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Rudiy27
3 years ago
9

A bat hasa mads of 2kg at the velocity of 45 m/s what is the kinectic energy could he give to a ball

Physics
1 answer:
il63 [147K]3 years ago
8 0

Answer:

the  kinetic energy the bat can give to a ball is 2,025 J.

Explanation:

Given;

mass of the bat, m = 2kg

velocity of the bat, v = 45 m/s

The kinetic energy the bat can give to a ball is calculated as;

K.E = \frac{1}{2} mv^2\\\\K.E = \frac{1}{2} \times \ 2 \ \times \ 45^2\\\\K.E = 2,025 \ J

Therefore, the  kinetic energy the bat can give to a ball is 2,025 J.

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The star Betelgeuse is 6.1 x 10^18 m away from Earth. How old is the light we see from that star when it reaches us? There are 3
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The light reaching the earth from the sun will travel at a speed called the speed of light, and this has a universal value of 3 × 10⁸ m/s. Bearing this in mind, let us calculate the age of the light reaching the Earth from the sun:

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We have distance and speed, let us calculate the time of travel of the light from the star to the earth.

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6.1 × 10⁸ = 3 × 10⁸ × time

time = \frac{6.1 \times 10^{18}}{3 \times 10^8}

In order to do the division above, we will divide the whole numbers normally, then we will apply the law of indices to the power that says:

Xᵃ ÷ Xᵇ = X⁽ᵃ⁻ᵇ⁾

\therefore time = \frac{6.1 \times 10^{18}}{3 \times 10^8}\\= \frac{2.03 \times 10^{(18-8)}}{1} \\= 2.03 \times 10^{10}}\ seconds

Next, we are told that there are 3.2 × 10⁷ seconds in a year.

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3.2\ \times 10^7\ seconds = 1\ year\\1\ second =  \frac{1}{3.2\ \times 10^7} \\\therefore 2.03 \times 10^{10}\ seconds = \frac{2.03 \times 10^{10}}{3.2\ \times 10^7}

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2.03 × 10¹⁰ = 20300000000

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\therefore \frac{2.03 \times 10^{10}}{3.2\ \times 10^7}\\= \frac{20300000000}{32000000} \\\\= \frac{20300}{32} \\= 634.347\ years\\

The closest answer in the option is 635 years, and we are short of this by some points due probably to approximations in the calculation.

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