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Elza [17]
3 years ago
5

A gray kangaroo can bound across a flat stretch of ground with each jump carrying it 11 m from the takeoff point. part a if the

kangaroo leaves the ground at a 21 ∘ angle, what is its takeoff speed
Physics
2 answers:
jeka943 years ago
8 0
<span>Each jump = 11 m
 Angle of the speed = 21 degrees.
Gravitational Acceleration g = 9.8 m/s^2
  Initial velocity V^2 = (Rg)/sin2A
 V^2 = 11 x 9.81 / sin (2 x 21)
 V^2 = 107. 91 / 0.669
 V^2 = 161.27
 V = 12.70
Now we got the velocity and calculate the horizontal angle, Velocity (horizontal) = vcosA = 12.70 x cos 21 Velocity = 11.85 m/s</span>
Svetllana [295]3 years ago
6 0
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
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Answer:

a

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a

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b

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c

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2 years ago
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3 years ago
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neonofarm [45]

Answer:

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From the question we are told that

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And this is mathematically represented as

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       \rho_n  =  \frac{m}{V_n }

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Explanation:

Crumple zones are sections in cars that are designed to crumple up when the car encounters a collision. Crumple zones minimize the effect of the force in an automobile collision in two ways. By crumpling, the car is less likely to rebound upon impact, thus minimizing the momentum change and the impulse.

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