Answer:
.
Explanation:
If the mass of an object is 
and the velocity of that object is 
, the linear momentum of that object would be 
.
Assume that the initial velocity of the mass is positive (
.) However, the direction of the velocity is reversed after the impact. Thus, the sign of the new velocity of the object would be negative- the opposite of that of the initial velocity. The new velocity would be 
.
Thus, the change in the velocity of the mass would be:
.
The change in the linear momentum of the mass would be:
.
Thus, the magnitude of the change of the linear momentum would be 
.
Full outer shell of electrons, they are colorless and odorless ,their melting and boiling points are close together which gives them very narrow liquid range
<span>Here, g is the acceleration due to gravity at the Earth's surface and is the standard gravity, defined as 9.8m/s....So, the force by that gravity is called g-force.....</span>
For the answer to the question above, if your question is a lever with a pivot and which sounds more likely ...then I would say from the principle of moments: F1*d1=F2*d2
<span>given F1=high and F2=low ---> d1 must be low and d2 high for the equations to be balanced. </span>
<span>So the answer is simply: Inversely</span>
Answer:
speed
=
12
m/s
Explanation:
We're asked to find the initial velocity of the box, given it constantly accelerated at
−
3
m/s
2
for
24
m
before coming to a stop.
To do this, we can use the equation
(
v
x
)
2
=
(
v
0
x
)
2
+
2
a
x
(
Δ
x
)
−−−−−−−−−−−−−−−−−−−−−−
where
v
x
is the final velocity of the box (which is
0
since it came to rest)
v
0
x
is the initial velocity of the box (what we're trying to find)
a
x
is the acceleration of the box, given as
−
3
m/s
2
(negative because it's directed opposite to motion; i.e. it is decelerating)
Δ
x
is the distance traveled by the box, given as
24
m
Plugging in knonw values, we have
0
=
(
v
0
x
)
2
+
2
(
−
3
l
m/s
2
)
(
24
l
m
)
v
0
x
=
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
∣
∣
12
l
m/s
∣
∣
−−−−−−−−−−−−
The truck was thus traveling at a speed of
12
m/s
.
