Answer:
1) t=1.743 sec
2)Vo=61.388 m/sec
3)the x component of its velocity just be- fore it strikes the ground is the same as the initial velocity of the ball that is=61.388 m/sec
4)Vf=17.08 m/s
Explanation:
1)From second equation of motion we get
h=Vit+(1/2)gt^2
here in case(a): Vi=0 m/s,h=14.9m,,put these values in above equation to find the time the ball is in motion
14.9=(0)*t+(1/2)(9.8)t^2
t^2=14.9/4.9
t^2=3.040 sec
t=1.743 sec
2) s=Vo*t
Putting values we get
107=Vo*1.743
Vo=61.388 m/sec
3)the x component of its velocity just be- fore it strikes the ground is the same as the initial velocity of the ball that is=61.388 m/sec
4)From third equation of motion we know that
Vf^2-Vi^2=2gh
here Vi=0 m/s,h=14.9 m
Vf^2=Vi^2+2gh=0+2(9.8)(14.9)
Vf^2=292.04
Vf=17.08 m/s
S orbital.
Group 1 elements have a general configuration
, where n represents the highest occupied Principal Energy Level. For example, Lithium has the valence configuration
whereas Cesium has
. Both of them belong to Group 1 of Periodic Table.
Group 2 elements have a general configuration of
. For example, Magnesium has
as its outer shell configuration while Strontium has the same as
.
We see that in both the cases, the outermost S orbital is being filled.
Answer:
1.045 m from 120 kg
Explanation:
m1 = 120 kg
m2 = 420 kg
m = 51 kg
d = 3 m
Let m is placed at a distance y from 120 kg so that the net force on 51 kg is zero.
By use of the gravitational force
Force on m due to m1 is equal to the force on m due to m2.



3 - y = 1.87 y
3 = 2.87 y
y = 1.045 m
Thus, the net force on 51 kg is zero if it is placed at a distance of 1.045 m from 120 kg.
Answer:
d = 10.076 m
Explanation:
We need to obtain the velocity of the ball in the y direction
Vy = 24.5m/s * sin(35) = 14.053 m/s
To obtain the distance, we use the formula
vf^2 = v0^2 -2*g*d
but vf = 0
d = -vo^2/2g
d = (14.053)^2/2*(9.8) = 10.076 m
Answer:
1.contact force
2. Non contact
Explanation:
1. Because bodies are in contact
2.Bodies are not in contact.