Lenny loves physics and math. In which Energy career pathway would these interests be the most helpful?
2 answers:
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Answer:
51.96 m/s^-1
Explanation:
a) see the attachment
b) As we know the velocity of the projectile has two component, horizontal velocity v_ox. and vertical velocity v_oy as shown in the figure. At the highest point of the trajectory, the projectile has only horizontal velocity and vertical velocity is zero. Therefore at the highest point of the trajectory, the velocity of the projectile will be
v_ox=v_o*cosФ
=60*cos (30)
= 51.96 m/s^-1
Answer:
The attached diagram explains the system,
Sum of Fy = 0
N=9.81
N - mgCos60 = 0
F= ukN= (0.53)(9.81) =
F= 5.12 N
So
F.d= 1/2(mv.v) - mgdsin60
-5.12*0.5 = 0.5*v^2 - 2*(9.81)*(0.5*sin60)
(a) v = 2.436 m/s
For deflection
-F.x = 1/2(mv.v) - mgxsin60 + 1/2 (k*x*x)
by solving for with values of v, m, g, F, k
800x^2 - 11.87 x - 5.938 = 0
by solving the quadratic equation
x = 0.093, -0.079
(b) x = 0.093 m
correct Answer is 0.093m
Explanation:
