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solong [7]
3 years ago
5

What happens to the car if it is traveling in a circular path suddenly encounters ice?

Physics
2 answers:
pychu [463]3 years ago
8 0

the car's friction is reduced to zero

the car's centripetal force drops to zero

the car continues in a straight path from the point at which it encountered the ice

dimulka [17.4K]3 years ago
5 0

When a car is travelling in a circular path then for the circular motion of the the centripetal force is provided by the friction force

This friction force will help to take turn in circular path

Now during the motion of car if friction suddenly disappear due to which there is no centripetal force will act on it

Since there is no centripetal force on the car so now it will not be able to take turn

so here is the possible outcomes

A.) the car's friction is reduced to zero.  

B.) the car's centripetal force drops to zero.  

C.) the car continues in a straight path from the point at which it encountered the ice.

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When a potential difference of 10 V is placed across a certain solid cylindrical resistor, the current through it is 2 A. If the
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Answer:

New Resistance = 0.5556 ohm

Explanation:

Resistance = resistivity * length /area

Here since resistivity and length are constant, we only need to see how the resistance increases or decreases with change in area.

New Area = pi * (3*D)^2 / 4

Old Area = pi * D^2 / 4

The ratio of new area / old area is :

\frac{New Area}{Old Area} = 9

Since area increases 9 times, and it is inversely proportional to resistance:

Resistance decreases by 9 times.

So, old resistance = Voltage / Current = 10 / 2 = 5 ohm

New Resistance = 5 / 9 = 0.5556 ohm    (decreases by 9 times)

8 0
3 years ago
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.

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3 years ago
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Answer:

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Explanation:

The velocity v₁ can be calculated with the kinematic formula:

v_1^{2} =v_0^{2} +2gh

Since the object is initially at rest, v₁ becomes:

v_1=\sqrt{2gh}

Where g is the acceleration due to gravity. Now, the velocity v₂ can be calculated with the same formula, but now the initial velocity is v₁:

v_2^{2}=v_1^{2} +2gh

Substituting v₁ in this expression and solving for v₂, we get:

v_2^{2}=(\sqrt{2gh} )^{2} +2gh=4gh\\\\\implies v_2=\sqrt{4gh}=2\sqrt{gh}

Now, dividing v₂ over v₁, we get the expression:

\frac{v_2}{v_1}=\frac{2\sqrt{gh} }{\sqrt{2gh}}=\sqrt{2}\\   \\\implies v_2=\sqrt{2}v_1

It means that v₂ is √2 times v₁.

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Explanation:

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