Answer:
W'=125.44 N
Explanation:
The weight of a person on the surface of Earth is 784 N
Weight is given by :
W = mg
m is mass of the person and g is acceleration due to gravity on the surface of Earth (10 m/s²)
The acceleration due to gravity on the surface of Moon, g' = 1.6 m/s²
Weight of the person on the moon is :
W'=mg'
Hence, the person would weigh 125.44 N on the Moon.
Answer:
a) p₀ = 1.2 kg m / s, b) p_f = 1.2 kg m / s, c) θ = 12.36, d) v_{2f} = 1.278 m/s
Explanation:
For this exercise we define a system formed by the two balls, which are isolated and the forces during the collision are internal, therefore the moment is conserved
a) the initial impulse is
p₀ = m v₁₀ + 0
p₀ = 0.6 2
p₀ = 1.2 kg m / s
b) as the system is isolated, the moment is conserved so
p_f = 1.2 kg m / s
we define a reference system where the x-axis coincides with the initial movement of the cue ball
we write the final moment for each axis
X axis
p₀ₓ = 1.2 kg m / s
p_{fx} = m v1f cos 20 + m v2f cos θ
p₀ = p_f
1.2 = 0.6 (-0.8) cos 20+ 0.6 v_{2f} cos θ
1.2482 = v_{2f} cos θ
Y axis
p_{oy} = 0
p_{fy} = m v_{1f} sin 20 + m v_{2f} cos θ
0 = 0.6 (-0.8) sin 20 + 0.6 v_{2f} sin θ
0.2736 = v_{2f} sin θ
we write our system of equations
0.2736 = v_{2f} sin θ
1.2482 = v_{2f} cos θ
divide to solve
0.219 = tan θ
θ = tan⁻¹ 0.21919
θ = 12.36
let's look for speed
0.2736 = v_{2f} sin θ
v_{2f} = 0.2736 / sin 12.36
v_{2f} = 1.278 m / s
Answer:
<u>Assuming b = 9.3i + 9.5j</u> <em>(b = 931 + 9.5 is wrong):</em>
a) a×b = 34.27k
b) a·b = 128.43
c) (a + b)·b = 305.17
d) The component of a along the direction of b = 9.66
Explanation:
<u>Assuming b = 9.3i + 9.5j</u> <em>(b = 931 + 9.5 is wrong)</em> we can proceed as follows:
a) The vectorial product, a×b is:
b) The escalar product a·b is:
c) <u>Asumming (a</u><u> </u><u>+ b)·b</u> <em>instead a+b·b</em> we have:
d) The component of a along the direction of b is:
I hope it helps you!