<span>speed of slower=x; distance in 6 hours is 6x
speed of larger=2x; distance in 6 hours is 12x
12x-6x=204 miles
</span><span> 6x=204
x=34 mph slower, 204 miles in 6 hours.
2x=68 mph, faster, 408 miles in 6 hours, and difference is 204 miles</span>
For number one the answer is Iodine because it is in the same group as fluorine. For number two the answer is Germanium for the same reason. For number three the answer is Aluminum for the same reason.
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Answer:
i) E = 269 [MJ] ii)v = 116 [m/s]
Explanation:
This is a problem that encompasses the work and principle of energy conservation.
In this way, we establish the equation for the principle of conservation and energy.
i)

![W_{1-2}= (F*d) - (m*g*h)\\W_{1-2}=(500000*2.5*10^3)-(40000*9.81*2.5*10^3)\\W_{1-2}= 269*10^6[J] or 269 [MJ]](https://tex.z-dn.net/?f=W_%7B1-2%7D%3D%20%28F%2Ad%29%20-%20%28m%2Ag%2Ah%29%5C%5CW_%7B1-2%7D%3D%28500000%2A2.5%2A10%5E3%29-%2840000%2A9.81%2A2.5%2A10%5E3%29%5C%5CW_%7B1-2%7D%3D%20269%2A10%5E6%5BJ%5D%20or%20269%20%5BMJ%5D)
At that point the speed 1 is equal to zero, since the maximum height achieved was 2.5 [km]. So this calculated work corresponds to the energy of the rocket.
Er = 269*10^6[J]
ii ) With the energy calculated at the previous point, we can calculate the speed developed.
![E_{k2}=0.5*m*v^2\\269*10^6=0.5*40000*v^2\\v=\sqrt{\frac{269*10^6}{0.5*40000} }\\ v=116[m/s]](https://tex.z-dn.net/?f=E_%7Bk2%7D%3D0.5%2Am%2Av%5E2%5C%5C269%2A10%5E6%3D0.5%2A40000%2Av%5E2%5C%5Cv%3D%5Csqrt%7B%5Cfrac%7B269%2A10%5E6%7D%7B0.5%2A40000%7D%20%7D%5C%5C%20v%3D116%5Bm%2Fs%5D)
Answer
given,
weight of the oak board = 600 N
Weight of Joe = 844 N
length of board = 4 m
Joe is standing at 1 m from left side
vertical wire is supporting at the end.
Assuming the system is in equilibrium
T₁ and T₂ be the tension at the ends of the wire
equating all the vertical force
T₁ + T₂ = 600 + 844
T₁ + T₂ = 1444...........(1)
taking moment about T₂
T₁ x 4 - 844 x 3 - 600 x 2 = 0
T₁ x 4 = 3732
T₁ = 933 N
from equation (1)
T₂ = 1444 - 933
T₂ = 511 N