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Nikitich [7]
2 years ago
12

We would like to know the velocity of the block when it reaches some position x. Finding this requires an integration. However,

acceleration is defined as a derivative with respect to time, which leads to integrals with respect to time, but the force is given as a function of position. To get around this, use the chain rule to find an alternative definition for the acceleration ax that can be written in terms of vx and dx/ dx.
Mathematics
1 answer:
Slav-nsk [51]2 years ago
4 0

Answer:

An alternative definition for the acceleration ax that can be written in terms of v_x and \frac{dv_x}{dx} is a_x=v_x \frac{dv_x}{dx}

Step-by-step explanation:

We know that :

a_x=\frac{dv_x}{dt}

Now we are supposed to find an alternative definition for the acceleration ax that can be written in terms of v_x and \frac{dv_x}{dx}

So, We will use chain rule over here :

a_x=\frac{dv_x}{dt}\\a_x=\frac{dv_x}{dt} \times \frac{dx}{dx}\\a_x=\frac{dv_x}{dx} \times \frac{dx}{dt}\\a_x=\frac{dv_x}{dx} \times \frac{dx}{dt}  [\frac{dx}{dt}=v_x]\\a_x=\frac{dv_x}{dx} \times v_x\\a_x=v_x\frac{dv_x}{dx}

Hence an alternative definition for the acceleration ax that can be written in terms of v_x and \frac{dv_x}{dx} is a_x=v_x \frac{dv_x}{dx}

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Using the exchange rate £1 = $1.33, calculate how many £ are in $6.50. Give your answer rounded to 2 dp.
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Answer:

Step-by-step explanation:

If the exchange rate between British pounds (£) and US dollars ($) is £1 = $1.40, how much is £200 worth in dollars?

Include the $ sign in your answer.

280

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What is the result of anding 255 and 15?
Tema [17]
The result of adding 255 and 15 is;

255 + 15 = 270


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2 years ago
Which expression shows the amount of tax on a T-shirt that costs x dollars, if tax is 7%?
alina1380 [7]

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Step-by-step explanation:

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1 year ago
Two processes are used to produce forgings used in an aircraft wing assembly. Of 200 forgings selected from process 1, 10 do not
tangare [24]

Answer:

a.

\bar p_1=0.05\\\bar p_2=0.067

b-Check illustration  below

c.(-0.0517,0.0177

Step-by-step explanation:

a.let p_1  \& p_2 denote processes 1 & 2.

For p_1: T1=10,n1=200

For p_2:T2=20,n2=300

Therefore

\bar p_1=\frac{t_1}{N_1}=\frac{10}{200}=0.05\\\bar p_2=\frac{t_2}{N_2}=\frac{20}{300}=0.067

b. To test for hypothesis:-

i.

H_0:p_1=p_2\\H_A=p_1\neq p_2\\\alpha=0.05

ii.For a two sample Proportion test

Z=\frac{\bar p_1-\bar p_2}{\sqrt(\bar p(1-\bar p)(\frac{1}{n_1}+\frac{1}{n_2})}\\

iii. for \frac{\alpha}{2}=(-1.96,+1.96) (0.5 alpha IS 0.025),

reject H_o if|Z|>1.96

iv. Do not reject H_o. The noncomforting proportions are not significantly different as calculated below:

z=\frac{0.050-0.067}{\sqrt {(0.06\times0.94)\times \frac{1}{500}}}

z=-0.78

c.(1-\alpha).100\% for the p1-p2 is given as:

(\bar p_1-\bar p_2)\pm Z_0_._5_\alpha \times \sqrt   \frac{ \bar p_1(1-\bar p_1)}{n_1}+\frac{\bar p_2(1-\bar p_2)}{n_2}\\\\=(0.05-0.067)\pm 1.645  \times \sqrt \ \frac{0.05+0.95}{200}+\frac{0.067+0.933}{300}\\

=(-0.0517,+0.0177)

*CI contains o, which implies that proportions are NOT significantly different.

4 0
2 years ago
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