Answer:
5.4 J.
Explanation:
Given,
mass of the object, m = 2 Kg
initial speed, u = 5 m/s
mass of another object,m' = 3 kg
initial speed of another orbit,u' = 2 m/s
KE lost after collusion = ?
Final velocity of the system
Using conservation of momentum
m u + m'u' = (m + m') V
2 x 5 + 3 x 2 = ( 2 + 3 )V
16 = 5 V
V = 3.2 m/s
Initial KE = 
= 
= 31 J
Final KE = 
Loss in KE = 31 J - 25.6 J = 5.4 J.
Ok. PEMDAS tells us to take care of the square first. When we do that, the denominator becomes
(6.4)^2 x 10^12
= 40.96 x 10^12 .
Now it's just a matter of mashing out the fraction.
The 'mantissa' (the number part) is
6/40.96 = 0.1465
and the order of magnitude is
10^24 / 10^12 = 10^12 .
Put it all together and you've got
1.465 x 10^11 .
By looking at the acceleration of the object.
In fact, Netwon's second law states that the resultant of the forces acting on an object is equal to the product between the mass m of the object and its acceleration:
So, when static friction is acting on the object, if the object is still not moving we know that all the forces are balanced: in fact, since the object is stationary, its acceleration is zero, and so the resultant of the forces (left term in the formula) must be zero as well (i.e. the forces are balanced).
It's kinda long but...
A tectonic setting where volcanic action occurs is called <span>a </span>hot-spot (intraplate<span>), which describes volcanic activity that occurs </span>within tectonic plates<span> and is generally NOT related to plate boundaries and plate movements.
</span>Hope this helps!!:)
Elastic collision is when kinetic energy before = kinetic energy after
Ek= 1/2mv^2
total before
Ek=1/2(2)(2.2^2) = 4.84 J
total after
Ek= 1/2(2+4)(v^2) = 3v^2
Before = after
4.84=3v^2 | divide by 3
121/75 = v^2 | square root both sides
v=1.27 m/s
