<span>5.3 cm/s
This is a matter of conservation of momentum. Since there's no mention of the puck rebounding, I will consider this to be a totally non-elastic collision. So, let's determine the starting momentum of the system.
Goalie is at rest, so his momentum is 0.
Puck is moving at 30.00 m/s with a mass of 0.16 kg, so:
30.00 m/s * 0.16 kg = 4.8 kg*m/s
So the starting momentum is 4.8 kg*m/s moving towards the goal. After the collision, the puck and goalie will have the same momentum. So figure out the mass of the new system:
90.00 kg + 0.16 kg = 90.16 kg
And divide the system momentum by the system mass:
4.8 kg*m/s / 90.16 kg = 0.053238687 m/s
Finally, round to the least precise datum, so the result to 2 significant figures is 0.053 m/s, or 5.3 cm/s.</span>
Answer:
Explanation:
Given that,
A point charge is placed between two charges
Q1 = 4 μC
Q2 = -1 μC
Distance between the two charges is 1m
We want to find the point when the electric field will be zero.
Electric field can be calculated using
E = kQ/r²
Let the point charge be at a distance x from the first charge Q1, then, it will be at 1 -x from the second charge.
Then, the magnitude of the electric at point x is zero.
E = kQ1 / r² + kQ2 / r²
0 = kQ1 / x² - kQ2 / (1-x)²
kQ1 / x² = kQ2 / (1-x)²
Divide through by k
Q1 / x² = Q2 / (1-x)²
4μ / x² = 1μ / (1 - x)²
Divide through by μ
4 / x² = 1 / (1-x)²
Cross multiply
4(1-x)² = x²
4(1-2x+x²) = x²
4 - 8x + 4x² = x²
4x² - 8x + 4 - x² = 0
3x² - 8x + 4 = 0
Check attachment for solution of quadratic equation
We found that,
x = 2m or x = ⅔m
So, the electric field will be zero if placed ⅔m from point charge A, OR ⅓m from point charge B.
Here, we are required to determine which combination of molecules will produce ammonia with no leftovers.
Option A: 2N2 and 6H2 is the correct combination of molecules that will produce ammonia with no leftovers.
First, it is important to know that both Nitrogen and Hydrogen used in the production of ammonia are diatomic.
Secondly, Nitrogen and Hydrogen are in the ratio 1 : 3.....
As such, the coefficient of hydrogen should be thrice that of Nitrogen to ensure that there are no leftovers.
Therefore, option A which has:
2N2 and 6H2 is the correct combination of molecules that will produce ammonia with no leftovers.
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