac{\rho}{A}L=KL

\frac{R}{L}=K

Where K=\frac{\rho}{A}=Constant

Using the formula

\frac{13}{L}=\frac{\frac{13}{8}}{L'}

\frac{L'}{L}=\frac{13}{8}\times \frac{1}{13}=\frac{1}{8}

L'=\frac{L}{8}

Length of short piece of wire=L'=\frac{L}{8}

Length of long piece of  wire=L-L'=L-\frac{L}{8}=\frac{8L-L}{8}=\frac{7}{8}L

% of length of short piece of   wire=\frac{\frac{L}{8}}{L}\times 100=12.5%%

The resistance of the short piece=\frac{13}{8}\Omega

The resistance of the long piece=\frac{91}{8}\Omega

8 0
3 years ago
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