Answer:
Due to the resistance of air, a drag force acts on a falling body (parachute) to slow down its motion. Without air resistance, or drag, objects would continue to increase speed until they hit the ground. The larger the object, the greater its air resistance. Parachutes use a large canopy to increase air resistance. Also, Once the parachute is opened, the air resistance overwhelms the downward force of gravity. The net force and the acceleration on the falling skydiver is upward. An upward net force on a downward falling object would cause that object to slow down. The skydiver thus slows down. Sorry if not helpful.
Answer:
x=2d
Explanation:
initial stretch in the spring is d
so using Hook's law
at equilibrium position
k×d=mg
where k= spring constant
m= mass of fish
g= acceleration due to gravity.
d=mg/k ................ (1)
in second case by energy conservation
1/2 kx^2=mgx
x=2mg/k
using equation 1
x=2d
Answer:
Option C, increases and decreases
Explanation:
When an object making noise approaches you, the wave frequency increases leading to a higher pitch. Conversely, when it moves away from you or retreats, the wave frequency decreases leading to a lower pitch. This can be observed in ambulance sirens.
Answer:
x=?
dt=?
vi=23m/s
vf=0m/s (it stops)
d=0.25m/s^2
time =
vf=vi+d: 0=23m/s+(0.25m/s^2)t
t=92s
displacement=
vf^2=vi^2+2a(dx)
23^2=0^2+2(0.25m/s^2)x =-1058m
Explanation:
you can find time from vf = vi + a(Dt): 0 = 23 m/s + (0.25 m/s/s)t so t = 92 s and you can find the displacement from vf2 = vi2 + 2a(Dx) and find the answer in one step: 232 = 02 + 2(0.25 m/s/s)x so x = -1058 m
V=IR
V=15x11
V=165ohms
I don’t quite remember the unit
Ohms law
