<span>Large intestine, small intestine, rectum is the correct order.</span>
Answer:
Explanation:
initial velocity, u = 0
final velocity, v = 80 ft/s
acceleration, a = 150 ft/s²
Let the time taken is t.
v = u + at
80 = 0 + 150 x t
t = 0.53 second
Answer:
10.4 m/s
Explanation:
First, find the time it takes for the projectile to fall 6 m.
Given:
y₀ = 6 m
y = 0 m
v₀ = 0 m/s
a = -9.8 m/s²
Find: t
y = y₀ + v₀ t + ½ at²
(0 m) = (6 m) + (0 m/s) t + ½ (-9.8 m/s²) t²
t = 1.11 s
Now find the horizontal position of the target after that time:
Given:
x₀ = 6 m
v₀ = 5 m/s
a = 0 m/s²
t = 1.11 s
Find: x
x = x₀ + v₀ t + ½ at²
x = (6 m) + (5 m/s) (1.11 s) + ½ (0 m/s²) (1.11 s)²
x = 11.5 m
Finally, find the launch velocity needed to travel that distance in that time.
Given:
x₀ = 0 m
x = 11.5 m
t = 1.11 s
a = 0 m/s²
Find: v₀
(11.5 m) = (0 m) + v₀ (1.11 s) + ½ (0 m/s²) (1.11 s)²
v₀ = 10.4 m/s
Answer:
a) 
b) 
c) 
d)
would be the same.
would decrease.
would be the same.
Explanation:
a) On an inclined plane the force of gravity is the sine component of the weight of the block.

b) The friction force is equal to the normal force times coefficient of friction.

c) The work done by the normal force is zero, since there is no motion in the direction of the normal force.
d) The relation between the vertical height and the distance on the ramp is

According to this relation, the work done by the gravity wouldn't change, since the force of gravity includes a term of
.
The work done by the friction force would decrease, because both the cosine term and the distance on the ramp would decline.
The work done by the normal force would still be zero.