Answer:

Explanation:
From the question we are told that
Weight of fireman 
Pole distance 
Final speed is 
Generally the equation for velocity is mathematically represented as

Therefore Acceleration a
Generally the equation for Frictional force
is mathematically given as



Therefore

Answer:
12.0 meters
Explanation:
Given:
v₀ = 0 m/s
a₁ = 0.281 m/s²
t₁ = 5.44 s
a₂ = 1.43 m/s²
t₂ = 2.42 s
Find: x
First, find the velocity reached at the end of the first acceleration.
v = at + v₀
v = (0.281 m/s²) (5.44 s) + 0 m/s
v = 1.53 m/s
Next, find the position reached at the end of the first acceleration.
x = x₀ + v₀ t + ½ at²
x = 0 m + (0 m/s) (5.44 s) + ½ (0.281 m/s²) (5.44 s)²
x = 4.16 m
Finally, find the position reached at the end of the second acceleration.
x = x₀ + v₀ t + ½ at²
x = 4.16 m + (1.53 m/s) (2.42 s) + ½ (1.43 m/s²) (2.42 s)²
x = 12.0 m
Answer:
The height of the object is 5007.4 miles.
Explanation:
Given that,
Weight of object = 200 lb
We need to calculate the value of 
Using formula of gravitational force

Put the value into the formula



We need to calculate the height of the object
Using formula of gravitational force

Put the value into the formula





Hence. The height of the object is 5007.4 miles.
Answer:
if the radius of the artery decreases by a factor of 1.5 because of the blockage by what factor has the volume flow rate decreased?