Kinetic energy = 1/2 m v²
If we reduce the mass by half > m/2
Kinetic energy = 1/2 m/2 v²
We should know that 1/2 × 1/2 = 1/4
So kinetic energy will be :
1/4 × m × v²
We have that the maximum rank of the kangaroo is given by:
R = v0 ^ 2 sin (2θ) / g
where,
v0 = initial velocity
θ = angle of the velocity vector formed from the horizontal
g = gravity
Clearing the speed we have:
v0 ^ 2 = (R * g) / (sin (2θ))
Substituting values
v0 = root (((11) * (9.8)) / (sin (2 (21 * (pi / 180)))))
v0 = 12.69 m / s
answer
its takeoff speed is 12.69 m / s
Answer:
the correct affirmation is the 3
Explanation:
Let's analyze the problem with Newton's second law before looking at the claims.
X axis parallel to the plane, positive down
F -fr + Wₓ = ma
Y Axis perpendicular to the plane
N -Wy = 0
With trigonometry
Wₓ = W sin θ
Wy = w cos θ
Let's multiply by the displacement along the plane, to relate to the work, which has as expression W = F d
F d -fr d + Wx d = ma d
Push W₁ = Fd
frictional force W₂ = -fr d
gravity W₃ = Wx d
W₁ + W₃ -W₂ = m a d
Analysis affirmations:
R1) false. The work of gravity is the subtraction
R2) false. Each force contributes according to its magnitude
R3) true. In the equation we see that, if the acceleration is zero, W2 = W1 + W3
R4) False. It equals the difference
the correct affirmation is the 3
Explanation:
Given that,
Fapp = 67.0 N at 30.0° above the horizontal (rightward and upward)
Fnorm = 64.5 N, up
Ffrict = 27.6 N, left
Fgrav = 98 N, down
(a) Resolution of the applied force:
Horizontal component,

Vertical component,

(b) Net horizontal force :

It is positive, it will act in right side
Net vertical force :

Hence, it is clear that the net force is in horizontal direction i.e. 30.42 N due right side.