Answer: part a: 19.62m
part b: 19.62 m/s
part a: 2.83 secs
Explanation:If the air resistance is ignored then the swimmer experience free fall under gravity hence
u=0
a=9.81 m/s2
t=2 secs
s=h
Part b
Part c
now we have h=2*19.62=39.24
In the diagram, a force of 20 newtons is applied to a block. The block is in dynamic equilibrium. What is the magnitude and dire
2 answers:
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Answer:
z = 3,737 10⁵ m
Explanation:
a) As they indicate that the atmosphere behaves like an ideal gas, we can use the equation
P V = n R T
P = (n r / V) T
We replace
P = (n R / V) T₀
b) Let's apply this equation in the points
Lower
.z = 0
P₀ = 610 Pa
P₀ = (nR / V) T₀
Higher.
P = 10 Pa
P = (n R / V) T₀ e^{- C z}
We replace
P = P₀ e^{- C z}
e^{- C z} = P / P₀
C z = ln P₀ / P
z = 1 / C ln P₀ / P
Let's calculate
z = 1 / 1.1 10⁻⁵ ln (610/10)
z = 3,737 10⁵ m
Answer:
a. 11 m/s at 76° with respect to the original direction of the lighter car.
Explanation:
In this exercise, since both cars make a right angle, let's assume that the lighter car only has a horizontal velocity component (vx) and that the heavier one only has a vertical velocity component (vy). The final velocities for both components for the system can be determined as:
Assume that the lighter car has a 1kg mass and that the heavier car has a 4 kg mass.
The magnitude of the final velocity of the wreck can be found as:
The final velocity has an intensity of roughly 11 m/s
As for the angle, it can be determined in respect to the lighter car (x axis) as follows:
Therefore, the wreck has a velocity with an intensity of 11 m/s at 76° with respect to the original direction of the lighter car.