Answer:
, the minus meaning west.
Explanation:
We know that linear momentum must be conserved, so it will be the same before (
) and after (
) the explosion. We will take the east direction as positive.
Before the explosion we have 
.
After the explosion we have pieces 1 and 2, so 
.
These equations must be vectorial but since we look at the instants before and after the explosions and the bomb fragments in only 2 pieces the problem can be simplified in one dimension with direction east-west.
Since we know momentum must be conserved we have:
Which means (since we want 
and 
):
So for our values we have:
Answer:
1793.7m
Explanation:
From the principle of conservation of energy; the kinetic energy substended by the object equals the potential energy sustain by the object when it gets to its maximum position.
Now the kinetic energy; is
K.E = 1/2 × m × v2
Where m is mass
v is velocity
Hence.
K.E = 1/2 × 2.25 × (187.5)^2
Now this should be same with the potential energy which is given as;
P.E = m× g× h
Where m is mass of object
g is acceleration of free fall due to gravity = 9.8m/S2
h is maximum height substain by the object.
Hence P.E = 2.25 × 9.8 × h
From the foregoing analysis of energy conversation it implies;
1/2 × 2.25 × (187.5)^2 =2.25 × 9.8 × h
=> 1/2 × (187.5)^2 = 9.8 × h
=>1/2 × (187.5)^2 / 9.8 = h
=> 1793.69m = h
h= 1793.69m
h =1793.7m to 1 decimal place
Answer:
solution given:
acceleration (a)=?
initial velocity (u)=3m/s
final velocity (v)=6m/s
distance (s)=90m
we have
v²=u²+2as
substituting value
6²=3²+2*a*90
36=9+180a
36-9=180a
a=25/180
<u>a=0.1388m/s²</u>
Hi there!
Impulse = Change in momentum
I = Δp = mΔv = m(vf - vi)
Where:
m = mass of object (kg)
vf = final velocity (m/s)
vi = initial velocity (m/s)
Begin by converting grams to kilograms:
1 kg = 1000g ⇒ 145g = .145kg
Now, plug in the given values. Remember to assign directions since velocity is a vector. Let the initial direction be positive and the opposite be negative.
I = (.145)(-20 - 17) = -5.365 Ns
The magnitude is the absolute value, so:
|-5.365| = 5.365 Ns
