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3 years ago

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Graphically determine the resultant of the following three vector displacements: (1) 34 m, 25º north of east; (2) 48 m, 33° east

of north; and (3) 22 m, 56° west of south.

1 answer:

Anni [7]3 years ago

8 0

Answer:

Explanation:

In the first attached image, we can see, the nomenclature used for Angle orientations.

To sum the vectors we must join the head of one of them with the tail of the next, at the end joins the tail of the first vector with the head of the last vector.

In the second attached image, we can find graphically the sum of each of the vectors.

The resultant vector has a length of 57.36[m], and goes 48° north of east.

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Answer:

The gain in velocity is 0.37m/s

Explanation:

We need solve this problem though the conservation of momentum. That is,

$m_1 v_1 = m_2 v_2$

$m_1=2260Kg\\m_2=15.5Kg\\v_2= 109m/s$

Using the equation to find $v_1$ ,

$v_1=\frac{m_2 v_2}{m_1}\\v_1=\frac{15.5*109}{2260}\\v_1= 0.7475$

Using the conservation of energy equation, we have,

$KE= \frac{1}{2}m*v^2$

$KE_{cball}=\frac{1}{2}(15.5)(109)^2=92077.75J$

$KE_{cannon}=\frac{1}{2}(2260)(0.7475)^2=631.39J$

$Total KE= 92077.75+13425530=92708.9J$

Now this energy over the cannonball

$KE=\frac{1}{2}m*v_2^2$

$92708.9=\frac{1}{2}15.5v_2^2$

$V_2 = 109.37m/s$

The gain in velocity is 0.37m/s

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Evaluate (x +y)0 for x= -3 and y=5.<br> 0 1<br> 2<br> 01

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Answer:2.01201

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The captain told the passengers that the plane is flying at 450 miles per hour. This information describes the plane's

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The speed of the plane

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A 0.877 mol sample of N2(g) initially at 298 K and 1.00 atm is held at constant volume while enough heat is applied to raise the

MissTica

Answer : The value of $q\text{ and }\Delta U$ is 286.2 J and 286.2 J respectively.

Explanation : Given,

Moles of sample = 0.877 mol

Change in temperature = 15.7 K

First we have to calculate the heat absorbed by the system.

Formula used :

$q=n\times c_v\times \Delta T$

where,

q = heat absorbed by the system = ?

n = moles of sample = 0.877 mol

$\Delta T$ = Change in temperature = 15.7 K

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Now put all the given value in the above formula, we get:

$q=0.877mol\times \frac{5}{2}\times 8.314J/mol.K\times 15.7K$

$q=286.2J$

Now we have to calculate the change in internal energy of the system.

$\Delta U=q+w$

As we know that, work done is zero at constant volume. So,

$\Delta U=q=286.2J$

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3 years ago

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