there are 3 atoms in each silver sulfide
<span>The diver is heading downwards at 12 m/s
Ignoring air resistance, the formula for the distance under constant acceleration is
d = VT - 0.5AT^2
where
V = initial velocity
T = time
A = acceleration (9.8 m/s^2 on Earth)
In this problem, the initial velocity is 2.5 m/s and the target distance will be -7.0 m (3.0 m - 10.0 m = -7.0 m)
So let's substitute the known values and solve for T
d = VT - 0.5AT^2
-7 = 2.5T - 0.5*9.8T^2
-7 = 2.5T - 4.9T^2
0 = 2.5T - 4.9T^2 + 7
We now have a quadratic equation with A=-4.9, B=2.5, C=7. Using the quadratic formula, find the roots, which are -0.96705 and 1.477251164.
Now the diver's velocity will be the initial velocity minus the acceleration due to gravity over the time. So
V = 2.5 m/s - 9.8 m/s^2 * 1.477251164 s
V = 2.5 m/s - 14.47706141 m/s
V = -11.97706141 m/s
So the diver is going down at a velocity of 11.98 m/s
Now the negative root of -0.967047083 is how much earlier the diver would have had to jump at the location of the diving board. And for grins, let's compute how fast he would have had to jump to end up at the same point.
V = 2.5 m/s - 9.8 m/s^2 * (-0.967047083 s)
V = 2.5 m/s - (-9.477061409 m/s)
V = 2.5 m/s + 9.477061409 m/s
V = 11.97706141 m/s
And you get the exact same velocity, except it's the opposite sign.
In any case, the result needs to be rounded to 2 significant figures which is -12 m/s</span>
Answer:The sled slides 16.875m before rest.
Explanation:

a=0.6 m/s²




Answer:Bruce is knocked backwards at
14
m
s
.
Explanation:
This is a problem of momentum (
→
p
) conservation, where
→
p
=
m
→
v
and because momentum is always conserved, in a collision:
→
p
f
=
→
p
i
We are given that
m
1
=
45
k
g
,
v
1
=
2
m
s
,
m
2
=
90
k
g
, and
v
2
=
7
m
s
The momentum of Bruce (
m
1
) before the collision is given by
→
p
1
=
m
1
v
1
→
p
1
=
(
45
k
g
)
(
2
m
s
)
→
p
1
=
90
k
g
m
s
Similarly, the momentum of Biff (
m
2
) before the collision is given by
→
p
2
=
(
90
k
g
)
(
7
m
s
)
=
630
k
g
m
s
The total linear momentum before the collision is the sum of the momentums of each of the football players.
→
P
=
→
p
t
o
t
=
∑
→
p
→
P
i
=
→
p
1
+
→
p
2
→
P
i
=
90
k
g
m
s
+
630
k
g
m
s
=
720
k
g
m
s
Because momentum is conserved, we know that given a momentum of
720
k
g
m
s
before the collision, the momentum after the collision will also be
720
k
g
m
s
. We are given the final velocity of Biff (
v
2
=
1
m
s
) and asked to find the final velocity of Bruce.
→
P
f
=
→
p
1
f
+
→
p
2
f
→
P
f
=
m
1
v
1
f
+
m
2
v
2
f
Solve for
v
1
:
v
1
f
=
→
P
f
−
m
2
v
2
f
m
1
Using our known values:
v
1
f
=
720
k
g
m
s
−
(
90
k
g
)
(
1
m
s
)
45
k
g
v
1
f
=
14
m
s
∴
Bruce is knocked backwards at
14
m
s
.
Explanation: