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2 years ago

13

How many complete revolutions are needed to draw the angle 725°?

2 answers:

uysha [10]2 years ago

8 0

C. A complete revolution is 360 degree. two revolution is 720.

fiasKO [112]2 years ago

4 0

Answer:

$725\ degree=2\ revolution$

Explanation:

We know that 1 complete revolution is equal to 360 degrees. We need to find the number of complete revolution needed to draw the angle of 725 degrees.

Since, 1 revolution = 360 degrees

or

$1\ revolution=2\pi \ radian$

$1\ degree=\dfrac{1}{360}\ revolution$

$725\ degree=\dfrac{725}{360}\ revolution$

$725\ degree=2.01\ revolution$

or

$725\ degree=2\ revolution$

So, to draw an angle of 725 degrees, there are 2 revolutions. Hence, this is the required solution.

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The ratio of time of flight for inelastic collision to elastic collision is 1:2

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<em>mass of the bullet, = m₁</em>
<em>mass of the block, = m₂</em>
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<em>initial velocity of the block, = u₂</em>

Considering inelastic collision, the final velocity of the system is calculated as;

$m_1u_1 + m_2u_2 = v(m_1 + m_2)\\\\m_1u_1 + 0 = v(m_1 + m_2)\\\\v= \frac{m_1u_1}{m_1 + m_2} \ -- (1)\\\\$

The time of motion of the system form top of the table is calculated as;

$v = u + gt\\\\v = 0 + gt\\\\v = gt\\\\t= \frac{v}{g} \\\\t_A = \frac{m_1u_1}{g(m_1 + m_2)} \ \ ---(2)$

Considering elastic collision, the final velocity of the system is calculated as;

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Apply one-directional velocity

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Substitute the value of $v_1$ into the above equation;

$m_1u_1 = m_1(v_2 - u_1) + m_2 v_2\\\\m_1u_1 = m_1v_2 -m_1u_1 + m_2v_2\\\\2m_1u_1 = m_1v_2 + m_2v_2\\\\2m_1u_1= v_2(m_1 + m_2)\\\\v_2 = \frac{2m_1u_1}{m_1+ m_2} \ --(3)$

where;

$v_2$ is the final velocity of the block after collision

<em>Since the</em><em> bullet bounces off</em><em>, we assume that </em><em>only the block fell </em><em>to the ground from the table.</em>

The time of motion of the block is calculated as follows;

$v_2 = v_0 + gt\\\\v_2 = 0 + gt\\\\t = \frac{v_2}{g} \\\\t_B = \frac{v_2}{g} \\\\ t_B = \frac{2m_1u_1}{g(m_1 + m_2)} \ \ ---(4)$

The ratio of time of flight for inelastic collision to elastic collision is calculated as follows;

$\frac{t_A}{t_B} = \frac{m_1u_1}{g(m_1 + m_2)} \times \frac{g(m_1 + m_2)}{2m_1u_1} \\\\\frac{t_A}{t_B} = \frac{1}{2} \\\\t_A:t_B = 1: 2$

Learn more about elastic and inelastic collision here: brainly.com/question/7694106

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2 years ago

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Answer:

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