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iren [92.7K]
3 years ago
15

Four objects are situated along the y axis as follows: a 1.99-kg object is at 2.99 m, a 2.96-kg object is at 2.57 m, a 2.43-kg o

bject is at the origin, and a 3.96-kg object is at -0.502 m. Where is the center of mass of these objects
Physics
1 answer:
Dominik [7]3 years ago
3 0

Answer:

The center of mass for the object is  y_c = 1.063 \  m from the origin

Explanation:

From the question we are told that

   The mass of the first object is  m_1 =  1.99 \  kg

   The position of first object with respect to origin y_1 =  2.99 \ m

   The mass of the second object is  m_2 =  2.96 \  kg

   The position of second object with respect to origin y_2 =  2.57 \ m

   The mass of the third object is  m_3 =  2.43  \  kg

   The position of third object with respect to origin y_3 =  0 \ m

   The mass of the fourth object is  m_3 =  3.96  \  kg

   The position of fourth object with respect to origin y_3 =  -0.502  \ m

Generally the center of mass of the object along the x-axis is  zero  because all the mass lie on the y axis

Generally the location of the center mass of the object is mathematically represented as

    y_c = \frac{m_1 * y_1 + m_2 * y_2 + m_3 * y_3 + m_4 * y_4}{m_1 + m_2 + m_3 + m_4}

=>y_c = \frac{1.99 * 2.99 + 2.96 * 2.57 + 2.43 * 0 + 3.96 * (-0.502)}{1.99+ 2.96  + 2.43 + 3.96}

=>y_c = 1.063 \  m

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0.78m (rounded to nearest hundredth of a meter)

explanation:
time taken for going up=time taken for drop down after reaching the highest point. at the highest point, the velocity becomes 0.

now all thats left is dropping an object from a height (h) and seeing how long it takes to reach the ground. then find out the flight’s total time divided by 2 (0.8/2=0.4)

lets say the velocity is v and the height she jumped to is h. we can make a kinematic expression:
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once we put it all together you should get this:

h=0×0.4+½(9.81) 0.4²


.
∴
Time taken for downward drop
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0.8
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Suppose that she jumped with initial velocity
=
u

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h
Using following kinematic expression
s
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and inserting various quantities we get
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Diano4ka-milaya [45]

Answer:

(a). The critical angle for the liquid when surrounded by air is 44.37°

(b). The angle of refraction is 26.17°.

Explanation:

Given that,

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(a). We need to calculate the refraction of liquid

Using Snell's law

n=\dfrac{\sin i}{\sin r}

Put the value into the formula

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Using formula of critical angle

C=\sin^{-1}(\dfrac{1}{n})

Put the value into the formula

C=\sin^{-1}(\dfrac{1}{1.43})

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(b). Given that,

Incidence angle = 37.5°

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We need to calculate the index of refraction

Using formula of index of refraction

n=\dfrac{c}{v}

Put the value into the formula

n=\dfrac{3\times10^{8}}{2.17\times10^{8}}

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Using Snell's law

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Put the value into the formula

\sin r=\dfrac{\sin 37.5}{1.38}

r=\sin^{-1}(\dfrac{\sin 37.5}{1.38})

r=26.17^{\circ}

Hence, (a). The critical angle for the liquid when surrounded by air is 44.37°

(b). The angle of refraction is 26.17°.

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