Responder:
Explicación:
Para que podamos calcular el tiempo que le tomará al esquiador permanecer en el aire, encontraremos el tiempo de vuelo como se muestra;
T = 2Usin theta / 2
theta = 90 grados
U = 25 m / s
T = 25sin90 / 2 (9,8)
T = 25 / 19,62
T = 1,27 segundos
Por lo tanto, los cielos usarán 1.27 segundos en el aire.
La distancia horizontal es el rango;
Rango R = U√2H / g
R = 25√2 (80) /9,8
R = 25√160 / 9,8
R = 25 * √16,326
R = 25 * 4.04
R = 101,02 m
Por tanto, la distancia horizontal recorrida por el esquiador es 101,02 m
Answer:
a) The student feel light
b) Nbottom = 758 N
c) N'top= 236 N
d) N'bottom= 1055 N
Explanation:
a) W= 659N , Ntop= 560N
W > Ntop ---> Student feel less weight
b) Top:
∑F= W - Ntop = m.v²/R
m.v²/R = 659N - 560 N = 99 N
Bottom:
∑F= Nbottom- W = m.v²/R
Nbottom= W + m.v²/R = 659N + 99 N = 758N
c) W= 659 N , Ntop= 560 N , v'=2.v
N'top= ?
∑F= W - N'top = m.v'²/R
N'top= W - 4.m.v²/R
N'top = 659 N - 4. 99 N = 263 N
d) N'bottom = ?
∑Fbottom= N'bottom- W = m.v'²/R
N'bottom = W + 4.m.v²/R = 659 N + 4. 99 N = 1055 N
The cyclist who travels 20 kilometers per hour for 15 kilometers
Answer:
Part a)
Part b)
Explanation:
Part a)
If block is sliding up then net force must be zero and friction will be in opposite to the direction of motion of the block
so we have
Part b)
If block is sliding down then net force must be zero and friction will be in opposite to the direction of motion of the block
so we have
