A parachute increases air friction, thus reducing falling speed.
Indeed, the air friction is roughly proportional to the surface of the object falling down; the parachute tremendously increases that surface.
The equation to be used is the derived formulas for rectilinear motion at a constant acceleration. The formula for acceleration is
a = (v - v₀)/t
where
v and v₀ are the initial and final velocities, respectively
t is the time
a is the acceleration
Since it started from rest, v₀ = 0. Using the formula:
0.15 m/s² = (v - 0)/[2 minutes*(60 s/1 min)]
Solving for v,
v = 18 m/s
Answer:
The correct answer is C. All three have equal non-zero pressure
Explanation:
Pressure is the relationship between the force and the area of a body, when the bodies are liquid the formula that
P = rho g h
Where rho is the density and h the height of the liquid
We see that for this expression the pressure does not depend on the shape of the container, but on its height, as the three vessels have the same height, the pressure at the bottom is the same.
The correct answer is C All three have equal non-zero pressure
They can't hear an echo in small room because in it the sound can't be reflected back. For an echo of a sound to be heard,the minimum distance between the source of sound and the walls of the roomshould be 17.2 m. Obviously,in asmall room echoes cannot beheard.
Answer:
x(t) = - 6 cos 2t
Explanation:
Force of spring = - kx
k= spring constant
x= distance traveled by compressing
But force = mass × acceleration
==> Force = m × d²x/dt²
===> md²x/dt² = -kx
==> md²x/dt² + kx=0 ------------------------(1)
Now Again, by Hook's law
Force = -kx
==> 960=-k × 400
==> -k =960 /4 =240 N/m
ignoring -ve sign k= 240 N/m
Put given data in eq (1)
We get
60d²x/dt² + 240x=0
==> d²x/dt² + 4x=0
General solution for this differential eq is;
x(t) = A cos 2t + B sin 2t ------------------------(2)
Now initially
position of mass spring
at time = 0 sec
x (0) = 0 m
initial velocity v= = dx/dt= 6m/s
from (2) we have;
dx/dt= -2Asin 2t +2B cost 2t = v(t) --- (3)
put t =0 and dx/dt = v(0) = -6 we get;
-2A sin 2(0)+2Bcos(0) =-6
==> 2B = -6
B= -3
Putting B = 3 in eq (2) and ignoring first term (because it is not possible to find value of A with given initial conditions) - we get
x(t) = - 6 cos 2t
==>
