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mariarad [96]
3 years ago
8

When jumping straight down, you can be seriously injured if you land stiff-legged. One way to avoid injury is to bend your knees

upon landing to reduce the force of the impact. A 63.5-kg man just before contact with the ground has a speed of 7.89 m/s. (a) In a stiff-legged landing he comes to a halt in 3.99 ms. Find the magnitude of the average net force that acts on him during this time. (b) When he bends his knees, he comes to a halt in 0.205 s. Find the magnitude of the average net force now. (c) During the landing, the force of the ground on the man points upward, while the force due to gravity points downward. The average net force acting on the man includes both of these forces. Taking into account the directions of the forces, find the magnitude of the force applied by the ground on the man in part (b).
Physics
1 answer:
choli [55]3 years ago
6 0

Answer:

a)  F = 1.26 10⁵ N, b)  F = 2.44 10³ N, c)   F_net = 1.82 10³ N  directed vertically upwards

Explanation:

For this exercise we must use the relationship between momentum and momentum

         I = Δp

         F t = p_f -p₀

a) It asks to find the force

as the man stops the final velocity is zero

         F = 0 - p₀ / t

the speed is directed downwards which is why it is negative, therefore the result is positive

         F = m v₀ / t

         F = 63.5 7.89 / 3.99 10⁻³

         F = 1.26 10⁵ N

b) in this case flex the knees giving a time of t = 0.205 s

          F = 63.5 7.89 / 0.205

          F = 2.44 10³ N

c) The net force is

         F_net = Sum F

         F_net = F - W

         F_net = F - mg

let's calculate

         F_net = 2.44 10³ - 63.5 9.8

         F_net = 1.82 10³ N

since it is positive it is directed vertically upwards

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two sinusoidal waves, which are identical except for a phase shift, travel along in the same direction. The wave equation of the resultant wave is yR (x, t) = 0.70 m sin⎛ ⎝3.00 m−1 x − 6.28 s−1 t + π/16 rad⎞ ⎠ . What are the angular frequency, wave number, amplitude, and phase shift of the individual waves?

ω = 6.28 s − 1 ,

k = 3.00 m− 1 ,

φ = π rad,

A R = 2 A cos (φ 2 ) ,

A = 0.37 m

Explanation:

y1 ( x , t ) = A sin( k x − ω t +φ ) ,

y 2 ( x , t ) = A sin ( k x − ω t ) .

from the principle of superposition which states that when two or more waves combine, there resultant wave is the algebriac sum of the individual waves

y1 ( x , t ) = A sin( k x − ω t +φ ) ,   is generaL form of thw wave eqaution

A=amplitude

k=angular wave number

ω=angular frequency

φ =phase constant

k=2π/lambda

ω=2π/T

yR (x, t) = 0.70 m sin{3.00 m−1 x − 6.28 s−1 t + π/16 rad}....................*

two waves superposed to give the above, assuming they are moving in the +x direction

y1 ( x , t ) = A sin( k x − ω t +φ ) , .....................1

y 2 ( x , t ) = A sin ( k x − ω t ) ...........................2

adding the two equation will give

A sin( k x − ω t +φ )+A sin ( k x − ω t ) .................3

A( sin( k x − ω t +φ )+ sin ( k x − ω t ) ),......................4

similar to the following trigonometry identity

sina+sinb=2cos(a-b)/2sin(a+b)/2

let a= ( k x − ω t

b=k x − ω t +φ )

y(x,t)=2Acos(φ/2)sin(k x − ω t +φ/2)

k=3m^-1

lambda=2π/k=2.09m

ω=6.28= T=2π/6.28

T=1s

φ/2=π/16

φ=π/8rad

amplitude

2Acos(φ/2)=0.70 m

A=0.7/2cos(π/8)

A=0.37 m

6 0
4 years ago
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