Acceleration = force / mass.
A = 100/50 = 2 m/s^2 .
I think D. It starts at (0.0) and goes to the correct points so it makes sense
The velocity of the boat after the package is thrown is 0.36 m/s.
<h3>
Final velocity of the boat</h3>
Apply the principle of conservation of linear momentum;
Pi = Pf
where;
- Pi is initial momentum
- Pf is final momentum
v(74 + 135) = 15 x 5
v(209) = 75
v = 75/209
v = 0.36 m/s
Thus, the velocity of the boat after the package is thrown is 0.36 m/s.
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Answer:
m = 105.37 kg
Explanation:
We are given;
Mass of man; m = 113 kg
Length of boat = 6.3m
Now, The position of the center of mass will not change during the motion of the man.
Thus,
X_g,i = X_g,f
So,
[113(6.3) + ma]/(113 + m) = [113(3.26) + m(a +3.26)]/(113 + m)
113 + m will cancel on both sides to give;
113(6.3) + ma = [113(3.26) + m(a +3.26)]
711.9 + ma = 368.38 + ma + 3.26m
ma will cancel out to give;
711.9 - 368.38 = 3.26m
343.52/3.26 = m
m = 105.37 kg
Answer:
R = 8.01 m
Explanation:
We can solve this problem using the projectile launch equations. The jump length is the throw range
R = v₀² sin 2θ / g
in the exercise they give us the initial speed of 9.14 m / s and in the launch angle 35º
let's calculate
R = 9.14² sin (2 35) / 9.8
R = 8.01 m
this is the jump length