Answer:
A. 1222,10 N
Explanation:
First, we determine the value of the acceleration that the hydraulic launching system must guarantee. Let’s remember that the definition of acceleration is the rate of change of velocity with respect to time. Therefore:
a_hyd = ∆v/∆t = (54 m/s-0)/(3,8 s-0) = 14,21 m/s^2
According to Newton’s second law:
∑F_net = m*a = 86 kg*14,21 m/s^2 = 1222,10 N
So, the net force required to accelerate an 86-kg man is 1222,10 N
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Answer:
3 j
Explanation:
12.0 kilo / 4.00 meters = 3 j
Answer:
Time = 40 s
It will take the spaceship 40s to coast that distance.
Explanation:
Using the impulse momentum equation;
Impulse = change in momentum
Ft = m(∆v) ........1
Given;
Force F = 1200kN
time of action t = 20s
Mass of spaceship m = 8.0×10^4 kg
∆v = change in velocity
Substituting the values into equation 1;
1200kN × 20s =8.0×10^4 kg × ∆v
∆v = 1200000×20/80000
∆v = 300 m/s
Since it was initially at rest, V0 = 0
Final velocity V2 = 300 + 0
V2 = 300m/s
Time To travel 12 km,
Time = distance/speed
Distance = 12 km = 12000m
Speed = 300m/s
Time = 12000/300
Time = 40 s
It will take the spaceship 40s to coast that distance.
A boiling pot of water (the water travels in a current throughout the pot), a hot air balloon (hot air rises, making the balloon rise) , and cup of a steaming, hot liquid (hot air rises, creating steam) are all situations where convection occurs.
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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.