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olasank [31]
3 years ago
12

Estimate the number of times the earth will rotate on its axis during a human’s lifetime

Physics
1 answer:
Fudgin [204]3 years ago
7 0

Here we have to calculate  the number of rotations made by earth around its own axis in the entire life time of a human being.

The average human life is 50 years .

Each year is 365.5 days

Hence 50 years =50 *365.5 days

                 =18,275 days

The earth rotates around its own axis in 23 hours 56 minutes and 4 second which is approximately equal to 24 hours or one days.

Hence one rotation takes one days.

⇒18,275 days =18,275 rotations

Hence the total number of rotations made by earth around its own axis in the life time of a average human life time will be 18,275 times

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the car is moving so that how it gos so fast

Explanation:

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If the 50 kg objects slows down to a velocity of 1 m/s how much kinetic energy does it have?
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3 years ago
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When cars are equipped with flexible bumpers, they will bounce off each other during low-speed collisions, thus causing less dam
Len [333]

Answer:

Explanation:

Given that,

Mass of the heavier car m_1 = 1750 kg

Mass of the lighter car m_2 = 1350 kg

The speed of the lighter car just after collision can be represented as follows

m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\v_2=\frac{m_1u_1+m_2u_2-m_1v_1}{m_2}

v_2=\frac{(1850)(1.4)+(1450)(-1.10)-(1850)(0.250)}{1450} \\\\=\frac{2590+(-1595)-(462.5)}{1450} \\\\=\frac{2590-1595-462.5}{1450} \\\\=\frac{532.5}{1450}\\\\=0.367m/s

b) the change in the combined kinetic energy of the two-car system during this collision

\Delta K.E=(\frac{1}{2} m_1v_1^2+\frac{1}{2} m_2v_2^2)-(\frac{1}{2} m_1u_1^2+\frac{1}{2} m_2u_2^2)\\\\=\frac{1}{2} (m_1(v_1^2-u_1^2)+m_2(v_2^2-u_2^2))

substitute the value in the equation above

=\frac{1}{2} (1850((0.250)^2-(1.4)^2)+(1450((0.3670)^2-(-1.10)^2)\\\\=\frac{1}{2}(11850(0.0625-1.96)+(1450(0.1347)-(1.21))\\\\= \frac{1}{2}(11850(-1.8975))+(1450(-1.0753))\\\\=\frac{1}{2} (-3510.375+(-1559.185)\\\\=\frac{1}{2} (-5069.56)\\\\=-2534.78J

Hence, the change in combine kinetic energy is -2534.78J

8 0
3 years ago
Point charges of 21.0 μC and 47.0 μC are placed 0.500 m apart. (a) At what point (in m) along the line connecting them is the el
rewona [7]

Answer:

a) x = 0.200 m

b)E = 3.84*10^{-4} N/C

Explanation:

q_1 = 21.0\mu C

q_1 = 47.0\mu C

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by relation for electric field we have following relation

E = \frac{kq}{x}^2

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FROM FIGURE

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6 0
3 years ago
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<span>Which of the following substance are not formed by chemical bonds? </span>A MIXTURE 
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