Answer:
The stretch cord stores potential energy as a result of stretching but due to kinetic energy, it will move back to its original state. Since air resistance is not being ignored in this case, it will experience a slight delay in stretching at first.
Explanation:
In case, where air resistance is being ignored the stretch cord will stretch as it normally does.
- Air resistance is a force that any object experiences as a result of its motion through the air.
There are various factors that affect air resistance like speed, the density of air, area, the shape of an object etc. Meanwhile, the density of air changes with temperature or altitude. <em>Hence this force is not constant but is thought to be constant during short time frames. </em>
Answer:
-3413 ft/s2
Explanation:
We need to know the velocity with which he landed on the snow.
He 'dropped' from 512 feet. This is the displacement. His initial velocity is 0 and the acceleration of gravity is 32 ft/s2.
We use the equation of mition

v and u are the initial and final velocities, a is the acceleration and s is the displacement. Putting the appropriate values


This is the final velocity of the fall and becomes the initial velocity as he goes into the snow.
In this second motion, his final velocity is 0 because he stops after a displacement of 4.8 ft. We use the same equation of motion but with different values. This time,
, v = 0 and s = 4.8 ft.


Note that this is negative because it was a deceleration, that is, his velocity was decreasing.
Is there a picture that I can see
At a point on the streamline, Bernoulli's equation is
p/ρ + v²/(2g) = constant
where
p = pressure
v = velocity
ρ = density of air, 0.075 lb/ft³ (standard conditions)
g = 32 ft/s²
Point 1:
p₁ = 2.0 lb/in² = 2*144 = 288 lb/ft²
v₁ = 150 ft/s
Point 2 (stagnation):
At the stagnation point, the velocity is zero.
The density remains constant.
Let p₂ = pressure at the stagnation point.
Then,
p₂ = ρ(p₁/ρ + v₁²/(2g))
p₂ = (288 lb/ft²) + [(0.075 lb/ft³)*(150 ft/s)²]/[2*(32 ft/s²)
= 314.37 lb/ft²
= 314.37/144 = 2.18 lb/in²
Answer: 2.2 psi
Answer:34 cm
Explanation:
Given
mass of meter stick m=80 gm
stick is balanced when support is placed at 51 cm mark
Let us take 5 gm tack is placed at x cm on meter stick so that balancing occurs at x=50 cm mark
balancing torque





