V^2-u^2=2as
v=final velocity=unkown
u=initial velocity=0 m/s, because freely falling
a=acceleration due to gravity=9.8 m/s^2
s=distance (here height) traveled=4.5m
therefore the final velocity,
v^2=2*9.8*4.5
v=<span>9.39m/s</span>
Answer:
Incident Command Structure, ICS or ICS-like EOC Structure is familiar and aligns with the on-scene incident organization. ICS or ICS-like EOC Structure is familiar and aligns with the on-scene incident organization.
Answer:
s=0.36m part c lands
Explanation:
from conservation of linear momentum
we know that
the momentum before Collision(explosion) is equal to the momentum after collision( explosion)
0=maua+mbub+mcuc
ma=.5kg
mb=.5kg
mc=9.8-(0.5+0.5)
mc=8.8kg
since the fragments all fall to the ground at the same time, we assume same time for all
t=0.27s
the velocity of a
v=u +at
since it falling stariaght down
v=0+9.81*.27
v=2.64m/s is the vertical side of its velocity
, since its falling straight down , the horizontal velocity is zero
the velocity of b
s=ut horizontal component of the distance
6.5=.27*ub
ub=24.07m/s
from
0=maua+mbub+mcuc
0=0.5*0+0.5*24.07+8.8*cu
uc=1.36
the distance from a will be
s=ut
s=1.36*.27
s=0.36m
Answer:
T = 5516.63 seconds
Explanation:
Given that,
The International Space Station is orbiting at an altitude of about 370 km above the earth's surface.
Mass of the Earth,
Radius of Earth,
We need to find the period of the International Space Station's orbit. It is a case of Kepler's third law. Its mathematical form is given by :
R = r + h
So, the period of the International Space Station's orbit is 5516.63 seconds.