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3 years ago

IBM has a fast computer that it calls the Blue Gene/L that can do '136.8

Physics

1 answer:

dmitriy555 [2]3 years ago

6 0

Answer:

138.6 megacalculations

Explanation:

This is a pretty straightforward one.

All it needs is to convert the degree of measurement.

Dimensions in physics are attributed names, which state the power to which they're are raised. Just as how

Kilo and Mega means the numbers are raised to the power of 3 and 6 respectively. There also exists the ones that indicates how small, such as milli and micro, which are to the powers of -3 & -6.

The question says the IBM computer calculates at an astonishing 136.8 teracalculations.

Tera in physics means it's raised to the power of 12. Thus, the IBM calculates at an astonishing rate of

136.8*10^12 calculations per second.

We're then asked how many calculations it does in 1 micro second. Like I had highlighted earlier, 1 micro second is 1 raised to the power of -6. Or succinctly put,

1 micro second = 1*10^-6.

If the IBM does

138.6*10^12 = 1 second,

Then it does

x = 1*10^-6 second.

When we cross multiply, we have

138.6*10^12 * 1*10^-6, and that is

138.6*10^6 calculations, or say, 138.6 megacalculations.

The IBM does 138.6 megacalculations in 1 micro second, which is still astonishing, by the way

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Until he was in his seventies, Henri LaMothe excited audiences by belly-flopping from a height of 9 m into 32 cm. of water. Assu

baherus [9]

Answer:

823.46 kgm/s

Explanation:

At 9 m above the water before he jumps, Henri LaMothe has a potential energy change, mgh which equals his kinetic energy 1/2mv² just as he reaches the surface of the water.

So, mgh = 1/2mv²

From here, his velocity just as he reaches the surface of the water is

v = √2gh

h = 9 m and g = 9.8 m/s²

v = √(2 × 9 × 9.8) m/s

v = √176.4 m/s

v₁ = 13.28 m/s

So his velocity just as he reaches the surface of the water is 13.28 m/s.

Now he dives into 32 cm = 0.32 m of water and stops so his final velocity v₂ = 0.

So, if we take the upward direction as positive, his initial momentum at the surface of the water is p₁ = -mv₁. His final momentum is p₂ = mv₂.

His momentum change or impulse, J = p₂ - p₁ = mv₂ - (-mv₁) = mv₂ + mv₁. Since m = Henri LaMothe's mass = 62 kg,

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3 years ago

Suppose electrons in a TV tube are accelerated through a potential difference of 2.00 104 V from the heated cathode (negative el

Ivanshal [37]

Answer:

83816746.4254 m/s

Explanation:

m = Mass of electron = $9.11\times 10^{-31}\ kg$

q = Charge of electron = $1.6\times 10^{-19}\ C$

V = Voltage = $2\times 10^4\ V$

The kinetic energy of the electron is

$K=\dfrac{1}{2}mv^2$

Energy is given by

$E=qV$

Balancing the energy

$qV=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{\dfrac{2qV}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 2\times 10^4}{9.11\times 10^{-31}}}\\\Rightarrow v=83816746.4254\ m/s$