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Naya [18.7K]
3 years ago
12

How can two strengths be added together

Physics
1 answer:
Ket [755]3 years ago
6 0
By using mind and dna
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Brendan drew a diagram to illustrate the centripetal force acting on a satellite.  

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Draw the Fg = Fc vector so it points at a 45° angle from the v vector.  

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A cube of each of length 3m is subjected to a normal force of 36N. What will be the stress on cube a)12. Ab)2 c)18. d)4
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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:
v=20 km/s=20000 m/s
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
A 1.34 kg ball is connected by means of two massless strings, each of length L = 1.70m, to a vertical, rotating rod. The strings
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I think it’s B I’m so right
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3 years ago
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