First I’ll show you this standard derivation using conservation of energy:
Pi=Kf,
mgh = 1/2 m v^2,
V = sqrt(2gh)
P is initial potential energy, K is final kinetic, m is mass of object, h is height from stopping point, v is final velocity.
In this case the height difference for the hill is 2-0.5=1.5 m. Thus the ball is moving at sqrt(2(10)(1.5))=
5.477 m/s.
A-200 kg
I hope this helped xx
Answer:
Solution:
we have given the equation of motion is x(t)=8sint [where t in seconds and x in centimeter]
Position, velocity and acceleration are all based on the equation of motion.
The equation represents the position. The first derivative gives the velocity and the 2nd derivative gives the acceleration.
x(t)=8sint
x'(t)=8cost
x"(t)=-8sint
now at time t=2pi/3,
position, x(t)=8sin(2pi/3)=4*squart(3)cm.
velocity, x'(t)=8cos(2pi/3)==4cm/s
acceleration, x"(t)==8sin(2pi/3)=-4cm/s^2
so at present the direction is in y-axis.
Answer:
(
)=1913.31 N/m^2
Explanation:
given:
=0.85
=90 m/s
γ∞=1.23 kg/m^3
solution:
since outside pressure is atm pressure vaccum can be defined by ()
=√2()/γ∞[
-1]
()=1913.31 N/m^2
Answer:
by using formula F=ma which is m stand for mass a stand for acceleration. so 500kg × 2 ms^-2
