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3 years ago

10

A circus performer walked up and to the right for a total displacement of 10\,\text m10m10, start text, m, end text along a diag

onal tightrope angled 30 \degree30°30, degree above the ground.

1 answer:

Tanya [424]3 years ago

8 0

Answer: 5m

(10m) sin 30

=5m

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On august 10, 1972, a large meteorite skipped across the atmosphere above the western united states and western canada, much lik

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(a)
The velocity of the meteorite just before hitting the ground is:
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The loss of energy of the meteorite corresponds to the kinetic energy the meteorite had just before hitting the ground, so:
$\Delta K = \frac{1}{2}mv^2= \frac{1}{2}(3.4 \cdot 10^6 kg)(20000 m/s)^2=6.8 \cdot 10^{14}J$

(b) 1 megaton of tnt is equal to $1 MT=4.2 \cdot 10^{15}J$
To find to how many megatons the meteorite energy loss $\Delta E$
corresponds, we can set the following proportion
$1 MT: 4.2 \cdot 10^{15}J=x: \Delta E$
And so we find
$x= \frac{\Delta E}{4.2 \cdot 10^{15}J} = \frac{6.8 \cdot 10^{14}J }{4.2 \cdot 10^{15}J} =0.162 MT$
So, 0.162 megatons.

(c) 1 Hiroshima bomb is equivalent to 13 kilotons (13 kT). The impact of the meteorite had an energy of $\Delta E=0.162 MT=162 kT$ . So, to find to how many hiroshima bombs it corresponds, we can set the following proportion:
$1:13 kT=x:162 kT$
And so we find
$x= \frac{162 kT}{13 kT}=12.46$
So, the energy released by the impact of the meteorite corresponds to the energy of 12.46 hiroshima bombs.

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4 years ago