Electricity which may efficient transportation if you develop your way depending contact metal basis , such as electric train .
-Water is from of heat-energy using solar power which run a turbine may be wheel on a railway path.
The gravitational force <em>F</em> between two masses <em>M</em> and <em>m</em> a distance <em>r</em> apart is
<em>F</em> = <em>G M m</em> / <em>r</em> ²
Decrease the distance by a factor of 7 by replacing <em>r</em> with <em>r</em> / 7, and decrease both masses by a factor of 8 by replacing <em>M</em> and <em>m</em> with <em>M</em> / 8 and <em>m</em> / 8, respectively. Then the new force <em>F*</em> is
<em>F*</em> = <em>G </em>(<em>M</em> / 8) (<em>m</em> / 8) / (<em>r</em> / 7)²
<em>F*</em> = (1/64 × <em>G M m</em>) / (1/49 × <em>r</em> ²)
<em>F*</em> = 49/64 × <em>G M m</em> / <em>r</em> ²
In other words, the new force is scaled down by a factor of 49/64 ≈ 0.7656, so the new force has magnitude approx. 76.56 N.
A closed system is a system that is completely isolated from its environment. The physical universe, as we currently understand it, appears to be a closed system. An open system is a system that has flows of information, energy, and/or matter between the system and its environment, and which adapts to the exchange.
D makes the most sense, but you just have to put two and two together and go with your gut feeling, first cross out the answers that don't make sense (A didnt make sense) and go from there! I hope the little bit above me helped you answer or decide :) Good luck!
Answer:
Explanation:
Given that,
At one instant,
Center of mass is at 2m
Xcm = 2m
And velocity =5•i m/s
One of the particle is at the origin
M1=? X1 =0
The other has a mass M2=0.1kg
And it is at rest at position X2= 8m
a. Center of mass is given as
Xcm = (M1•X1 + M2•X2) / (M1+M2)
2 = (M1×0 + 0.1×8) /(M1 + 0.1)
2 = (0+ 0.8) /(M1 + 0.1)
Cross multiply
2(M1+0.1) = 0.8
2M1 + 0.2 =0.8
2M1 = 0.8-0.2
2M1 = 0.6
M1 = 0.6/2
M1 = 0.3kg
b. Total momentum, this is an inelastic collision and it momentum after collision is given as
P= (M1+M2)V
P = (0.3+0.1)×5•i
P = 0.4 × 5•i
P = 2 •i kgm/s
c. Velocity of particle at origin
Using conversation of momentum
Momentum before collision is equal to momentum after collision
P(before) = M1 • V1 + M2 • V2
We are told that M2 is initially at rest, then, V2=0
So, P(before) = 0.3V1
We already got P(after) = 2 •i kgm/s in part b of the question
Then,
P(before) = P(after)
0.3V1 = 2 •i
V1 = 2/0.3 •i
V1 = 6 ⅔ •i m/s
V1 = 6.667 •i m/s
